Category: Class 11

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 1 Angle and its Measurement Ex 1.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.2

Question 1.
Find the length of an arc of a circle which subtends an angle of 108° at the centre, if the radius of the circle is 15 cm.
Solution:
Here, r = 15cm and
θ = 108° = \(\left(108 \times \frac{\pi}{180}\right)^{\mathrm{c}}=\left(\frac{3 \pi}{5}\right)^{\mathrm{c}}\)
Since S = r.θ
S = 15 x \(\frac{3 \pi}{5}\)
= 9π cm.

Question 2.
The radius of a circle is 9 cm. Find the length of an arc of this circle which cuts off a chord of length equal to length of radius.
Solution:
Here, r = 9cm
Let the arc AB cut off a chord equal to the radius of the circle.
Since OA = OB = AB,
ΔOAB is an equilateral triangle.
m∠AOB = 60°
θ = 60°
= \(\left(60 \times \frac{\pi}{180}\right)^{c}=\left(\frac{\pi}{3}\right)^{c}\)
Since S = r.θ,
S = 9 x \(\frac{\pi}{3}\) = 3π cm.

Question 3.
Find the angle in degree subtended at the centre of a circle by an arc whose length is 15 cm, if the radius of the circle is 25 cm.
Solution:
Here, r = 25 cm and S = 15 cm
Since S = r.θ,
15 = 25 x θ

∴ The required angle in degree is \(\left(\frac{108}{\pi}\right)^{0}\) or (34.40)°(approx.).

Question 4.
A pendulum of length 14 cm oscillates through an angle of 18°. Find the length of its path.
Solution:

Question 5.
Two arcs of the same length subtend angles of 60° and 75° at the centres of the two circles. What is the ratio of radii of two circles?
Solution:
Let r₁, and r₂ be the radii of the two circles and let their arcs of same length S subtend angles of 60° and 75° at their centres.
Angle subtended at the centre of the first circle,
θ₁ = 60° = \(\left(60 \times \frac{\pi}{180}\right)^{c}=\left(\frac{\pi}{3}\right)^{c}\)
∴ S = r₁θ₁ = r₁(\(\left(\frac{\pi}{3}\right)\))
Angle subtended at the centre of the second circle,

Question 6.
The area of the circle is 2571 sq.cm. Find the length of its arc subtending an angle of 144° at the centre. Also find the area of the corresponding sector.
Solution:
Area of circle = πr²
But area is given to be 25 π sq.cm
∴ 25π = πr²
∴ r² = 25
∴ r = 5 cm
θ = 144° = \(=\left(144 \times \frac{\pi}{180}\right)^{c}=\left(\frac{4 \pi}{5}\right)^{\mathrm{c}}\)
Since s = rθ
S = 5(\(\frac{4 \pi}{5}\)) = 4π
Also, A(sector) = \(\frac{1}{2}\) x r x S = \(\frac{1}{2}\) x 5 x 4π
= 10π sq. cm

Question 7.
OAB is a sector of the circle having centre at O and radius 12 cm. If m∠AOB = 45°, find the difference between the area of sector OAB and ΔAOB.
Solution:
Here, r = 12 cm
\(\theta=45^{\circ}=\left(45 \times \frac{\pi}{180}\right)^{c}=\left(\frac{\pi}{4}\right)^{c}\)

Draw AM ⊥ OB
In ΔOAM,

[Note: The question has been modified.]

Question 8.
OPQ is the sector of a circle having centre at O and radius 15 cm. If m∠POQ = 30°, find the area enclosed by arc PQ and chord PQ.
Solution:
Here, r = 15 cm
m∠POQ = 30°
\(\left(30 \times \frac{\pi}{180}\right)^{c}[/larex]

∴ θ = [latex]\left(\frac{\pi}{6}\right)^{c}\)
Draw QM ⊥ OP
In ΔOQM,
sin 30° = \(\frac{\text { QM }}{15}\)
QM= 15 x \(\frac{1}{2}=\frac{15}{2}\)
Shaded portion indicates the area enclosed by arc PQ and chord PQ.
∴ A(shaded portion)
= A(sector OPQ) – A(ΔOPQ)

Question 9.
The perimeter of a sector of the circle of area 25π sq.cm is 20 cm. Find the area of sector.
Solution:
Area of circle = πr²
But area is given to be 25π sq.cm.
∴ 25π = πr²
∴ r² = 25
∴ r = 5 cm
Perimeter of sector = 2r + S
But perimeter is given to be 20 cm.
∴ 20 = 2(5) + S
∴ 20 = 10 + S
∴ S = 10 cm
Area of sector = \(\frac{1}{2}\) x r x S
= \(\frac{1}{2}\) x 5 x 10
= 25sq.cm.

Question 10.
The perimeter of a sector of the circle of area 64 7i sq.cm is 56 cm. Find the area of the sector.
Solution:
Area of circle = πr²
But area is given to be 25π sq.cm.
∴ 64π = πr²
∴ r² = 64
∴ r = 8 cm
Perimeter of sector = 2r + S
But perimeter is given to be 20 cm.
∴ 56 = 2(5) + S
∴ 56 = 16 + S
∴ S = 40 cm
Area of sector = \(\frac{1}{2}\) x r x S
= \(\frac{1}{2}\) x 8 x 40
= 160sq.cm.

Balbharti Maharashtra State Board Hindi Yuvakbharati 11th Digest Chapter 2 लघु कथाएँ Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board 11th Hindi Yuvakbharati Solutions Chapter 2 लघु कथाएँ

11th Hindi Digest Chapter 2 लघु कथाएँ Textbook Questions and Answers

आकलन

1. लिखिए :

प्रश्न अ.
दावत में होने वाली अन्न की बरबादी पर उषा की प्रतिक्रिया –
उत्तर :
उषा की आँखों में आँसू आ गए।

प्रश्न आ.
संवादों का उचित घटनाक्रम –
(i) “रुपये खर्च हो गए मालिक।”
(ii) “स्कूल नहीं जाता तू? अजीब है….!”
(iii) “अरे क्या हुआ ! जाता क्यों नहीं?”
(iv) “माँ, बाल-मजदूरी अपराध है न?’
उत्तर :
(i) “स्कूल नहीं जाता तू? अजीब है….!”
(ii) “माँ, बाल-मजदूरी अपराध है न?’
(iii) “अरे क्या हुआ ! जाता क्यों नहीं?”
(iv) “रुपये खर्च हो गए मालिक।”

शब्द संपदा

2.

प्रश्न अ.
समूह में से विसंगति दर्शानेवाला कृदंत/तद्धित शब्द चुनकर लिखिए –
(i) मानवता, हिंदुस्तानी, ईमानदारी, पढ़ाई
(ii) थकान, लिखावट, सरकारी, मुस्कुराहट
(iii) बुढ़ापा, पितृत्व, हँसी, आतिथ्य
(iv) कमाई, अच्छाई, सिलाई, चढ़ाई
उत्तर :
(i) पढ़ाई, (कृदंत शब्द)
(ii) सरकारी – (तद्धित शब्द)
(iii) हँसी – (कृदंत शब्द)
(iv) अच्छाई – (तद्धित शब्द)

प्रश्न आ.
निम्नलिखित वाक्यों में आए हुए शब्दों के वचन परिवर्तन करके वाक्य फिर से लिखिए –

(i) पेड़ पर सुंदर फूल खिला है।
उत्तर :
पेड़ों पर सुंदर फूल खिले हैं।

(ii) कला के बारे में उनकी भावना उदात्त थी।
उत्तर :
कलाओं के बारे में उनकी भावनाएँ उदात्त थीं।

(iii) दीवारों पर टँगे हुए विशाल चित्र देखे।
उत्तर :
दीवार पर टँगा हुआ विशाल चित्र देखा।

(iv) वे बहुत प्रसन्न हो जाते थे।
उत्तर :
वह बहुत प्रसन्न हो जाता था।

(v) हमारी-तुम्हारी तरह इनमें जड़ें नहीं होती।
उत्तर :
हमारी-तुम्हारी तरह इसमें जड़ नहीं होती।

उत्तर :
(vi) ये आदमी किसी भयानक वन की बात कर रहे थे।
उत्तर :
यह आदमी किसी भयानक वन की बात कर रहा था।

(vii) वह कोई बनावटी सतह की चीज है।
उत्तर :
वे कोई बनावटी सतह की चीजें हैं।

(इ) निम्नलिखित शब्दों का लिंग परिवर्तन करके प्रत्येक का वाक्य में प्रयोग कीजिए –
अध्यापक, रानी, नायिका, देवर, पंडित, यक्ष, बुद्धिमान, श्रीमती, दुखियारा, विद्वान

परिवर्तित शब्द	वाक्य में प्रयोग
(1) ……………………………………….	(1) ……………………………………….
(2) ……………………………………….	(2) ……………………………………….
(3) ……………………………………….	(3) ……………………………………….
(4) ……………………………………….	(4) ……………………………………….
(5) ……………………………………….	(5) ……………………………………….
(6) ……………………………………….	(6) ……………………………………….
(7) ……………………………………….	(7) ……………………………………….
(8) ……………………………………….	(8) ……………………………………….
(9)  ……………………………………….	(9) ……………………………………….
(10) ……………………………………….	(10) ……………………………………….

उत्तर :

परिवर्तित शब्द	वाक्य में प्रयोग
अध्यापिका	मेरा सपना था कि एक दिन अध्यापिका बन जाऊँ।
राजा	राजा प्रजाहित दक्ष था।
नायक	वह उस चित्रपट का नायक था।
देवरानी	देवरानी ने चूड़ियाँ पहनी।
पंडिताइन	पंडिताइन मौसी ने मुझे पुकारा।
यक्षिनी	यक्षिणी और अप्सराएँ विहार कर रही थीं।
बुद्धिमती	वह एक बुद्धिमती नारी है।
श्रीमान	आइए, श्रीमान जी थोड़ा आराम कीजिए।
दुखियारी	बुढ़िया बेचारी दुखियारी लग रही है।
विदुषी	उस विदुषी नारी ने सभा में तात्विक चर्चा की।

अभिव्यक्ति
3.
प्रश्न अ.
‘अन्न बैंक की आवश्यकता’, इसपर अपने विचार लिखिए।
उत्तर :
‘वित्त बैंक’, ‘ब्लड बैंक’ यह शब्द हम सुनते हैं किंतु ‘अन्न बैंक’ शब्द कभी सुना नहीं है। सचमुच ऐसा बैंक अगर खुल जाए तो गरीबों के जीवन में भूखा सोने की नौबत नहीं आएगी। आज अमीरों के घरों का बहुत सारा खाना कूड़े-कचरे के हवाले हो जाता है।

होटलों का, शादी-ब्याह में लोगों के प्लेटों का बचा-खुचा खाना अगर जरूरतमंदों को मिल जाए तो बेचारों की जिंदगी खुशी से भर जाएगी। अत: ‘अन्न बैंक’ खुलवाकर वहाँ अगर ऐसा अन्न दिया जाए तो इस अन्न को सुरक्षित रखने की व्यवस्था हो जाएगी और आवश्यकता के अनुसार यह अन्न गरीबों को दे दिया जाए तो देश की भूख की समस्या हल हो पाएगी।

प्रश्न आ.
‘शिक्षा से वंचित बालकों की समस्याएँ’, इस विषय पर अपना मत लिखिए।
उत्तर :
भारत सरकार ने सर्व शिक्षा अभियान के अंतर्गत 06 से 14 साल तक के बच्चों के लिए मुफ्त और अनिवार्य शिक्षा का कानून बनाया है। फिर भी आज तक अनेक बालक हशिये पर हैं। निरक्षरता सभी समस्याओं की नींव है। इसी कारण सरकार ने शिक्षा अभियान का प्रारंभ किया है।

बंजारे, आदिवासी, गड़रिया, भटकते मजदूरों की टोलियाँ इन लोगों के बच्चे शिक्षा से वंचित रहते हैं। झुग्गी-झोपड़ियों में रहनेवाले गरीबों के बच्चे भी पेट के पीछे दौड़ते स्कूल से वंचित रहते हैं। इन समस्याओं पर सरकार के साथ हम सबका योगदान भी आवश्यक है।

आज सरकार मुक्त विद्यालय की स्थापना कर चुकी है। जिसके माध्यम से नियमित स्कूल न जानेवाले बच्चे भी शिक्षा से जुड़े रह सकते हैं। हर पढ़े-लिखे व्यक्ति ने स्कूल से वंचित बच्चों को पढ़ाने के लिए दिल से प्रयास किया तो संभव है कि समस्या कुछ हद तक मिट पाएगी।

पाठ पर आधारित लघूत्तरी प्रश्न

4.
प्रश्न अ.
‘उषा की दीपावली लघुकथा द्वारा प्राप्त संदेश लिखिए।
उत्तर :
‘उषा की दीपावली’ यह श्रीमती संतोष श्रीवास्तव जी द्वारा लिखित एक सुंदर मर्मस्पर्शी (heart touching) लघुकथा है।

इस पाठ की छोटी उषा एक संवेदनशील (sensitive) लड़की है। दीपावली के अवसर पर वह देखती है कि सफाई का काम करने वाला बबन ‘नरक चौदस’ पर जलाए हुए आटे के दीपक कूड़े-कचरे के डिब्बे में न फेंकते हुए अपनी जेब में रख रहा है। बबन इतना गरीब था कि ये दीपक सेंककर खाना चाहता था। ये आटे के दीप जिसे लोग कचरे में फेंकते हैं वे किसी का पेट भरने के भी काम आते हैं। यह सुनकर उषा को तकलीफ होती है।

शादी-ब्याह में लोग प्लेटों में जरूरत से ज्यादा खाना लेकर बाद में बचा हुआ खाना फेंकते हैं। यह दृश्य उषा को याद आया और उषा दीपावली के पकवान बबन को देकर सच्ची खुशी महसूस करती है। इस कहानी से अन्न की बरबादी टालकर बचा-खुचा अन्न गरीबों तक पहुँचाने का संदेश मिलता है। ‘देने की खुशी महसूस करने का अनोखा संदेश इस कहानी से मिलता है।

प्रश्न आ.
‘मुस्कुराती चोट’ शीर्षक की सार्थकता स्पष्ट कीजिए।
उत्तर :
‘मुस्कुराती चोट’ एक प्रेरणादायी लघुकथा है। इस कथा का बबलू अभाव में जीता है। ‘पिता की बीमारी’ माँ का संघर्ष देखकर खुद भी घर-घर जाकर रद्दी इकट्ठा करता है। किताबें न मिलने से उसकी पढ़ाई रुक गई है। बबूल एक दिन एक घर में रद्दी लेने के लिए जाता है तो उसकी बाल-मजदूरी पर बिना वजह उसके माँ-बाप को दोष देकर मालकिन ताने मारती है।

मालकिन की लड़की जब रद्दी की किताबें उसकी पढ़ाई के लिए मुफ्त में देना चाहती है, तब मालकिन विरोध करती है। किताबें लेकर वह पढ़ाई करेगा इस पर अविश्वास प्रकट करती है। ये बातें बबलू के मन को चोट पहुँचाती हैं। परंतु बाद में जब मालकिन को पता चलता है कि उन किताबों को रद्दी में बेचने के बजाय उसने खुद की पढ़ाई के लिए किताबें अलग रखी हैं तो उसे अपने अपशब्दों पर पछतावा होता है और वह बबलू की आगे की पढ़ाई का सारा खर्चा स्वयं उठाने का निश्चय करती है। इससे बबलू की चोट मुस्कुराहट में परिवर्तित होती है।

दिल की चोट अब खुशी में बदलती है। अत: सुखांत वाली इस लघुकथा को ‘मुस्कुराती चोट’ यह शीर्षक अत्यंत सार्थक लगता है।

साहित्य संबंधी सामान्य ज्ञान

5. जानकारी दीजिए:

प्रश्न अ.
संतोष श्रीवास्तव जी लिखित साहित्यिक विधाएँ –
……………………………………………………………………………………….
……………………………………………………………………………………….
उत्तर :
कहानी, उपन्यास, लघुकथा, ललित निबंध तथा यात्रा संस्मरण इन अनेक गद्य विधाओं में संतोष जी का संचार हुआ

प्रश्न आ.
अन्य लघुकथाकारों के नाम –
……………………………………………………………………………………….
……………………………………………………………………………………….
उत्तर :
डॉ. कमल किशोर गोयनका, डॉ. सतीश दुबे, संतोष सुपेकर, कमल चोपड़ा आदि।

Yuvakbharati Hindi 11th Textbook Solutions Chapter 2 लघु कथाएँ Additional Important Questions and Answers

कृतिपत्रिका

(अ) निम्नलिखित गद्यांश पढ़कर दी गई सूचनाओं के अनुसार कृतियाँ कीजिए :

गद्यांश : आटे के दीपक कंपाउंड की मुंडेर पर जलकर ……………………………………….. आँसू छलक आए। (पाठ्यपुस्तक पृष्ठ क्र. 5)

प्रश्न 1.
संजाल पूर्ण कीजिए :

उत्तर :

उत्तर लिखिए :

प्रश्न 1.
उषा की आँखों के सामने दावतों में दिखाई देनेवाला यह दृश्य आया …………………………………………..
उत्तर :
जरा-सा ट्रॅगने वाले मेहमान भरी प्लेटें कचरे के हवाले करते हैं।

प्रश्न 2.
उषा ने बबन को यह दे दी …………………………………………..
उत्तर :
दीपावली के लिए बने पकवानों की थैली।

प्रश्न 4.
कोष्ठक में दिए गए शब्द उचित स्थान पर लिखिए : (दीपक, जेब, आँखें, पटाखा)
उत्तर :
पुल्लिंग शब्द – स्त्रीलिंग शब्द
दीपक – आँखें
पटाखा – जेब

प्रश्न 5.
‘शादी में अन्न की बरबादी’ इस विषय पर अपने विचार लिखिए।
उत्तर :
आज समाज में अमीर और गरीब के बीच एक बड़ी खाई है। आज-कल शादी मतलब बड़प्पन दिखाने का एक जरिया बन गया है। शादी में होने वाला खर्चा एक अलग चिंता का विषय है। बड़े लोग शादी में जो भोजन खिलाते हैं, उनमें इतनी विविधता होती है कि खाने वाला परेशान होता है कि क्या खाया जाए और क्या न खाया जाए। खड़खाना (बुफे पद्धति) आज काफी लोकप्रिय है।

इसमें लोग कतार में खड़े होने से बचने के लिए प्लेटों में एक ही बार ढेर सारा खाना ले लेते हैं। इतना ज्यादा खाना खा नहीं पाने से आखिर जूठा फेंका जाता है। यह सारा अन्न कूड़े-कचरे में जाकर बरबाद होता है। दूसरी ओर दिन-रात परिश्रम करके भी गरीबों को पेटभर खाना नसीब नहीं होता। एक वक्त की रोटी पाने के लिए वे तरसते हैं। यह बरबाद होने वाला अन्न गरीबों तक पहुँचाने की सुविधा हो तो शादी में दुल्हा-दुल्हन को सच्ची दुआएँ मिलेंगी।

(आ) निम्नलिखित गद्यांश पढ़कर दी गई सूचनाओं के अनुसार कृतियाँ कीजिए :

गद्यांश : घर में बाबा बीमार थे ……………………………………….. इसलिए पढ़ाई रुक गई। (पाठ्यपुस्तक पृष्ठ क्र. 6)

प्रश्न 1.
तालिका पूर्ण कीजिए :
उत्तर :
बबलू की जानकारी –
पिता – बीमार
माता – चौका-बर्तन का काम करना
पढ़ाई – वीच में ही छूटना
कार्य – बाल मजदूरी / रद्दी इकट्ठा करना

प्रश्न 2.
कारण लिखिए :

(i) बबलू की पढ़ाई रुक गई थी क्योंकि
उत्तर :
किताबों के लिए पैसे नहीं थे।

(ii) रद्दी तौलते समय बबलू की नजर स्कूल की किताबों पर थीं क्योंकि ….
उत्तर :
वह चाह रहा था कि वे किताबें उसे मिल जाएँ।

प्रश्न 3.
(क) गद्यांश से शव्दयुग्म ढूँढ़कर लिखिए :
(i) ……………………………
उत्तर :
(i) चौका-बर्तन

(ii) ……………………………
उत्तर :
(ii) घर-घर

उदा. –
माँ-बाप

(ख) निम्न शब्दों के लिए हिंदी मानक शब्द लिखिए :
(i) स्कूल –
उत्तर :
(i) स्कूल – पाठशाला (विद्यालय)

(ii) कॉलेज – …………………………………………
उत्तर :
(ii) कॉलेज – महाविद्यालय

प्रश्न 4.
‘वाल-मजदूरी : कारण और उपाय’ इस विषय पर अपने विचार लिखिए।
उत्तर :
भारत में ‘बाल-मजदूरी’ करवाना एक गुनाह मान लिया जाता है, किंतु दुकान हो या खेती, होटल हो या ठेला अनेक जगहों पर छोटी उम्र के बच्चे काम करते हुए दिखाई देते हैं। बाल-मजदूरी कानूनन अपराध होने पर भी इसपर रोक नहीं लगा पा रहे हैं।

इसके पीछे अनेक कारण हैं। गरीबी, व्यसनी पिता, बीमार माता-पिता, माँ-बाप का अभाव। विपन्नावस्था (poverty) के कारण जिस उम्र में बच्चे को स्कूल जाना जरूरी है उस उम्र में उन्हें परिवार के लिए काम करना पड़ता है। इसमें न माँबाप को खुशी मिलती है न बच्चों को, परंतु दोनों ओर मजबूरी होती है।

इस समस्या को मिटाने के लिए देश में बढ़ रही अमीरी और गरीबी की खाई का मिटना बहुत जरूरी है। यह संभव नहीं तो हर अमीर परिवार द्वारा कुछ बच्चों का खर्चा चलाकर उनकी पढ़ाई का बोझ उठाने पर उनका भविष्य सुधर जाएगा।

(इ) निम्नलिखित गद्यांश पढ़कर दी गई सूचनाओं के अनुसार कृतियाँ कीजिए :

गद्यांश : बबलू ने रद्दी के पैसे ……………………………………………… बबलू की खुशी का ठिकाना न था। (पाठ्यपुस्तक पृष्ठ क्र. 6)

प्रश्न 1.
घटनाक्रम के अनुसार वाक्यों का उचित क्रम लगाइए :
(i) डाँट खाने के बावजूद बबलू मुस्कुरा रहा था।
(ii) बबलू ने बोरे में से किताबें निकालकर अलग रख दीं।
(iii) रास्ते में बबलू को मालकिन और उनकी लड़की मिल गई।
(iv) दुकानदार बबलू पर झल्ला उठा।
उत्तर :
(i) बबलू ने बोरे में से किताबें निकालकर अलग रख दीं।
(ii) दुकानदार बबलू पर झल्ला उठा।
(iii) डाँट खाने के बावजूद बबलू मुस्कुरा रहा था।
(iv) रास्ते में बबलू को मालकिन और उनकी लड़की मिल गई।

प्रश्न 2.
कारण लिखिए :
(i) मालकिन ने बबलू की आगे की पढ़ाई का खर्चा उठाने का निश्चय किया ……………………………………
उत्तर :
मालकिन ने बबलू की आगे की पढ़ाई का खर्चा उठाने का निश्चय किया क्योंकि उसने बबलू की पढ़ाई के प्रति लालसा को देखा।

(ii) बबलू अब स्कूल जा सकेगा ……………………………………
उत्तर :
बबलू अब स्कूल जा सकेगा क्योंकि उसके पास किताबें थीं।

प्रश्न 3.
(क) कृदंत रूप लिखिए :

(i) मुस्कुराना – ……………………………………
उत्तर :
मुस्कुराहट

(ii) झुकना – ……………………………………
उत्तर :
(ii) झुकाव

(ख) अपशब्द शब्द में ‘अप’ उपसर्ग लगा है, ‘अप’ उपसर्ग लगाकर नए शब्द बनाकर लिखिए :
(i) …………………………………..
(ii) …………………………………..
उत्तर :
(i) अपहरण
(iii) अपयश

उदा. – अपमान

प्रश्न 4.
शिक्षा से वंचित वालकों की सहायता हेतु उपाय मुझाइए।
उत्तर :
यह बात सत्य है कि आज शिक्षा प्राप्ति के प्रति समाज के सभी वर्गों में जागरुकता आई है लेकिन आज भी कुछ बच्चे शिक्षा से वंचित रह जाते हैं। अनुसूचित जाति जनजाति के बच्चों तक सरकार द्वारा चलाई जा रही योजनाओं की जानकारी पहुँचाना हमारा कर्तव्य है।

अध्ययनरत विद्यार्थियों के लिए छात्रावास तथा छात्रवृत्ति का प्रावधान सरकारी तथा निजी संस्थाओं द्वारा, एन्.जी.ओ. द्वारा होना चाहिए। विशिष्ट आवश्यकता वाले बच्चों के लिए जो शारीरिक रूप से अक्षम है उनके लिए विशिष्ट शिक्षा की व्यवस्था होनी चाहिए और उनके लिए विशेष अध्यापकों की नियुक्ति सरकार द्वारा होनी चाहिए। सर्वशिक्षा अभियान के साथ-साथ ‘समावेशन’ भी आवश्यक है ताकि शिक्षा से कोई भी बालक वंचित न रहें।

लघु कथाएँ Summary in Hindi

लघु कथाएँ लेखक परिचय :

श्रीमती संतोष श्रीवास्तव जी का जन्म मध्यप्रदेश के मंडला नामक गाँव में 23 नवंबर 1952 को हुआ। आधुनिक नारी जीवन के विविध आयाम तथा सामाजिक जीवन की विसंगतियाँ (discrepancy) आपके साहित्य द्वारा चित्रित है।

लघु कथाएँ रचनाएँ :

आपकी बहुविध रचनाएँ प्रकाशित हैं : जैसे ‘दबे पाँव प्यार’ ‘टेम्स की सरगम’ ‘ख्वाबों के पैरहन (उपन्यास)’ ‘बहके बसंत तुम’, ‘बहते ग्लेशियर’ (कहानी संग्रह), ‘फागुन का मन’ (ललित निबंध संग्रह) ‘नीली पत्तियों का शायराना हरारत’ (यात्रा संस्मरण)

लघु कथाएँ विधा परिचय :

‘लघुकथा’ यह एक गद्य विधा है। कथा तत्त्वों से परिपूर्ण किंतु आकार से लघुता यह उसकी विशेषता है। अत्यंत कम शब्दों में जीवन की पीड़ा, संवेदना, आनंद की गहराई को प्रकट करने की क्षमता लघुकथा में होती है। कोई भी छोटी घटना, प्रसंग, परिस्थिति का आधार लेकर लघुकथा लिखी जाती है। हिंदी साहित्य में डॉ. कमल किशोर गोयनका, डॉ. सतीश दुबे, संतोष सुपेकर, कमल चोपड़ा आदि को प्रमुख लघुकथाकार के रूप में पहचाना जाता है।

लघु कथाएँ विषय प्रवेश :
(अ) उषा की दीपावली : ‘उषा की दीपावली’ यह एक मर्मस्पर्शी (touching) लघुकथा है। अनाज की बरबादी पर बालमन की संवेदनशील प्रतिक्रिया का सुंदर चित्रण है। एक ओर शादी-ब्याह में थालियों में जरूरत से ज्यादा खाना लेकर उसे जूठा छोड़ने वाले श्रीमान लोग तो दूसरी ओर एक वक्त की रूखी-सूखी रोटी के लिए भी तरसने वाले लोग, इस सामाजिक विरोधाभास (paradox) का चित्रण इस लघुकथा में है।

(आ) ‘मुस्कुराती चोट’ : ‘मुस्कुराती चोट’ इस दूसरी लघुकथा में गरीबी के कारण इच्छा होकर भी पढ़ाई जारी न रख पानेवाला एक छोटा लड़का ‘बबलू’ की उथल-पुथल भरी जिंदगी है। बबलू की पढ़ाई की अदम्य (indomitable) इच्छा, उसकी बाधाएँ, छोटी-छोटी चोटें और अंत में मुस्कुराहट फैलानेवाली सुखद बात पाठकों को लुभाती है। दोनों लघुकथाएँ ‘आशावाद’ को जगाती है।

लघु कथाएँ महावरा :

टस से मस न होना – बिल्कुल प्रभाव न पड़ना, कुछ परिणाम न होना।

लघु कथाएँ सारांश :

(अ) उषा की दीपावली : दीपावली का त्योहार : दीपावली का त्योहार था। नरक-चौदस के दिन लोगों ने कंपाउंड के मुंडेर पर आटे के दीप जलाए थे। सुबह तक वे दीप बुझ गए थे।

बबन की गरीबी : सुबह बबन सफाई के लिए आया था, जो एक गरीब इन्सान था। बुझे दीप उठाकर कूड़े में फेंकने के बजाय वह जेब में रख रहा था। दस साल की उषा ने यह दृश्य देखा। उसे पूछने पर पता चला कि बबन वे आटे के दीपक सेंककर खाना चाहता है।

उषा की संवेदना : उषा की आँखों के सामने वह दृश्य घूमने लगा, जो उसने अनेक बार देखा था। अमीर लोग शादी-ब्याह में ढेर सारा खाना प्लेटों में लेकर बचा-खुचा फेंक देते हैं। उषा की आँखें भर आईं। घर में जाकर उसने दीपावली के मौके पर बनाए हुए पकवान लाकर बबन को दे दिए जिसकी वजह से बबन की आँखों में खुशी के हजारों दीप जगमगा उठे। उषा की दीपावली भी इससे खुशहाल हो गई।

(आ) मुस्कुराती चोट : ‘बबलू की गरीबी’ – इस लघुकथा का बबलू, गरीब परिवार का लड़का है। माँ चौका-बर्तन करके कुछ पैसे कमाती है। पिता जी बीमार है। माँ का हाथ बँटाने के लिए बबलू घर-घर जाकर रद्दी इकट्ठा करके रद्दी वाले को बेचता है। बबलू के पास किताबें खरीदने के लिए पैसे न होने से उसका स्कूल छूट गया था।

अविश्वास की चोट : एक दिन वह एक घर में रद्दी लेने गया था। उस रददी में किताबें थीं। घर मालकिन की बेटी वे किताबें बबलू को पढ़ने के लिए मुफ्त में देना चाहती थी। परंतु घर मालकिन का बबलू पर भरोसा नहीं था। वह किताबें बेचकर चैन करेगा ऐसा उसे शक था। बबलू ने रद्दी में से किताबें अलग रखवा दीं। रद्दी वाले को किताबों के पैसे खर्च करने की झूठ बात बताकर उससे डाँट सुनकर भी चुप रहा।

पछतावा : बबलू किताबें लेकर वापस आ रहा था। मालकिन ने उसे देखा तो अपने अपशब्दों पर उसे पछतावा हुआ। उसे पता चला कि बबलू पढ़ने की अदम्य इच्छा रखता है। बबलू की आगे की सारी पढ़ाई का खर्चा उठाने का उसने निश्चय किया। सुखद अंत वाली यह कहानी आशावादी है।

लघु कथाएँ शब्दार्थ :

• सैलाब = बाढ़ (Flood),
• देहरी = दहलीज (threshold),
• तरबतर = भीगा हुआ, गीला (wet),
• लालसा = इच्छा (desire),
• मुंडेर = दीवार का ऊपरी भाग जो छत के चारों ओर कुछ उठा होता है। (parapet),
• झल्लाना = परेशान होना (irritated)
• देहरी = दहलीज
• तरबतर = गीला, भीगा
• कृशकाय = दुबला-पतला शरीर
• बेरहमी = निर्दयता, दयाहीनता

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B)

(I) Select the correct option from the given alternatives.

Question 1.
Given A = \(\left[\begin{array}{ll}
1 & 3 \\
2 & 2
\end{array}\right]\), I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) if A – λI is a singular matrix, then ___________
(a) λ = 0
(b) λ² – 3λ – 4 = 0
(c) λ² + 3λ – 4 = 0
(d) λ² – 3λ – 6 = 0
Answer:
(b) λ² – 3λ – 4 = 0
Hint:

Question 2.
Consider the matrices A = \(\left[\begin{array}{ccc}
4 & 6 & -1 \\
3 & 0 & 2 \\
1 & -2 & 5
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & 4 \\
0 & 1 \\
-1 & 2
\end{array}\right]\), C = \(\left[\begin{array}{l}
3 \\
1 \\
2
\end{array}\right]\). Out of the given matrix products, ___________
(i) (AB)^TC
(ii) C^TC(AB)^T
(iii) C^TAB
(iv) A^TABB^TC
(a) Exactly one is defined
(b) Exactly two are defined
(c) Exactly three are defined
(d) all four are defined
Answer:
(c) Exactly three are defined
Hint:
A is of order 3 × 3, B is of order 3 × 2 and C is of order 3 × 1.
(AB)^TC is of order 2 × 1.
C^TC and (AB)^T are of different orders.
C^TC (AB)^T is not defined.
C^TAB is of order 1 × 2.
A^TABB^TC is of order 3 × 1.

Question 3.
If A and B are square matrices of equal order, then which one is correct among the following?
(a) A + B = B + A
(b) A + B = A – B
(c) A – B = B – A
(d) AB = BA
Answer:
(a) A + B = B + A
Hint:
Matrix addition is commutative.
∴ A + B = B + A

Question 4.
If A = \(\left[\begin{array}{ccc}
1 & 2 & 2 \\
2 & 1 & -2 \\
a & 2 & b
\end{array}\right]\) is a matrix satisfying the equation AA^T = 9I, where I is the identity matrix of order 3, then the ordered pair (a, b) is equal to ___________
(a) (2, -1)
(b) (-2, 1)
(c) (2, 1)
(d) (-2, -1)
Answer:
(d) (-2, -1)

Question 5.
If A = \(\left[\begin{array}{ll}
\alpha & 2 \\
2 & \alpha
\end{array}\right]\) and |A³| = 125, then α = ___________
(a) ±3
(b) ±2
(c) ±5
(d) 0
Answer:
(a) ±3
Hint:
|A³| = 125
|A|³ = 53 …….[∵ |Aⁿ| = |A|ⁿ, n ∈ N]
∴ |A| = 5
\(\left|\begin{array}{ll}
\alpha & 2 \\
2 & \alpha
\end{array}\right|=5\)
α² – 4 = 5
α² = 9
∴ α = ± 3

Question 6.
If \(\left[\begin{array}{ll}
5 & 7 \\
x & 1 \\
2 & 6
\end{array}\right]-\left[\begin{array}{cc}
1 & 2 \\
-3 & 5 \\
2 & y
\end{array}\right]=\left[\begin{array}{cc}
4 & 5 \\
4 & -4 \\
0 & 4
\end{array}\right]\), then ___________
(a) x = 1, y = -2
(b) x = -1, y = 2
(c) x = 1, y = 2
(d) x = -1, y = -2
Answer:
(c) x = 1, y = 2

Question 7.
If A + B = \(\left[\begin{array}{ll}
7 & 4 \\
8 & 9
\end{array}\right]\) and A – B = \(\left[\begin{array}{ll}
1 & 2 \\
0 & 3
\end{array}\right]\), then the value of A is ___________
(a) \(\left[\begin{array}{ll}
3 & 1 \\
4 & 3
\end{array}\right]\)
(b) \(\left[\begin{array}{ll}
4 & 3 \\
4 & 6
\end{array}\right]\)
(c) \(\left[\begin{array}{ll}
6 & 2 \\
8 & 6
\end{array}\right]\)
(d) \(\left[\begin{array}{cc}
7 & 6 \\
8 & 12
\end{array}\right]\)
Answer:
(b) \(\left[\begin{array}{ll}
4 & 3 \\
4 & 6
\end{array}\right]\)

Question 8.
If \(\left[\begin{array}{cc}
x & 3 x-y \\
z x+z & 3 y-w
\end{array}\right]=\left[\begin{array}{ll}
3 & 2 \\
4 & 7
\end{array}\right]\), then ___________
(a) x = 3, y = 7, z = 1, w = 14
(b) x = 3, y = -5, z = -1, w = -4
(c) x = 3, y = 6, z = 2, w = 7
(d) x = -3, y = -7, z = -1, w = -14
Answer:
(a) x = 3, y = 7, z = 1, w = 14

Question 9.
For suitable matrices A, B, the false statement is ___________
(a) (AB)^T = A^TB^T
(B) (A^T)^T = A
(C) (A – B)^T = A^T – B^T
(D) (A + B)^T = A^T + B^T
Answer:
(a) (AB)^T = A^TB^T
Hint:
(AB)^T = B^TA^T

Question 10.
If A = \(\left[\begin{array}{cc}
-2 & 1 \\
0 & 3
\end{array}\right]\) and f(x) = 2x² – 3x, then f(A) = ___________
(a) \(\left[\begin{array}{cc}
14 & 1 \\
0 & -9
\end{array}\right]\)
(b) \(\left[\begin{array}{cc}
-14 & 1 \\
0 & 9
\end{array}\right]\)
(c) \(\left[\begin{array}{cc}
14 & -1 \\
0 & 9
\end{array}\right]\)
(d) \(\left[\begin{array}{cc}
-14 & -1 \\
0 & -9
\end{array}\right]\)
Answer:
(c) \(\left[\begin{array}{cc}
14 & -1 \\
0 & 9
\end{array}\right]\)
Hint:

(II) Answer the following questions.

Question 1.
If A = diag[2, -3, -5], B = diag[4, -6, -3] and C = diag [-3, 4, 1], then find
i. B + C – A
ii. 2A + B – 5C.
Solution:

Question 2.
If f(α) = A = \(\left[\begin{array}{ccc}
\cos \alpha & -\sin \alpha & 0 \\
\sin \alpha & \cos \alpha & 0 \\
0 & 0 & 1
\end{array}\right]\), find
i. f(-α)
ii. f(-α) + f(α)
Solution:

Question 3.
Find matrices A and B, where
(i) 2A – B = \(\left[\begin{array}{cc}
1 & -1 \\
0 & 1
\end{array}\right]\) and A + 3B = \(\left[\begin{array}{cc}
1 & -1 \\
0 & 1
\end{array}\right]\)
(ii) 3A – B = \(\left[\begin{array}{ccc}
-1 & 2 & 1 \\
1 & 0 & 5
\end{array}\right]\) and A + 5B = \(\left[\begin{array}{ccc}
0 & 0 & 1 \\
-1 & 0 & 0
\end{array}\right]\)
Solution:
i. Given equations are
2A – B = \(\left[\begin{array}{cc}
1 & -1 \\
0 & 1
\end{array}\right]\) …….(i)

Question 4.
If A = \(\left[\begin{array}{cc}
2 & -3 \\
3 & -2 \\
-1 & 4
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
-3 & 4 & 1 \\
2 & -1 & -3
\end{array}\right]\), verify
i. (A + 2B^T)^T = A^T + 2B
ii. (3A – 5B^T)^T = 3A^T – 5B.
Solution:

Question 5.
If A = \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\) and A + A^T = I, where I is a unit matrix of order 2 × 2, then find the value of α.
Solution:

Question 6.
If A = \(\left[\begin{array}{cc}
1 & 2 \\
3 & 2 \\
-1 & 0
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
1 & 3 & 2 \\
4 & -1 & -3
\end{array}\right]\), show that AB is singular.
Solution:

Question 7.
If A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 4 & 6 \\
1 & 2 & 3
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & -1 & 1 \\
-3 & 2 & -1 \\
-2 & 1 & 0
\end{array}\right]\), show that AB and BA are both singular matrices.
Solution:

Question 8.
If A = \(\left[\begin{array}{ccc}
1 & -1 & 0 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
2 & 2 & -4 \\
-4 & 2 & -4 \\
2 & -1 & 5
\end{array}\right]\), show that BA = 6I.
Solution:

Question 9.
If A = \(\left[\begin{array}{ll}
2 & 1 \\
0 & 3
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & 2 \\
3 & -2
\end{array}\right]\), verify that |AB| = |A|.|B|.
Solution:

Question 10.
If Aα = \(\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\), show that Aα . Aβ = Aα+β
Solution:

Question 11.
If A = \(\left[\begin{array}{cc}
1 & \omega \\
\omega^{2} & 1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
\omega^{2} & 1 \\
1 & \omega
\end{array}\right]\), where ω is a complex cube root of unity, then show that AB + BA + A – 2B is a null matrix.
Solution:
ω is the complex cube root of unity.
ω³ = 1
ω³ – 1 = 0
(ω – 1) (ω² + ω + 1) = 0
ω = 1 or ω² + ω + 1 = 0
But, ω is a complex number.
1 + ω + ω² = 0 …….(i)
AB + BA + A – 2B

which is a null matrix.

Question 12.
If A = \(\left[\begin{array}{lrr}
2 & -2 & -4 \\
-1 & 3 & 4 \\
1 & -2 & -3
\end{array}\right]\), show that A² = A.
Solution:

Question 13.
If A = \(\left[\begin{array}{ccc}
4 & -1 & -4 \\
3 & 0 & -4 \\
3 & -1 & -3
\end{array}\right]\), show that A² = I.
Solution:

Question 14.
If A = \(\left[\begin{array}{cc}
3 & -5 \\
-4 & 2
\end{array}\right]\), show that A² – 5A – 14I = 0.
Solution:

Question 15.
If A = \(\left[\begin{array}{cc}
2 & -1 \\
-1 & 2
\end{array}\right]\), show that A – 4A + 3I = 0.
Solution:

Question 16.
If A = \(\left[\begin{array}{cc}
-3 & 2 \\
2 & -4
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & x \\
y & 0
\end{array}\right]\) and (A + B)(A – B) = A² – B², find x and y.
Solution:
(A + B)(A – B) = A² – B²
A² – AB + BA – B² = A² – B²
-AB + BA = 0
AB = BA

By equality of matrices, we get
2 – 4x = -3x
∴ x = 2 and 2y = 2x
y = x
∴ y = 2
∴ x = 2, y = 2

Question 17.
If A = \(\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}\right]\), show that (A + B)(A – B) ≠ A² – B².
Solution:
We have to prove that
(A + B) . (A – B) ≠ A² – B²
i.e., to prove that A(A – B) + B(A – B) ≠ A² – B²
i.e., to prove that A² – AB + BA – B² ≠ A² – B²
i.e., to prove that AB ≠ BA.

∴ AB ≠ BA

Question 18.
If A = \(\left[\begin{array}{ll}
2 & -1 \\
3 & -2
\end{array}\right]\), find A³.
Solution:

∴ A² = I
Multiplying throughout by A, we get
A³ = A . I
∴ A³ = A

Question 19.
Find x, y if,

Solution:

Question 20.
Find x, y, z if

Solution:

Question 21.
If A = \(\left[\begin{array}{ccc}
2 & 1 & -3 \\
0 & 2 & 6
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & 0 & -2 \\
3 & -1 & 4
\end{array}\right]\), find AB^T and A^TB.
Solution:

Question 22.
If A = \(\left[\begin{array}{cc}
2 & -4 \\
3 & -2 \\
0 & 1
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
-2 & 1 & 0
\end{array}\right]\), show that (AB)^T = B^TA^T.
Solution:

Question 23.
If A = \(\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right]\), prove that Aⁿ = \(\left[\begin{array}{cc}
1+2 n & -4 n \\
n & 1-2 n
\end{array}\right]\), for all n ∈ N.
Solution:

Question 24.
If A = \(\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\), prove that Aⁿ = \(\left[\begin{array}{cc}
\cos n \theta & \sin n \theta \\
-\sin n \theta & \cos n \theta
\end{array}\right]\), for all n ∈ N.
Solution:

Question 25.
Two farmers Shantaram and Kantaram cultivate three crops rice, wheat, and groundnut. The sale (in Rupees) of these crops by both the farmers for the month of April and may 2008 is given below,

Find
(i) the total sale in rupees for two months of each farmer for each crop.
(ii) the increase in sales from April to May for every crop of each farmer.
Solution:
(i) Total sale for Shantaram:
For rice = 15000 + 18000 = ₹ 33000.
For wheat = 13000 + 15000 = ₹ 28000.
For groundnut = 12000 + 12000 = ₹ 24000.
Total sale for Kantaram:
For rice = 18000 + 21000 = ₹ 39000
For wheat = 15000 + 16500 = ₹ 31500
For groundnut = 8000 + 16000 = ₹ 24000

Alternate method:
Matrix form

∴ The total sale of April and May of Shantaram in ₹ is ₹ 33000 (rice), ₹ 28000 (wheat), ₹ 24000 (groundnut), and that of Kantaram in ₹ is ₹ 39000(rice), ₹ 31500(wheat), and ₹ 24000 (groundnut).

(ii) Increase in sale from April to May for Shantaram:
For rice = 18000 – 15000 = ₹ 3000
For wheat = 15000 – 13000 = ₹ 2000
For groundnut = 12000 – 12000 = ₹ 0
Increase in sale from April to May for Kantaram:
For rice = 21000 – 18000 = ₹ 3000
For wheat = 16500 – 15000 = ₹ 1500
For groundnut = 16000 – 8000 = ₹ 8000

Alternate method:
Matrix form

∴ The increase in sales for Shantaram from April to May in each crop is ₹ 3000 (rice), ₹ 2000(wheat), 0 (groundnut), and that for Kantaram is ₹ 3000 (rice), ₹ 1500 (wheat), and ₹ 8000 (groundnut).

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.7 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.7

Question 1.
Find A^T, if
i. A = \(\left[\begin{array}{cc}
1 & 3 \\
-4 & 5
\end{array}\right]\)
ii. A = \(\left[\begin{array}{ccc}
2 & -6 & 1 \\
-4 & 0 & 5
\end{array}\right]\)
Solution:
i. A = \(\left[\begin{array}{cc}
1 & 3 \\
-4 & 5
\end{array}\right]\)
∴ A^T = \(\left[\begin{array}{rr}
1 & -4 \\
3 & 5
\end{array}\right]\)

ii. A = \(\left[\begin{array}{ccc}
-4 & 0 & 5
2 & -6 & 1 \\
\end{array}\right]\)
∴ A^T = \(\left[\begin{array}{cc}
2 & -4 \\
-6 & 0 \\
1 & 5
\end{array}\right]\)

[Note: Answer given in the textbook is A^T = \(\left[\begin{array}{cc}
2 & -4 \\
6 & 0 \\
1 & 5
\end{array}\right]\). However, as per our calculation it is A^T = \(\left[\begin{array}{cc}
2 & -4 \\
-6 & 0 \\
1 & 5
\end{array}\right]\). ]

Question 2.
If [a_ij]_3×3 where a_ij = 2(i – j), find A and
A^T. State whether A and A^T are symmetric or skew-symmetric matrices?
Solution:
A = [a_ij]_3×3 = \(\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right]\)
Given a_ij = 2 (i — j)
∴ a₁₁ = 2(1-1) = 0,
a₁₂ = 2(1-2) = -2,
a₁₃ = 2(1-3) = -4,
a₂₁ = 2(2-1) = 2,
a₂₂ = 2(2-2) = 0,
a₂₃=2(2-3) = -2,
a₃₁ = 2(3-1) = 4,
a₃₂ = 2(3-2) = 2,
a₃₃=2(3-3) = 0

∴ A^T = -A and A = -A^T
∴ A and A^T both are skew-symmetric matrices.

Questionn 3.
If A = \(\left[\begin{array}{cc}
5 & -3 \\
4 & -3 \\
-2 & 1
\end{array}\right]\), prove that (2A)^T = 2A^T.
Solution:

From (i) and (ii), we get
(2A)^T = 2A^T

Question 4.
If A = \(\left[\begin{array}{ccc}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\), prove that (3A)^T = 3A^T.
Solution:

From (i) and (ii), we get
(3A)^T = 3A^T

Question 5.
If A = \(\left[\begin{array}{ccc}
0 & 1+2 i & 1-2 \\
-1-2 i & 0 & -7 \\
2-i & 7 & 0
\end{array}\right]\),
where i = \(\sqrt{-1}\), prove that A^T = – A.
Solution:

Question 6.
If A = \(\left[\begin{array}{cc}
2 & -3 \\
5 & -4 \\
-6 & 1
\end{array}\right]\) , B = \(\left[\begin{array}{cc}
2 & 1 \\
4 & -1 \\
-3 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
1 & 2 \\
-1 & 4 \\
-2 & 3
\end{array}\right]\) then show that
i. (A + B)^T = AT + B^T
ii. (A – C)^T = A^T – C^T
Solution:

From (i) and (ii), we get
(A + B)^T = AT + B^T
[Note: The question has been modified.]

From (i) and (ii), we get
(A – C)^T = A^T – C^T</sup

Question 7.
If A = \(\left[\begin{array}{cc}
5 & 4 \\
-2 & 3
\end{array}\right]\) and \(\left[\begin{array}{cc}
-1 & 3 \\
4 & -1
\end{array}\right]\) then find C^T, such that 3A – 2B + C = I, where I is the unit matrix of order 2.
Solution:
3A – 2B + C = I
∴ C = I + 2B – 3A

Question 8.
If A = \(\left[\begin{array}{ccc}
7 & 3 & 0 \\
0 & 4 & -2
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
0 & -2 & 3 \\
2 & 1 & -4
\end{array}\right]\), then find
i. A^T + 4B^T
ii. 5A^T – 5B^T
Solution:

ii. ii. 5A^T – 5B^T = 5(A^T – B^T)

Question 9.
If A = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
3 & 1 & 2
\end{array}\right]\), B = \(\left[\begin{array}{rrr}
2 & 1 & -4 \\
3 & 5 & -2
\end{array}\right]\) and C = \(\left[\begin{array}{ccc}
0 & 2 & 3 \\
-1 & -1 & 0
\end{array}\right]\), verify that (A + 2B + 3C)^T = A^T + 2B^T + 3C^T.
Solution:
A + 2B + 3C

∴ A^T + 2B^T + 3C^T = \(\left[\begin{array}{cc}
5 & 6 \\
8 & 8 \\
2 & -2
\end{array}\right]\)
From (i) and (ii), we get
(A + 2B + 3C)^T = A^T + 2B^T + 3C^T

Question 10.
If A = \(\left[\begin{array}{ccc}
-1 & 2 & 1 \\
-3 & 2 & -3
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
2 & 1 \\
-3 & 2 \\
-1 & 3
\end{array}\right]\), prove that (A + B^T)^T = A^T + B.
prove that (A + B^T)^T = A^T + B
Solution:


From (i) and (ii), we get
(A + B^T)^T = A^T + B

Question 11.
Prove that A + A^T is a symmetric and A – A^T is a skew symmetric matrix, where
i. A = \(\left[\begin{array}{ccc}
1 & 2 & 4 \\
3 & 2 & 1 \\
-2 & -3 & 2
\end{array}\right]\)
ii. A = \(\left[\begin{array}{ccc}
5 & 2 & -4 \\
3 & -7 & 2 \\
4 & -5 & -3
\end{array}\right]\)
Solution:

∴ (A + A^T)^T = A + A^T, i.e., A + A^T = (A + A^T)^T
∴ A + A^T is a symmetric matrix.


∴ (A – A^T)^T = – (A – A^T),
i.e., A – A^T = -(A – A^T)^T
∴ A – A^T is skew symmetric matrix.

∴ (A + A^T)^T = A + A^T, i.e., A + A^T = (A + A^T)^T
∴ A + A^T is a symmetric matrix.

∴ (A – A^T)^T = – (A – A^T),
i.e., A – A^T = -(A – A^T)^T
∴ A – A^T is skew symmetric matrix.

Question 12.
Express the following matrices as the sum of a symmetric and a skew symmetric matrix.
i. \(\left[\begin{array}{cc}
4 & -2 \\
3 & -5
\end{array}\right]\)
ii. \(\left[\begin{array}{ccc}
3 & 3 & -1 \\
-2 & -2 & 1 \\
-4 & -5 & 2
\end{array}\right]\)
Solution:
A square matrix A can be expressed as the sum of a symmetric and a skew symmetric matrix as
A = \(\frac{1}{2}\) (A + A^T) + \(\frac{1}{2}\) (A – A^T)

P is symmetric matrix …[∵ a_ij = a_ji]
and Q is a skew symmetric matrix [∵ -a_ij = -a_ji]
A = P + Q
A = \(\left[\begin{array}{cc}
4 & \frac{1}{2} \\
\frac{1}{2} & -5
\end{array}\right]+\left[\begin{array}{ll}
0 & \frac{-5}{2} \\
\frac{5}{2} & 0
\end{array}\right]\)

∴ P is symmetric matrix …[∵ a_ij = a_ji]
and Q is a skew symmetric matrix [∵ -a_ij = -a_ji]
∴ A = P + Q
∴ A = \(\left[\begin{array}{cc}
4 & \frac{1}{2} \\
\frac{1}{2} & -5
\end{array}\right]+\left[\begin{array}{ll}
0 & \frac{-5}{2} \\
\frac{5}{2} & 0
\end{array}\right]\)

Question 13.
If A = \(\left[\begin{array}{cc}
2 & -1 \\
3 & -2 \\
4 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
0 & 3 & -4 \\
2 & -1 & 1
\end{array}\right]\), verify that
i. (AB)^T = B^TA^T
ii. (BA)^T = A^TB^T
Solution:


From (i) and (ii), we get
(AB)^T = B^TA^T

From (i) and (ii) we get
(BA)^T = A^TB^T

Question 14.
If A = \(\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\), show that A^TA = I, where I is the unit matrix of order 2.
Solution:


∴ A^TA = I, where I is the unit matrix of order 2.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.6 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6

Question 1.
Evaluate:
i. \(\left[\begin{array}{l}
3 \\
2 \\
1
\end{array}\right]\left[\begin{array}{lll}
{[2} & -4 & 3
\end{array}\right]\)
ii. \(\left[\begin{array}{lll}
2 & -1 & 3
\end{array}\right]\left[\begin{array}{l}
4 \\
3 \\
1
\end{array}\right]\)
Solution:
i. \(\begin{aligned}
\left[\begin{array}{l}
3 \\
2 \\
1
\end{array}\right]\left[\begin{array}{lll}
2 & -4 & 3
\end{array}\right] &=\left[\begin{array}{lll}
3(2) & 3(-4) & 3(3) \\
2(2) & 2(-4) & 2(3) \\
1(2) & 1(-4) & 1(3)
\end{array}\right] \\
&=\left[\begin{array}{ccc}
6 & -12 & 9 \\
4 & -8 & 6 \\
2 & -4 & 3
\end{array}\right]
\end{aligned}\)

ii. \(\left[\begin{array}{lll}
2 & -1 & 3
\end{array}\right]\left[\begin{array}{l}
4 \\
3 \\
1
\end{array}\right]\)
= [2(4)-1(3)+ 3(1)]
= [8 – 3 + 3] = [8]

Question 2.
If A = \(\left[\begin{array}{cc}
1 & -3 \\
4 & 2
\end{array}\right]\) B = \(\left[\begin{array}{cc}
4 & 1 \\
3 & -2
\end{array}\right]\), = show that AB ≠ BA.
Solution:

From (i) and (ii), we get
AB ≠ BA

Question 3.
If A = \(\left[\begin{array}{ccc}
-1 & 1 & 1 \\
2 & 3 & 0 \\
1 & -3 & 1
\end{array}\right]\) ,B = \(\left[\begin{array}{lll}
2 & 1 & 4 \\
3 & 0 & 2 \\
1 & 2 & 1
\end{array}\right]\) state whether AB = BA? Justify your answer.
Solution:


From (i) and (ii), we get
AB ≠ BA

Question 4.
Show that AB = BA, where
i. A = \(\left[\begin{array}{rrr}
-2 & 3 & -1 \\
-1 & 2 & -1 \\
-6 & 9 & -4
\end{array}\right]\) , B = \(\left[\begin{array}{rrr}
1 & 3 & -1 \\
2 & 2 & -1 \\
3 & 0 & -1
\end{array}\right]\)
ii. A = \(\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\), B = \(\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]\)
Solution:

From (i) and (ii), we get
AB = BA

From (i) and (ii), we get
AB = BA
[Note: The question has been modified.]

Question 5.
If A = \(\left[\begin{array}{cc}
4 & 8 \\
-2 & -4
\end{array}\right]\), prove that A² = 0.
Solution:
A² = A.A
= \(\left[\begin{array}{cc}
4 & 8 \\
-2 & -4
\end{array}\right]\left[\begin{array}{cc}
4 & 8 \\
-2 & -4
\end{array}\right]\)
= \(\left[\begin{array}{cc}
16-16 & 32-32 \\
-8+8 & -16+16
\end{array}\right] \)
= \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\) = 0

Question 6.
Verify A(BC) = (AB)C in each of the following cases:
i. A = \(=\left[\begin{array}{cc}
4 & -2 \\
2 & 3
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-1 & 1 \\
3 & -2
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
4 & 1 \\
2 & -1
\end{array}\right]\)
ii. A = \(\left[\begin{array}{ccc}
1 & -1 & 3 \\
2 & 3 & 2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & 0 \\
-2 & 3 \\
4 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
1 & 2 \\
-2 & 0 \\
4 & -3
\end{array}\right]\)
Solution:



From (i) and (ii), we get
A(BC) = (AB)C.

Question 7.
Verify that A(B + C) = AB + AC in each of the following matrices:
i. A = \(\left[\begin{array}{cc}
4 & -2 \\
2 & 3
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-1 & 1 \\
3 & -2
\end{array}\right]\) and C = \(=\left[\begin{array}{cc}
4 & 1 \\
2 & -1
\end{array}\right]\)
ii. A = \(\left[\begin{array}{ccc}
1 & -1 & 3 \\
2 & 3 & 2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & 0 \\
-2 & 3 \\
4 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
1 & 2 \\
-2 & 0 \\
4 & -3
\end{array}\right]\)
Solution:

From (i) and (ii), we get
A(B + C) = AB + AC.
[Note: The question has been modified.]

Question 8.
If A = \(\left[\begin{array}{cc}
1 & -2 \\
5 & 6
\end{array}\right]\), B = \(\left[\begin{array}{cc}
3 & -1 \\
3 & 7
\end{array}\right]\), find AB – 2I, where I is unit matrix of order 2.
Solution:

Question 9.
If A = \(\left[\begin{array}{ccc}
4 & 3 & 2 \\
-1 & 2 & 0
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & 2 \\
-1 & 0 \\
1 & -2
\end{array}\right]\), show that matrix AB is non singular.
Solution:
im
∴ AB is non-singular matrix.

Question 10.
If A = \(\), find the product (A + I)(A – I).
Solution:

[Note : Answer given in the textbook is \(\left[\begin{array}{ccc}
9 & 6 & 4 \\
15 & 32 & -2 \\
35 & -7 & 29
\end{array}\right]\)
However, as per our calculation it is \(\left[\begin{array}{ccc}
10 & 10 & 4 \\
25 & 39 & 2 \\
35 & 7 & 22
\end{array}\right]\). ]

Question 11.
If A = \(\left[\begin{array}{ll}
\alpha & 0 \\
1 & 1
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & 0 \\
2 & 1
\end{array}\right]\), find α, if A² = B.
Solution:
A² = B

∴ By equality of matrices, we get
α² = 1 and α + 1 = 2
∴ α = ± 1 and α = 1
∴ α = 1

Question 12.
If A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\), show that A² – 4A is scalar matrix.
Solution:
A² – 4A = A.A – 4A

Question 13.
If A = \(\left[\begin{array}{cc}
1 & 0 \\
-1 & 7
\end{array}\right]\), find k so that A² – 8A – kI = O, where I is a unit matrix and O is a null matrix of order 2.
Solution:
A² – 8A – kI = O
∴ A.A – 8A – kI = O

∴ by equality of matrices, we get
1 – 8 – k = 0
∴ k = -7

Question 14.
If A = \(\left[\begin{array}{cc}
8 & 4 \\
10 & 5
\end{array}\right]\), B = \(\left[\begin{array}{cc}
5 & -4 \\
10 & -8
\end{array}\right]\), show that (A+B)² = A² + AB + B².
Solution:
We have to prove that (A + B)² = A² + AB + B²,
i.e., to prove A² + AB + BA + B² = A² + AB + B²,
i.e., to prove BA = 0.
BA = \(\left[\begin{array}{cc}
5 & -4 \\
10 & -8
\end{array}\right]\left[\begin{array}{cc}
8 & 4 \\
10 & 5
\end{array}\right]\)
\(\left[\begin{array}{cc}
40-40 & 20-20 \\
80-80 & 40-40
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)

Question 15.
If A = \(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\), prove that A² – 5A + 7I = 0, where I is unit matrix of order 2.
Solution:
A² – 5A + 7I = 0 = A.A – 5A + 7I = 0

Question 16.
If A = \(\left[\begin{array}{cc}
3 & 4 \\
-4 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
2 & 1 \\
-1 & 2
\end{array}\right]\), show that (A + B)(A – B) = A² – B².
Solution:
We have to prove that (A + B)(A – B) = A² – B²,
i.e., to prove A² – AB + BA – B² = A² – B²,
i.e., to prove – AB + BA = 0,
i.e., to prove AB – BA.

From (i) and (ii), we get AB = BA

Question 17.
If A = \(\left[\begin{array}{cc}
1 & 2 \\
-1 & -2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & a \\
-1 & b
\end{array}\right]\) and (A + B)² = A² + B², find the values of a and b.
Solution:
Given, (A + B)² = A² + B²
∴ A² + AB + BA + B² = A² + B²
∴ AB + BA = 0
∴ AB = -BA

∴ by equality of matrices, we get
– 2 + a = 0 and 1 + b = 0
a = 2 and b = -1
[Note: The question has been modified.]

Question 18.
Find matrix X such that AX = B,
where A = \(\left[\begin{array}{cc}
1 & -2 \\
-2 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{c}
-3 \\
-1
\end{array}\right]\)
Solution:
Let X = \(\left[\begin{array}{c}
a \\
b
\end{array}\right]\)
But AX = B
∴ \(\left[\begin{array}{cc}
1 & -2 \\
-2 & 1
\end{array}\right]\left[\begin{array}{l}
\mathrm{a} \\
\mathrm{b}
\end{array}\right]=\left[\begin{array}{r}
-3 \\
-1
\end{array}\right]\)
∴ \(\left[\begin{array}{c}
a-2 b \\
-2 a+b
\end{array}\right]=\left[\begin{array}{l}
-3 \\
-1
\end{array}\right]\)
By equality of matrices, we get
a – 2b = -3 …(i)
-2a + b = -l …(ii)
By (i) x 2 + (ii), we get
-3b =-7
∴ b = \(\frac{7}{3}\)
Substituting b = \(\frac{7}{3}\) in (i), we get
a – 2 (\(\frac{7}{3}\)) = -3
∴ a = -3 + \(\frac{14}{3}=\frac{5}{3}\)
∴ X = \(\left[\begin{array}{l}
\frac{5}{3} \\
\frac{7}{3}
\end{array}\right]\)

Question 19.
Find k, if A = \(\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\) and A² = KA – 2I
Solution:
A² = kA – 2I
∴ AA + 2I = kA

∴ \(\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]=\left[\begin{array}{ll}
3 k & -2 k \\
4 k & -2 k
\end{array}\right]\)
∴ By equality of matrices, we get
3k = 3
∴ k = 1

Question 20.
Find x, if \(\left[\begin{array}{lll}
1 & x & 1
\end{array}\right]\left[\begin{array}{ccc}
1 & 2 & 3 \\
4 & 5 & 6 \\
3 & 2 & 5
\end{array}\right]\left[\begin{array}{c}
1 \\
-2 \\
3
\end{array}\right]\) = 0
Solution:

∴ [6 + 12x + 14] =[0]
∴ By equality of matrices, we get
∴ 12x + 20 = 0
∴ 12x =-20
∴ x = \(\frac{-5}{3}\)

Question 21.
Find x and y, if \(\left\{4\left[\begin{array}{ccc}
2 & -1 & 3 \\
1 & 0 & 2
\end{array}\right]-\left[\begin{array}{ccc}
3 & -3 & 4 \\
2 & 1 & 1
\end{array}\right]\right\}\left[\begin{array}{c}
2 \\
-1 \\
1
\end{array}\right]=\left[\begin{array}{l}
x \\
y
\end{array}\right]\)
Solution:

∴ By equality of matrices, we get
x = 19 andy = 12

Question 22.
Find x, y, z if
\(\left\{3\left[\begin{array}{ll}
2 & 0 \\
0 & 2 \\
2 & 2
\end{array}\right]-4\left[\begin{array}{cc}
1 & 1 \\
-1 & 2 \\
3 & 1
\end{array}\right]\right\}\left[\begin{array}{l}
1 \\
2
\end{array}\right]=\left[\begin{array}{c}
x-3 \\
y-1 \\
2 z
\end{array}\right]\)
Solution:

∴ By equality of matrices, we get
x – 3 = -6,y – 1 = 0, 2z = -2
∴ x = – 3, y = 1, z = – 1

Question 23.
If A = \(\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\) show that A² = \(=\left[\begin{array}{cc}
\cos 2 \alpha & \sin 2 \alpha \\
-\sin 2 \alpha & \cos 2 \alpha
\end{array}\right]\)
Solution:

Question 24.
If A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 5
\end{array}\right]\), B = \(\left[\begin{array}{cc}
0 & 4 \\
2 & -1
\end{array}\right]\)
show that AB ≠ BA, but |AB| = |A| . |B|.
Solution:
AB = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 5
\end{array}\right]\left[\begin{array}{cc}
0 & 4 \\
2 & -1
\end{array}\right]\)

Now, |AB| = \(\left|\begin{array}{cc}
4 & 2 \\
10 & 7
\end{array}\right|\) = 28 – 20 = 8
|A| = \(\left|\begin{array}{ll}
1 & 2 \\
3 & 5
\end{array}\right|\) = 5 – 6 = -1
|B| = \(\left|\begin{array}{cc}
0 & 4 \\
2 & -1
\end{array}\right|\) = 0 – 8 = -8
∴ |A| . |B| = (-1).(-8) = 8 = |AB|
∴ AB ≠ BA, but |AB| = |A|.|B|

Question 25.
Jay and Ram are two friends in a class. Jay wanted to buy 4 pens and 8 notebooks, Ram wanted to buy 5 pens and 12 notebooks. Both of them went to a shop. The price of a pen and a notebook which they have selected was 6 and ₹ 10. Using matrix multiplication, find the amount required from each one of them.
Solution:
Let A be the matrix of pens and notebooks and B be the matrix of prices of one pen and one notebook.

The total amount required for each one of them is obtained by matrix AB.

∴ Jay needs ₹ 104 and Ram needs ₹ 150.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.5 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.5

Question 1.
If A = \(\left[\begin{array}{cc}
2 & -3 \\
5 & -4 \\
-6 & 1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-1 & 2 \\
2 & 2 \\
0 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
4 & 3 \\
-1 & 4 \\
-2 & 1
\end{array}\right]\)
Show that
i. A+B=B+A
ii. (A + B) + C = A + (B + C)
Solution:

From (i) and (ii), we get
A + B = B + A

Question 2.
If A = \(\left[\begin{array}{cc}
1 & -2 \\
5 & 3
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & -3 \\
4 & -7
\end{array}\right]\) then find the matrix A – 2B + 6I, where I is the unit matrix of order 2.
Solution:

Question 3.
If A = \(\), B = \(\) then find the matrix C such that A + B + C is a zero matrix.
Solution:
A+ B + C is a zero matrix.
∴ A + B + C = O
C = -(A + B)

Question 4.
If A = \(\left[\begin{array}{cc}
1 & -2 \\
3 & -5 \\
-6 & 0
\end{array}\right]\) B = \(\left[\begin{array}{cc}
-1 & -2 \\
4 & 2 \\
1 & 5
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
2 & 4 \\
-1 & -4 \\
-3 & 6
\end{array}\right]\) , find the matrix X such that 3A – 4B + 5X = C.
Solution:
3A-4B + 5X = C
∴ 5X = C + 4B – 3A

Question 5.
Solve the following equations for X and Y, if 3X – Y = \(=\left[\begin{array}{cc}
1 & -1 \\
-1 & 1
\end{array}\right]\) and X – 3Y = \(\left[\begin{array}{ll}
0 & -1 \\
0 & -1
\end{array}\right]\)
Solution:
Given equations are
\(=\left[\begin{array}{cc}
1 & -1 \\
-1 & 1
\end{array}\right]\)……………….. (i)
and X – 3Y = \(\left[\begin{array}{ll}
0 & -1 \\
0 & -1
\end{array}\right]\) ………………(ii)
By (i) x 3 – (ii) we get

Question 6.
Find the matrices A and B, if 2A – B = \(=\left[\begin{array}{ccc}
6 & -6 & 0 \\
-4 & 2 & 1
\end{array}\right]\) and A – 2B = \(\left[\begin{array}{ccc}
3 & 2 & 8 \\
-2 & 1 & -7
\end{array}\right]\)
Solution:
Given equations are
2A – B = \(=\left[\begin{array}{ccc}
6 & -6 & 0 \\
-4 & 2 & 1
\end{array}\right]\) ……………….. (i)
and A – 2B = \(\left[\begin{array}{ccc}
3 & 2 & 8 \\
-2 & 1 & -7
\end{array}\right]\) ……………….(ii)
By (i) – (ii) x 2, we get

Question 7.
Simplify \(\cos \theta\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]+\sin \theta\left[\begin{array}{cc}
\sin \theta & -\cos \theta \\
\cos \theta & \sin \theta
\end{array}\right]\)
Solution:

Quesiton 8.
If A = \(\left[\begin{array}{cc}
1 & 2 i \\
-3 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
2 i & 1 \\
2 & -3
\end{array}\right]\) where i =\(\sqrt{-1}\), find A + B and A – B. Show that A + B is singular. Is A – B singular? Justify your answer.
Solution:

Question 9.
Find x and y, if \(\left[\begin{array}{ccc}
2 x+y & -1 & 1 \\
3 & 4 y & 4
\end{array}\right]+\left[\begin{array}{ccc}
-1 & 6 & 4 \\
3 & 0 & 3
\end{array}\right]=\left[\begin{array}{ccc}
3 & 5 & 5 \\
6 & 18 & 7
\end{array}\right]\)
Solution:

∴ By equality of matrices, we get
2x + y – 1 = 3 and 4y = 18
∴ 2x + y = 4 and y = \(\frac{18}{4}=\frac{9}{2}\)
∴ 2x + \(\frac{9}{2}\) = 4
∴ 2x = 4 – \(\frac{9}{2}\)
∴ 2x = \(\frac{1}{2}=\)
∴ x = –\(\frac{1}{4}=\) and y = \(\frac{9}{2}=\)

Question 10.
If \(\left[\begin{array}{ll}
2 a+b & 3 a-b \\
c+2 d & 2 c-d
\end{array}\right]=\left[\begin{array}{cc}
2 & 3 \\
4 & -1
\end{array}\right]\), find a, b, c and d.
Solution:
\(\left[\begin{array}{ll}
2 a+b & 3 a-b \\
c+2 d & 2 c-d
\end{array}\right]=\left[\begin{array}{cc}
2 & 3 \\
4 & -1
\end{array}\right]\)

∴ By equality of matrices, we get
2a + b = 2 ….(i)
3a – b = 3 ….(ii)
c + 2d = 4 ….(iii)
2c – d = -1 ….(iv)
Adding (i) and (ii), we get
5a = 5
∴ a = 1
Substituting a = 1 in (i), we get
2(1) + b = 2
∴ b = 0
By (iii) + (iv) x 2, we get
5c = 2
∴ c = \(\frac{2}{5}\)
Substituting c = \(\frac{2}{5}\) in (iii), we get
\(\frac{2}{5}\) + 2d = 4
∴ 2d = 4 – \(\frac{2}{5}\)
∴ 2d = \(\frac{18}{5}\)
∴ d = \(\frac{9}{5}\)
[Note: Answer given in the textbook is d = \(\frac{3}{5}\).
However, as per our calculation it is d = \(\frac{9}{5}\).]

Question 11.
There are two book shops owned by Suresh and Ganesh. Their sales (in Rupees) for books in three subjects – Physics, Chemistry and Mathematics for two months, July and August 2017 are given by two matrices A and B.

i. Find the increase in sales in Rupees from July to August 2017.
ii. If both book shops got 10% profit in the month of August 2017, find the profit for each bookseller in each subject in that month.
Solution:
i. Increase in sales in rupees from July to August 2017
For Suresh:
Increase in sales for Physics books
= 6650 – 5600= ₹ 1050
Increase in sales for Chemistry books
= 7055 – 6750 = ₹ 305
Increase in sales for Mathematics books
= 8905 – 8500 = ₹ 405
For Ganesh:
Increase in sales for Physics books
= 7000 – 6650 = ₹ 350
Increase in sales for Chemistry books
= 7500 – 7055 = ₹ 445
Increase in sales for Mathematics books
= 10200 – 8905 = ₹ 1295
[Note: Answers given in the textbook are 1760, 2090. However, as per our calculation, they are 1050, 305, 405, 350, 445, 1295.]

ii. Both book shops got 10% profit in the month of August 2017.
For Suresh:
Profit for Physics books = \(\frac{6650 \times 10}{100}\) = ₹ 665
Profit for Chemistry books = \(\frac{7055 \times 10}{100}\) = ₹ 705.50
Profit for Mathematics books = \(\frac{8905 \times 10}{100}\) = ₹ 890.50

For Ganesh:
Profit for Physics books = \(\frac{7000 \times 10}{100}\) = ₹ 700
Profit for Chemistry books = \(\frac{7500 \times 10}{100}\) = ₹ 750
Profit for Mathematics books = \(\frac{10200 \times 10}{100}\) = ₹ 1020
[Note: Answers given in the textbook for Suresh’s profit in Chemistry and Mathematics books are ? 675 and ?850 respectively. However, as per our calculation profit amounts are ₹ 705.50 and ₹ 890.50 respectively.]

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.4 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.4

Question 1.
Construct a matrix A = [a_ij]_{3 x 2} whose elements ay are given by
i. a_ij = \(\frac{(\mathbf{i}-\mathbf{j})^{2}}{5-\mathbf{i}}\)
ii. a_ij = i – 3j
iii. a_ij \(\frac{(i+j)^{3}}{5}\)
Solution:

[Note: Answer given in the textbook is A = \(\left[\begin{array}{ll}
0 & \frac{1}{4} \\
\frac{1}{2} & 0 \\
2 & \frac{1}{2}
\end{array}\right]\)
However, as per our calculation it is \(\left[\begin{array}{ll}
0 & \frac{1}{4} \\
\frac{1}{3} & 0 \\
2 & \frac{1}{2}
\end{array}\right]\) ].

ii. a_ij = i – 3j
∴ a₁₁ = 1 – 3(1) = 1 – 3 = -2,
a₁₂= 1 – 3(2) = 1 – 6 = -5,
a₂₁ = 2 – 3(1) = 2 – 3 =-1,
a₂₂ = 2 – 3(2) = 2 – 6 = – 4
a₃₁ = 3 – 3(1) = 3-3 = 0,
a₃₂ = 3 – 3(2) = 3 – 6 = -3
∴ A = \(\left[\begin{array}{cc}
-2 & -5 \\
-1 & -4 \\
0 & -3
\end{array}\right]\)

iii. a_ij = \(\frac{(i+j)^{3}}{5}\)

Question 2.
Classify the following matrices as a row, a column, a square, a diagonal, a scalar, a unit, an upper triangular, a lower triangular, a symmetric or a skew- symmetric matrix.
i. \(\left[\begin{array}{ccc}
3 & -2 & 4 \\
0 & 0 & -5 \\
0 & 0 & 0
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
3 & -2 & 4 \\
0 & 0 & -5 \\
0 & 0 & 0
\end{array}\right]\)
As every element below the diagonal is zero in matrix A.
∴ A is an upper triangular matrix.

ii. \(\left[\begin{array}{ccc}
0 & 4 & 7 \\
-4 & 0 & -3 \\
-7 & 3 & 0
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
0 & 4 & 7 \\
-4 & 0 & -3 \\
-7 & 3 & 0
\end{array}\right]\)
∴ A^T = \(\left[\begin{array}{ccc}
0 & -4 & -7 \\
4 & 0 & 3 \\
7 & -3 & 0
\end{array}\right]\)
∴ A^T = \(-\left[\begin{array}{ccc}
0 & 4 & 7 \\
-4 & 0 & -3 \\
-7 & 3 & 0
\end{array}\right]\)
∴ A^T = -A, i.e., A = -A^T
∴ A is a skew-symmetric matrix.

iii. \(\left[\begin{array}{c}
5 \\
4 \\
-3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{c}
5 \\
4 \\
-3
\end{array}\right]\)
∴ As matrix A has only one column.
∴ A is a column matrix.

iv. \(\left[\begin{array}{lll}
9 & \sqrt{2} & -3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
9 & \sqrt{2} & -3
\end{array}\right]\)
As matrix A has only one row.
∴ A is a row matrix.

v. \(\left[\begin{array}{ll}
6 & 0 \\
0 & 6
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
6 & 0 \\
0 & 6
\end{array}\right]\)
As matrix A has all its non-diagonal elements zero and diagonal elements same.
∴ A is a scalar matrix.

vi. \(\left[\begin{array}{ccc}
2 & 0 & 0 \\
3 & -1 & 0 \\
-7 & 3 & 1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
2 & 0 & 0 \\
3 & -1 & 0 \\
-7 & 3 & 1
\end{array}\right]\)
As every element above the diagonal is zero in matrix A.
∴ A is a lower triangular matrix.

vii. \(\left[\begin{array}{ccc}
3 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & \frac{1}{3}
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
3 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & \frac{1}{3}
\end{array}\right]\)
As matrix A has all its non-diagonal elements zero.
∴ A is a diagonal matrix.

viii. \(\left[\begin{array}{ccc}
10 & -15 & 27 \\
-15 & 0 & \sqrt{34} \\
27 & \sqrt{34} & \frac{5}{3}
\end{array}\right]\)
Solution:

∴ A^T = A, i/e., A = A^T
∴ A is a symmetric matrix.

ix. \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
Solution:
A = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
In matrix A, all the non-diagonal elements are zero and diagonal elements are one.
∴ A is a unit (identity) matrix.

x. \(\left[\begin{array}{lll}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{array}\right]\)
∴ A^T = A, i/e., A = A^T
∴ A is a symmetric matrix.
∴ A is a symmetric matrix.

Question 3.
Which of the following matrices are singular or non-singular?
i. \(\left[\begin{array}{ccc}
\mathbf{a} & \mathbf{b} & \mathbf{c} \\
\mathbf{p} & \mathbf{q} & \mathbf{r} \\
\mathbf{2 a}-\mathbf{p} & \mathbf{2 b}-\mathbf{q} & \mathbf{2 c}-\mathbf{r}
\end{array}\right]\)
ii. \(\left[\begin{array}{ccc}
5 & 0 & 5 \\
1 & 99 & 100 \\
6 & 99 & 105
\end{array}\right]\)
iii. \(\left[\begin{array}{ccc}
3 & 5 & 7 \\
-2 & 1 & 4 \\
3 & 2 & 5
\end{array}\right]\)
iv. \(\left[\begin{array}{cc}
7 & 5 \\
-4 & 7
\end{array}\right]\)
Solution:

ii. Let A = \(\left[\begin{array}{ccc}
5 & 0 & 5 \\
1 & 99 & 100 \\
6 & 99 & 105
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ccc}
5 & 0 & 5 \\
1 & 99 & 100 \\
6 & 99 & 105
\end{array}\right|\)
Applying C₂ → C₂ + C₁
|A| = \(\left|\begin{array}{ccc}
5 & 5 & 5 \\
1 & 100 & 100 \\
6 & 105 & 105
\end{array}\right|\)
= 0 … [∵ C₂ and C₃ are identical]
∴ A is a singular matrix.

iii. Let A = \(\left[\begin{array}{ccc}
3 & 5 & 7 \\
-2 & 1 & 4 \\
3 & 2 & 5
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ccc}
3 & 5 & 7 \\
-2 & 1 & 4 \\
3 & 2 & 5
\end{array}\right| \)
= 3(5 – 8) – 5(-10 – 12) + 7(-4 – 3)
= -9 + 110 – 49 = 52 ≠ 0
∴ A is a non-singular matrix.

iv. Let A = \(\left[\begin{array}{cc}
7 & 5 \\
-4 & 7
\end{array}\right]\)
∴ |A| = \(\left[\begin{array}{cc}
7 & 5 \\
-4 & 7
\end{array}\right]\) = 49 + 20 = 69 ≠ 0

Question 4.
Find k, if the following matrices are singular.
i. \(\left[\begin{array}{cc}
7 & 3 \\
-2 & k
\end{array}\right]\)
ii. \(\left[\begin{array}{ccc}
4 & 3 & 1 \\
7 & k & 1 \\
10 & 9 & 1
\end{array}\right]\)
iii. \(\left[\begin{array}{ccc}
k-1 & 2 & 3 \\
3 & 1 & 2 \\
1 & -2 & 4
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
7 & 3 \\
-2 & k
\end{array}\right]\)
Since A is a singular matrix,
|A|=0
∴ \(\left|\begin{array}{cc}
7 & 3 \\
-2 & \mathrm{k}
\end{array}\right|\) = o
∴ 7k + 6 = 0
∴ 7k = -6
k = -6/7

ii. Let A = \(\left[\begin{array}{ccc}
4 & 3 & 1 \\
7 & k & 1 \\
10 & 9 & 1
\end{array}\right]\)
Since A is a singular matrix,
|A|= 0
∴ \(\left|\begin{array}{ccc}
4 & 3 & 1 \\
7 & \mathrm{k} & 1 \\
10 & 9 & 1
\end{array}\right|\) = 0
∴ 4(k – 9) – 3(7 – 10) + 1(63 – 10k) = 0
∴ 4k – 36 + 9 + 63 – 10k = 0
∴ -6k + 36 = 0
∴ 6k = 36
∴ k = 6

iii. Let A = \(\left[\begin{array}{ccc}
\mathbf{k}-1 & 2 & 3 \\
3 & 1 & 2 \\
1 & -2 & 4
\end{array}\right]\)
Since A is a singular matrix
|A| = 0
∴ \(\left|\begin{array}{ccc}
k-1 & 2 & 3 \\
3 & 1 & 2 \\
1 & -2 & 4
\end{array}\right|\)
∴ (k – 1)(4 + 4) – 2(12 – 2) + 3 (-6 – 1) = 0
∴ 8k-8-20-21 =0
∴ 8k = 49
∴ k = 49/8

Question 5.
If A = \(\left[\begin{array}{lll}
5 & 1 & -1 \\
3 & 2 & 0
\end{array}\right]\), find (A^T)^T.
Solution:

Question 6.
If A = \(\), find (A^T)^T.
Solution:

Question 7.
Find a, b, c, if \(\left[\begin{array}{ccc}
1 & \frac{3}{5} & a \\
b & -5 & -7 \\
-4 & c & 0
\end{array}\right]\) is a symmetric matrix.
Solution:

Question 8.
Find x, y, z, if \(\) is a symmetric matrix.
Solution:

Question 9.
For each of the following matrices, using its transpose, state whether it is symmetric, skew-symmetric or neither.
i. \(\left[\begin{array}{ccc}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\)
ii. \(\left[\begin{array}{ccc}
2 & 5 & 1 \\
-5 & 4 & 6 \\
-1 & -6 & 3
\end{array}\right]\)
iii. \(\left[\begin{array}{ccc}
0 & 1+2 \mathbf{i} & \mathbf{i}-2 \\
-1-2 \mathbf{i} & 0 & -7 \\
2-\mathbf{i} & 7 & 0
\end{array}\right]\)
Solution:
i. Let A = \(\left[\begin{array}{ccc}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\)
∴ A^T =\(\left[\begin{array}{ccc}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\)
∴ A^T = A, i.e., A = A^T
∴ A is a symmetric matrix.

ii.

∴ A ≠ A^T, i.e., A ≠ -A^T
∴ A is neither a symmetric nor skew-symmetric matrix.

iii.

∴ A^T = -A, i.e., A = -A^T
∴ A is a skew-symmetric matrix.

Question 10.
Construct the matrix A = [a_ij]_{3 x 3}, where a_ij = i – j. State whether A is symmetric or skew-symmetric.
Solution:
A = [a_ij]_{3 x 3}
∴ A = \(\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right]\)
Given, a_ij = i – j
a₁₁ = 1-1 = 0, a₁₂ = 1-2 = – 1, a₁₃ = 1 – 3 = – 2,
a₂₁ – 2 – 1 = 1, a₂₂ = 2 – 2 = 0, a₂₃ =2 – 3 = – 1,
a₃₁ = 3 – 1 = 2, a₃₂ = 3 – 2 = 1, a₃₃ = 3 – 3 = 0

∴ A^T = -A, i.e., A = -A^T
∴ A is a skew-symmetric matrix.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(A) Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(A)

I. Select the correct option from the given alternatives.

Question 1.
The determinant D = \(\left|\begin{array}{ccc}
a & b & a+b \\
b & c & b+c \\
a+b & b+c & 0
\end{array}\right|\) = 0, if
(a) a, b, c are in A.P.
(b) a, b, c are in G.P.
(c) a, b, c are in H.P.
(d) α is a root of ax² + 2bx + c = 0
Answer:
(b) a, b, c are in G.P.
Hint:
Applying R₃ → R₃ – (R₁ + R₂), we get
\(\left|\begin{array}{llc}
a & b & a+b \\
b & c & b+c \\
0 & 0 & -(a+2 b+c)
\end{array}\right|=0\)
∴ a[-c(a + 2b + c) – 0] – b[-b(a + 2b + c) – 0] + (a + b) (0 – 0) = 0
∴ (-ac + b²) (a + 2b + c) = 0
∴ -ac + b² = 0 or a + 2b + c = 0
∴ b² = ac
∴ a, b, c are in G.P.

Question 2.
If \(\left|\begin{array}{lll}
x^{k} & x^{k+2} & x^{k+3} \\
y^{k} & y^{k+2} & y^{k+3} \\
z^{k} & z^{k+2} & z^{k+3}
\end{array}\right|\) = (x – y) (y – z) (z – x) \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\) then
(a) k = -3
(b) k = -1
(c) k = 1
(d) k = 3
Answer:
(b) k = -1
Hint:

Question 3.
Let D = \(\left|\begin{array}{ccc}
\sin \theta \cdot \cos \phi & \sin \theta \cdot \sin \phi & \cos \theta \\
\cos \theta \cdot \cos \phi & \cos \theta \cdot \sin \phi & -\sin \theta \\
-\sin \theta \cdot \sin \phi & \sin \theta \cdot \cos \phi & 0
\end{array}\right|\) then
(a) D is independent of θ
(b) D is independent of φ
(c) D is a constant
(d) \(\frac{d D}{d}\) at θ = \(\frac{\pi}{2}\) is equal to 0
Answer:
(b) D is independent of φ

Question 4.
The value of a for which the system of equations a³x + (a + 1)y + (a + 2)³ z = 0, ax + (a + 1)y + (a + 2)z = 0 and x + y + z = 0 has a non zero solution is
(a) 0
(b) -1
(c) 1
(d) 2
Answer:
(b) -1
Hint:
The given system of equations will have a non-zero solution, if

Question 5.
\(\left|\begin{array}{lll}
b+c & c+a & a+b \\
q+r & r+p & p+q \\
y+z & z+x & x+y
\end{array}\right|=\)
(a) 2 \(\left|\begin{array}{lll}
c & b & a \\
r & q & p \\
z & y & x
\end{array}\right|\)
(b) 2 \(\left|\begin{array}{lll}
b & a & c \\
q & p & r \\
y & x & z
\end{array}\right|\)
(c) 2 \(\left|\begin{array}{lll}
a & b & c \\
p & q & r \\
x & y & z
\end{array}\right|\)
(d) 2 \(\left|\begin{array}{lll}
a & c & b \\
p & r & q \\
x & z & y
\end{array}\right|\)
Answer:
(c) 2 \(\left|\begin{array}{lll}
a & b & c \\
p & q & r \\
x & y & z
\end{array}\right|\)
Hint:

Question 6.
The system 3x – y + 4z = 3, x + 2y – 3z = -2 and 6x + 5y + λz = -3 has atleast one solution when
(a) λ = -5
(b) λ = 5
(c) λ = 3
(d) λ = -13
Answer:
(a) λ = -5
Hint:
The given system of equations will have more than one solution if

Question 7.
If x = -9 is a root of \(\left|\begin{array}{lll}
x & 3 & 7 \\
2 & x & 2 \\
7 & 6 & x
\end{array}\right|=0\), has other two roots are
(a) 2, -7
(b) -2, 7
(c) 2, 7
(d) -2, -7
Answer:
(c) 2, 7
Hint:

Question 8.
If \(\left|\begin{array}{ccc}
6 i & -3 i & 1 \\
4 & 3 i & -1 \\
20 & 3 & i
\end{array}\right|\) = x + iy, then
(a) x = 3, y = 1
(b) x = 1, y = 3
(c) x = 0, y = 3
(d) x = 0, y = 0
Answer:
(d) x = 0, y = 0

Question 9.
If A(0, 0), B(1, 3) and C(k, 0) are vertices of triangle ABC whose area is 3 sq.units, then the value of k is
(a) 2
(b) -3
(c) 3 or -3
(d) -2 or 2
Answer:
(d) -2 or 2

Question 10.
Which of the following is correct?
(a) Determinant is a square matrix
(b) Determinant is number associated to matrix
(c) Determinant is a number associated with a square matrix
(d) None of these
Answer:
(c) Determinant is a number associated with a square matrix

II. Answer the following questions.

Question 1.
Evaluate:
(i) \(\left|\begin{array}{ccc}
2 & -5 & 7 \\
5 & 2 & 1 \\
9 & 0 & 2
\end{array}\right|\)
(ii) \(\left|\begin{array}{ccc}
1 & -3 & 12 \\
0 & 2 & -4 \\
9 & 7 & 2
\end{array}\right|\)
Solution:
(i) \(\left|\begin{array}{ccc}
2 & -5 & 7 \\
5 & 2 & 1 \\
9 & 0 & 2
\end{array}\right|\)

= 2(4 – 0) + 5(10 – 9) + 7(0 – 18)
= 2(4) + 5(1) + 7(-18)
= 8 + 5 – 126
= -113

(ii) \(\left|\begin{array}{ccc}
1 & -3 & 12 \\
0 & 2 & -4 \\
9 & 7 & 2
\end{array}\right|\)

= 1(4 + 28) + 3(0 + 36) + 12(0 – 18)
= 1(32) + 3(36) + 12(-18)
= 32 + 108 – 216
= -76

Question 2.
Evaluate determinant along second column \(\left|\begin{array}{ccc}
1 & -1 & 2 \\
3 & 2 & -2 \\
0 & 1 & -2
\end{array}\right|\)
Solution:

Question 3.
Evaluate:
(i) \(\left|\begin{array}{ccc}
2 & 3 & 5 \\
400 & 600 & 1000 \\
48 & 47 & 18
\end{array}\right|\)
(ii) \(\left|\begin{array}{ccc}
101 & 102 & 103 \\
106 & 107 & 108 \\
1 & 2 & 3
\end{array}\right|\)
by using properties.
Solution:

Question 4.
Find the minors and cofactors of elements of the determinants.
(i) \(\left|\begin{array}{ccc}
-1 & 0 & 4 \\
-2 & 1 & 3 \\
0 & -4 & 2
\end{array}\right|\)
(ii) \(\left|\begin{array}{ccc}
1 & -1 & 2 \\
3 & 0 & -2 \\
1 & 0 & 3
\end{array}\right|\)
Solution:

Question 5.
Find the values of x, if
(i) \(\left|\begin{array}{ccc}
1 & 4 & 20 \\
1 & -2 & -5 \\
1 & 2 x & 5 x^{2}
\end{array}\right|=0\)
(ii) \(\left|\begin{array}{ccc}
1 & 2 x & 4 x \\
1 & 4 & 16 \\
1 & 1 & 1
\end{array}\right|=0\)
Solution:
(i) \(\left|\begin{array}{ccc}
1 & 4 & 20 \\
1 & -2 & -5 \\
1 & 2 x & 5 x^{2}
\end{array}\right|=0\)
⇒ 1(-10x² + 10x) – 4(5x² + 5) + 20(2x + 2) = 0
⇒ -10x² + 10x – 20x² – 20 + 40x + 40 = 0
⇒ -30x² + 50x + 20 = 0
⇒ 3x² – 5x – 2 = 0 …..[Dividing throughout by (-10)]
⇒ 3x² – 6x + x – 2 = 0
⇒ 3x(x – 2) + 1(x – 2) = 0
⇒ (x – 2) (3x + 1) = 0
⇒ x – 2 = 0 or 3x + 1 = 0
⇒ x = 2 or x = \(-\frac{1}{3}\)

(ii) \(\left|\begin{array}{ccc}
1 & 2 x & 4 x \\
1 & 4 & 16 \\
1 & 1 & 1
\end{array}\right|=0\)
⇒ 1(4 – 16) – 2x(1 – 16) + 4x(1 – 4) = 0
⇒ 1(-12) – 2x(-15) + 4x(-3) = 0
⇒ -12 + 30x – 12x = 0
⇒ 18x = 12
⇒ x = \(\frac{12}{18}=\frac{2}{3}\)

Question 6.
By using properties of determinant, prove that \(\left|\begin{array}{ccc}
x+y & y+z & z+x \\
z & x & y \\
1 & 1 & 1
\end{array}\right|=0\)
Solution:

Question 7.
Without expanding the determinants, show that

Solution:

Question 8.
If \(\left|\begin{array}{lll}
a & 1 & 1 \\
1 & b & 1 \\
1 & 1 & c
\end{array}\right|=0\) then show that \(\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1\)
Solution:

Question 9.
Solve the following linear equations by Cramer’s Rule.
(i) 2x – y + z = 1, x + 2y + 3z = 8, 3x + y – 4z = 1
(ii) \(\frac{1}{x}+\frac{1}{y}=\frac{3}{2}, \quad \frac{1}{y}+\frac{1}{z}=\frac{5}{6}, \quad \frac{1}{z}+\frac{1}{x}=\frac{4}{3}\)
(iii) 2x + 3y + 3z = 5, x – 2y + z = -4, 3x – y – 2z = 3
(iv) x + y + 2z = 7, 3x + 4y – 5z = 5, 2x – y + 3z = 12
Solution:
(i) Given equations are
2x – y + z = 1
x + 2y + 3z = 8
3x + y – 4z = 1
D = \(\left|\begin{array}{ccc}
2 & -1 & 1 \\
1 & 2 & 3 \\
3 & 1 & -4
\end{array}\right|\)
= 2(-8 – 3) – (-1)(-4 – 9) + 1(1 – 6)
= 2(-11) + 1(-13) + 1(-5)
= -22 – 13 – 5
= -40 ≠ 0

Dx = \(\left|\begin{array}{ccc}
1 & -1 & 1 \\
8 & 2 & 3 \\
1 & 1 & -4
\end{array}\right|\)
= 1(-8 – 3) – (-1)(-32 – 3) + 1(8 – 2)
= 1(-11) + 1(-35) + 1(6)
= -11 – 35 + 6
= -40

Dy = \(\left|\begin{array}{ccc}
2 & 1 & 1 \\
1 & 8 & 3 \\
3 & 1 & -4
\end{array}\right|\)
= 2(-32 – 3) -1(-4 – 9) + 1(1 – 24)
= 2(-35) – 1(-13) + 1(-23)
= -70 + 13 – 23
= -80

Dz = \(\left|\begin{array}{ccc}
2 & -1 & 1 \\
1 & 2 & 8 \\
3 & 1 & 1
\end{array}\right|\)
= 2(2 – 8) – (-1)(1 – 24) + 1(1 – 6)
= 2(-6) + 1(-23) + 1(-5)
= -12 – 23 – 5
= -40
By Cramer’s Rule,

∴ x = 1, y = 2 and z = 1 are the solutions of the given equations.

(ii) Let \(\frac{1}{x}\) = p, \(\frac{1}{y}\) = q, \(\frac{1}{z}\) = r
∴ The given equations become
p + q = \(\frac{3}{2}\)
i.e., 2p + 2q = 3
i.e., 2p + 2q + 0 = 3
q + r = \(\frac{5}{6}\)
i.e., 6q + 6r = 5,
i.e., 0p + 6q + 6r = 5
r + p = \(\frac{4}{3}\)
i.e., 3r + 3p = 4,
i.e., 3p + 0q + 3r = 4
D = \(\left|\begin{array}{lll}
2 & 2 & 0 \\
0 & 6 & 6 \\
3 & 0 & 3
\end{array}\right|\)
= 2(18 – 0) -2(0 – 18) + 0
= 2(18) – 2(-18)
= 36 + 36
= 72 ≠ 0

Dp = \(\left|\begin{array}{lll}
3 & 2 & 0 \\
5 & 6 & 6 \\
4 & 0 & 3
\end{array}\right|\)
= 3(18 – 0) – 2(15 – 24) + 0
= 3(18) – 2(-9)
= 54 + 18
= 72

Dq = \(\left|\begin{array}{lll}
2 & 3 & 0 \\
0 & 5 & 6 \\
3 & 4 & 3
\end{array}\right|\)
= 2(15 – 24) – 3(0 – 18) + 0
= 2(-9) – 3(-18)
= -18 + 54
= 36

Dr = \(\left|\begin{array}{lll}
2 & 2 & 3 \\
0 & 6 & 5 \\
3 & 0 & 4
\end{array}\right|\)
= 2(24 – 0) – 2(0 – 15) + 3(0 – 18)
= 2(24) – 2(-15) + 3(-18)
= 48 + 30 – 54
= 24
By Cramer’s Rule,

∴ x = 1, y = 2 and z = 3 are the solutions of the given equations.

(iii) Given equations are
2x + 3y + 3z = 5
x – 2y + z = -4
3x – y – 2z = 3
D = \(\left|\begin{array}{ccc}
2 & 3 & 3 \\
1 & -2 & 1 \\
3 & -1 & -2
\end{array}\right|\)
= 2(4 + 1) – 3(-2 – 3) + 3(-1 + 6)
= 2(5) – 3(-5) + 3(5)
= 10 + 15 + 15
= 40 ≠ 0

Dx = \(\left|\begin{array}{ccc}
5 & 3 & 3 \\
-4 & -2 & 1 \\
3 & -1 & -2
\end{array}\right|\)
= 5(4 + 1) – 3(8 – 3) + 3(4 + 6)
= 5(5) – 3(5) + 3(10)
= 25 – 15 + 30
= 40

Dy = \(\left|\begin{array}{ccc}
2 & 5 & 3 \\
1 & -4 & 1 \\
3 & 3 & -2
\end{array}\right|\)
= 2(8 – 3) – 5(-2 – 3) + 3(3 + 12)
= 2(5) – 5(-5) + 3(15)
= 10 + 25 + 45
= 80

Dz = \(\left|\begin{array}{ccc}
2 & 3 & 5 \\
1 & -2 & -4 \\
3 & -1 & 3
\end{array}\right|\)
= 2(-6 – 4) – 3(3 + 12) + 5(-1 + 6)
= 2(-10) – 3(15) + 5(5)
= -20 -45 + 25
= -40
By Cramer’s Rule,

∴ x = 1, y = 2 and z = -1 are the solutions of the given equations.

(iv) Given equations are
x – y + 2z = 7
3x + 4y – 5z = 5
2x – y + 3z = 12
D = \(\left|\begin{array}{ccc}
1 & -1 & 2 \\
3 & 4 & -5 \\
2 & -1 & 3
\end{array}\right|\)
= 1(12 – 5) – (-1)(9 + 10) + 2(-3 – 8)
= 1(7) + 1(19) + 2(-11)
= 7 + 19 – 22
= 4 ≠ 0

Dx = \(\left|\begin{array}{ccc}
7 & -1 & 2 \\
5 & 4 & -5 \\
12 & -1 & 3
\end{array}\right|\)
= 7(12 – 5) – (-1)(15 + 60) + 2(-5 – 48)
= 7(7) + 1(75) + 2(-53)
= 49 + 75 – 106
= 18

Dy = \(\left|\begin{array}{ccc}
1 & 7 & 2 \\
3 & 5 & -5 \\
2 & 12 & 3
\end{array}\right|\)
= 1(15 + 60) – 7(9 + 10) + 2(36 – 10)
= 1(75) – 7(19) + 2(26)
= 75 – 133 + 52
= -6

Dz = \(\left|\begin{array}{ccc}
1 & -1 & 7 \\
3 & 4 & 5 \\
2 & -1 & 12
\end{array}\right|\)
= 1(48 + 5) – (-1)(36 – 10) + 7(-3 – 8)
= 1(53) + 1(26) + 7(-11)
= 53 + 26 – 77
= 2
By Cramer’s Rule,

∴ x = \(\frac{9}{2}\), y = \(\frac{-3}{2}\) and z = \(\frac{1}{2}\) are the solutions of the given equations.

Question 10.
Find the value of k, if the following equations are consistent.
(i) (k + 1)x + (k – 1)y + (k – 1) = 0
(k – 1)x + (k + 1)y + (k – 1) = 0
(k – 1)x + (k – 1)y + (k + 1) = 0
(ii) 3x + y – 2 = 0, kx + 2y – 3 = 0 and 2x – y = 3
(iii) (k – 2)x + (k – 1)y = 17, (k – 1)x +(k – 2)y = 18 and x + y = 5
Solution:
(i) Given equations are
(k + 1)x + (k – 1)y + (k – 1) = 0
(k – 1)x + (k + 1)y + (k – 1) = 0
(k – 1)x + (k – 1)y + (k + 1) = 0
Since these equations are consistent,

⇒ 2(2k + 2 + 2k – 2) – 0 + (k – 1) (4 – 0) = 0
⇒ 2(4k) + (k – 1)4 = 0
⇒ 8k + 4k – 4 = 0
⇒ 12k – 4 = 0
⇒ k = \(\frac{4}{12}=\frac{1}{3}\)

(ii) Given equations are
3x + y – 2 = 0
kx + 2y – 3 = 0
2x – y = 3, i.e., 2x – y – 3 = 0.
Since these equations are consistent,
\(\left|\begin{array}{rrr}
3 & 1 & -2 \\
k & 2 & -3 \\
2 & -1 & -3
\end{array}\right|=0\)
⇒ 3(-6 – 3) – 1(-3k + 6) – 2(-k – 4) = 0
⇒ 3(-9) – 1(-3k + 6) – 2(-k – 4) = 0
⇒ -27 + 3k – 6 + 2k + 8 = 0
⇒ 5k – 25 = 0
⇒ k = 5

(iii) Given equations are
(k – 2)x + (k – 1)y = 17
⇒ (k – 2)x + (k – 1)y – 17 = 0
(k – 1)x + (k – 2)y = 18
⇒ (k – 1)x + (k – 2)y – 18 = 0
x + y = 5
⇒ x + y – 5 = 0
Since these equations are consistent,

⇒ -1(-5k + 10 + 18) – 1(-5k + 5 + 18) + 1(k – 1 – k + 2) = 0
⇒ -1(-5k + 28) – 1(-5k + 23) + 1(1) = 0
⇒ 5k – 28 + 5k – 23 + 1 = 0
⇒ 10k – 50 = 0
⇒ k = 5

Question 11.
Find the area of triangle whose vertices are
(i) A(-1, 2), B(2, 4), C(0, 0)
(ii) P(3, 6), Q(-1, 3), R(2, -1)
(iii) L(1, 1), M(-2, 2), N(5, 4)
Solution:
(i) Here, A(x₁, y₁) = A(-1, 2)
B(x₂, y₂) = B(2, 4)
C(x₃, y₃) = C(0, 0)

Since area cannot be negative,
A(ΔABC) = 4 sq.units

(ii) Here, P(x₁, y₁) = P(3, 6)
Q(x₂, y₂) = Q(-1, 3)
R(x₃, y₃) = R(2, -1)

A(ΔPQR) = \(\frac{25}{2}\) sq.units

(iii) Here, L(x₁, y₁) = L(1, 1)
M(x₂, y₂) = M(-2, 2)
N(x₃, y₃) = N(5, 4)

Since area cannot be negative,
A(ΔLMN) = \(\frac{13}{2}\) sq.units

Question 12.
Find the value of k,
(i) if the area of a triangle is 4 square units and vertices are P(k, 0), Q(4, 0), R(0, 2).
(ii) if area of triangle is \(\frac{33}{2}\) square units and vertices are L(3, -5), M(-2, k), N(1, 4).
Solution:
(i) Here, P(x₁, y₁) = P(k, 0)
Q(x₂, y₂) = Q(4, 0)
R(x₃, y₃) = R(0, 2)
A(ΔPQR) = 4 sq.units

(ii) Here, L(x₁, y₁) = L(3, -5), M(x₂, y₂) = M(-2, k), N(x₃, y₃) = N(1, 4)
A(ΔLMN) = \(\frac{33}{2}\) sq. units
Area of a triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
\(\pm \frac{33}{2}=\frac{1}{2}\left|\begin{array}{ccc}
3 & -5 & 1 \\
-2 & k & 1 \\
1 & 4 & 1
\end{array}\right|\)
⇒ \(\pm \frac{33}{2}=\frac{1}{2}\) [3(k – 4) – (-5) (-2 – 1) + 1 (-8 – k)]
⇒ ±33 = 3k – 12 – 15 – 8 – k
⇒ ±33 = 2k – 35
⇒ 2k – 35 = 33 or 2k – 35 = -33
⇒ 2k = 68 or 2k = 2
⇒ k = 34 or k = 1

Question 13.
Find the area of quadrilateral whose vertices are A(0, -4), B(4, 0), C(-4,0), D (0, 4).
Solution:
A(0, -4), B(4, 0), C(-4, 0), D(0, 4)


∴ A(ABDC) = A(ΔABC) + A(ΔBDC)
= 16 + 16
= 32 sq.units

Question 14.
An amount of ₹ 5000 is put into three investments at the rate of interest of 6%, 7%, and 8% per annum respectively. The total annual income is ₹ 350. If the combined income from the first two investments is ₹ 70 more than the income from the third, find the amount of each investment.
Solution:
Let the amount of each investment be ₹ x, ₹ y and ₹ z.
According to the given conditions,
x + y + z = 5000,
6% x + 7% y + 8% z = 350




∴ The amounts of investments are ₹ 1750, ₹ 1500, and ₹ 1750.

Question 15.
Show that the lines x – y = 6, 4x – 3y = 20 and 6x + 5y + 8 = 0 are concurrent. Also, find the point of concurrence.
Solution:
Given equations of the lines are
x – y = 6, i.e., x – y – 6 = 0 ……(i)
4x – 3y = 20, i.e., 4x – 3y – 20 = 0 …..(ii)
6x + 5y + 8 = 0 ……(iii)
The given lines will be concurrent, if

= 1(-24 + 100) – (-1) (32 + 120) – 6(20 + 18)
= 1(76) + 1(152) – 6(38)
= 76 + 152 – 228
= 0
∴ The given lines are concurrent.
To find the point of concurrence, solve any two equations.
Multiplying (i) by 5, we get
5x – 5y – 30 = 0 …….(iv)
Adding (iii) and (iv), we get
11x – 22 = 0
∴ x = 2
Substituting x = 2 in (i), we get
2 – y – 6 = 0
∴ y = -4
∴ The point of concurrence is (2, -4).

Question 16.
Show that the following points are collinear using determinants:
(i) L(2, 5), M(5, 7), N(8, 9)
(ii) P(5,1), Q(1, -1), R(11, 4)
Solution:
(i) Here, L(x₁, y₁) = L(2, 5)
M(x₂, y₂) = M(5, 7)
N(X₃ y₃) = N(8, 9)
If A(ΔLMN) = 0, then the points L, M, N are collinear.

∴ The points L, M, N are collinear.

(ii) Here, P(x₁, y₁) = P(5, 1)
Q(x₂, y₂) = Q(1, -1)
R(x₃, y₃) = R(11, 4)
If A(ΔPQR) = 0, then the points P, Q, R are collinear.

∴ The points P, Q, R are collinear.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.3

Question 1.
Solve the following linear equations by using Cramer’s Rule.
x+y + z = 6, x – y + z = 2,.x + 2y – z = 2
x + y – 2z = -10,
2x +y – 3z = -19, 4x + 6y + z = 2
x + z = 1, y + z = 1, x + y = 4
\(\frac{-2}{x}-\frac{1}{y}-\frac{3}{z}\) = 3, \(\frac{2}{x}-\frac{3}{y}+\frac{1}{z}\) = -13 and \(\frac{2}{x}-\frac{3}{z}\) = -11
Solution:
Given equations are
x + y + z = 6,
x – y + z = 2,
x + 2y – z = 2.
D = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -1 & 1 \\
1 & 2 & -1
\end{array}\right|\)
1 2 -1 = 1(1 -2) – 1(-1 – 1) + 1(2 + 1)
= 1 (-1)-1 (-2)+ 1(3)
= -1 + 2 + 3
= 4 ≠ 0

D_x = \(\left|\begin{array}{ccc}
6 & 1 & 1 \\
2 & -1 & 1 \\
2 & 2 & -1
\end{array}\right|\)
= 6(1 – 2) – 1(-2 – 2) + 1(4 + 2)
= 6(-1) -1 (-4) + 1(6)
= -6 + 4 + 6
= 4

D_y = \(\left|\begin{array}{ccc}
1 & 6 & 1 \\
1 & 2 & 1 \\
1 & 2 & -1
\end{array}\right|\)
= 1(-2 – 2) – 6(-l – 1) + 1(2 = 1 (- 4) – 6 (- 2) + 1(0)
= -4+12 + 0 = 8

D_z = \(\left|\begin{array}{ccc}
1 & 1 & 6 \\
1 & -1 & 2 \\
1 & 2 & 2
\end{array}\right|\)
= l(-2 – 4) – 1(2 – 2) + 6(2 + 1)
= l(-6)-l(0) + 6(3)
= -6 + 0+18 = 12

By Cramer’s Rule,

∴ x = 1, y = 2 and z = 3 are the solutions of the given equations.

ii. Given equations are x+y- 2z = -10,
2x + y – 3z = -19,
Ax + 6y + z = 2.
D = \(\left|\begin{array}{ccc}
1 & 1 & -2 \\
2 & 1 & -3 \\
4 & 6 & 1
\end{array}\right|\)
= 1(1 + 18)- 1(2+ 12)-2(12-4)
= 1(19)-1(14)-2(8)
= 19-14-16 = -11 ≠ 0

D_x = \(\left|\begin{array}{ccc}
-10 & 1 & -2 \\
-19 & 1 & -3 \\
2 & 6 & 1
\end{array}\right|\)
= -10(1 + 18) – 1(-19 + 6) – 2(- 114 – 2)
= -10(19)- 1(-13) -2(-l 16)
= -190+ 13 + 232 = 55

D_y = \(\left|\begin{array}{ccc}
1 & -10 & -2 \\
2 & -19 & -3 \\
4 & 2 & 1
\end{array}\right|\)
= 1(-19 + 6) – (-10)(2 + 12) – 2(4 + 76) = 1(-13) + 10(14) – 2(80)
= -13 + 140-160 = -33

D_z = \(\left|\begin{array}{ccc}
1 & 1 & -10 \\
2 & 1 & -19 \\
4 & 6 & 2
\end{array}\right|\)
= 1(2+ 114)-1(4+ 76)-10(12-4)
= 1(116)-1(80)-10(8)
= 116-80-80 .
= -44
By Cramer’s Rule,

∴ x = -5, y = 3 and z = 4 are the solutions of the given equations.
[Note: The question has been modified]

iii. Given equations are
x + z = 1, i.e.,x + 0y + z = 1,
y + z = 1, i.e., 0x + y + z = 1,
x + y = 4, i.e., x + y + 0z = 4.
D = \(\left|\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 1 \\
1 & 1 & 0
\end{array}\right|\)
= 1(0 – 1) – 0 + 1(0 – 1)
= 1(-1)+1(-1)
= -1-1 = -2 ≠ 0
D_x = \(\left|\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 1 \\
1 & 1 & 0
\end{array}\right|\)
= 1(0 – 1) – 0 + 1(1 -4) = l(-l)+l(-3)
= -1 – 3
= -4

D_y = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
0 & 1 & 1 \\
1 & 4 & 0
\end{array}\right|\)
= 1(0 – 4) – 1(0 – 1) + 1(0 – 1)
= 1(-4) – 1(-1) + 1(-1)
= -4 + 1 – 1
= -4

D_z = \(\left|\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 1 \\
1 & 1 & 4
\end{array}\right|\)
= 1(4 – 1) – 0 + 1(0 – 1)
= 1(3) + 1(-1)
= 3 – 1
= 2

By Cramer’s Rule,

∴ x = 2, y = 2 and z = -1 are the solutions of the given equations.

Let \(\frac{1}{x}\) = p, \(\frac{1}{y}\) = q, \(\frac{1}{z}\) = r
∴ The given equations become
-2p – q – 3r = 3, i.e., 2p + q + 3r = -3,
2p-3q + r = -13,
2p – 3r = -11, i.e., 2p + 0q – 3r = -11.
D = \(\left|\begin{array}{ccc}
2 & 1 & 3 \\
2 & -3 & 1 \\
2 & 0 & -3
\end{array}\right|\)
= 2(9 – 0) – 1(-6 – 2) + 3(0 + 6)
= 2(9) – 1(-8) + 3(6)
= 18 + 8 + 18
= 44 ≠ 0

D_P = \(\left|\begin{array}{ccc}
-3 & 1 & 3 \\
-13 & -3 & 1 \\
-11 & 0 & -3
\end{array}\right|\)
= -3(9 – 0) – 1(39 + 11) + 3(0 – 33)
= -3(9) – 1(50) + 3(-33)
= -27 – 50 – 99
= -176

D_q = \(\left|\begin{array}{ccc}
2 & -3 & 3 \\
2 & -13 & 1 \\
2 & -11 & -3
\end{array}\right|\)
= 2(39 + 11)- (-3)(-6 – 2) + 3(-22 + 26)
= 2(50) + 3(-8) + 3(4)
= 100 – 24 + 12
= 88

D_r = \(\left|\begin{array}{ccc}
2 & 1 & -3 \\
2 & -3 & -13 \\
2 & 0 & -11
\end{array}\right|\)
= 2(33 – 0) – 1(-22 + 26) – 3(0 + 6)
= 2(33) – 1(4) – 3(6)
= 66 – 4 – 18
= 44

By Cramer’s Rule,

∴ x = \(\frac/{-1}{4}\), y = \(\frac{1}{2}\) z = 1 are the solutions of the given equations.

Question 2.
The sum of three numbers is 15. If the second number is subtracted from the sum of first and third numbers, then we get 5. When the third number is subtracted from the sum of twice the first number and the second number, we get 4. Find the three numbers.
Solution:
Let the three numbers be x, y and z.
According to the given conditions, x + y + z = 15,
x + z-y = 5, i.e., x – y + z = 5,
2x + y – z = 4.
D = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -1 & 1 \\
2 & 1 & -1
\end{array}\right|\)
= 1(1 – 1) – 1(-1 – 2) + 1(1 + 2)
= 1(0) – 1(-3) + 1(3)
= 0 + 3 + 3
= 6 ≠ 0

D_x = \(\left|\begin{array}{ccc}
15 & 1 & 1 \\
5 & -1 & 1 \\
4 & 1 & -1
\end{array}\right|\)
= 15(1 – 1) – 1(-5 – 4) + 1(5 + 4)
= 15(0) – 1(-9) + 1(9)
= 0 + 9 + 9
= 18

D_y = \(\left|\begin{array}{ccc}
1 & 15 & 1 \\
1 & 5 & 1 \\
2 & 4 & -1
\end{array}\right|\)
= 1(-5 – 4) – 15(-1 – 2) + 1(4 – 10)
= 1(-9) – 15(-3) + 1(-6)
= -9 + 45 – 6 = 30

D_z = \(\left|\begin{array}{ccc}
1 & 1 & 15 \\
1 & -1 & 5 \\
2 & 1 & 4
\end{array}\right|\)
= 1(-4 – 5) – 1(4 – 10) + 15(1 + 2)
= 1(-9) – 1(-6) + 15(3)
= -9 + 6 + 45
= 42
By Cramer’s Rule,

∴ The three numbers are 3, 5 and 7.

Question 3.
Examine the consistency of the following equations.
i. 2x – y + 3 = 0, 3x + y – 2 = 0, 11x + 2y – 3 = 0
ii. 2x + 3y – 4 = 0, x + 2y = 3, 3x + 4y + 5 = 0
iii. x + 2y – 3 = 0,7x + 4y – 11 = 0,2x + 4y – 6 = 0
Solution:
i. Given equations are 2x – y + 3 = 0,
3x + y – 2 = 0,
11x + 2y – 3 = 0.
D = \(\left|\begin{array}{ccc}
2 & -1 & 3 \\
3 & 1 & -2 \\
11 & 2 & -3
\end{array}\right|\)
= 2(-3 + 4) – (-l)(-9 + 22) + 3(6-11)
= 2(1)+1(13)+ 3(-5)
= 2 + 13-15 = 0
∴ The given equations are consistent.

ii. Given equations are 2x + 3y – 4 = 0,
x + 2y = 3, i.e., x + 2y – 3 = 0,
3x + 4y + 5 = 0.
\(\left|\begin{array}{ccc}
2 & 3 & -4 \\
1 & 2 & -3 \\
3 & 4 & 5
\end{array}\right|\)
= 2(10 + 12) – 3(5 + 9) – 4(4 – 6)
= 2 (22) – 3(14) – 4(-2)
= 44 – 42 + 8
= 10 ≠ 0
∴ The given equations are not consistent.

iii. Given equations are x + 2y – 3 =
7x + 4y – 11 =0,
2x + 4y – 6 = 0.
\(\left|\begin{array}{ccc}
1 & 2 & -3 \\
7 & 4 & -11 \\
2 & 4 & -6
\end{array}\right|\)
= 1(-24 + 44) – 2(-42 + 22) – 3(28 – 8)
= 1(20) – 2(-20) – 3(20)
= 20 + 40 – 60
= 0
∴ The given equations are consistent.

Question 4.
Find k, if the following equations are consistent.
i. 2x + 3y-2 = 0,2x + 4y-k = 0,x-2j + 3k = 0
ii. kx + 3,y +1 = 0, x + 2y+1 = 0, x + y = 0
Solution:
i. Given equations are 2x + 3y – 2 = 0,
2x + 4y – k = 0,
x – 2y + 3k = 0.
Since these equations are consistent,
\(\left|\begin{array}{ccc}
2 & 3 & -2 \\
2 & 4 & -k \\
1 & -2 & 3 k
\end{array}\right|\) = 0
∴ 2(12k – 2k) – 3(6k + k) – 2(- 4 – 4) = 0
∴ 2(10k) – 3(7k) – 2(- 8) = 0
∴ 20k – 21k + 16 = 0
∴ k = 16

Given equations are are
kx + 3y + 1 = 0,
x + 2y +1=0,
x + y = 0, i.e., x + y + 0 = 0.
Since these equations are consistent,
\(\left|\begin{array}{lll}
k & 3 & 1 \\
1 & 2 & 1 \\
1 & 1 & 0
\end{array}\right|\) = 0
∴ k(0 – 1) – 3(0 – 1) + 1(1 – 2) = 0
∴ k(-1) – 3(-1) + 1(-1) = 0
∴ -k + 3 – 1 = 0
∴ k = 2.

Question 5.
Find the area of triangle whose vertices are
i. A (5,8), B (5,0), C (1,0)
ii. P(3/2, 1), Q(4,2), R(4, -1/2)
iii. M (0, 5), N (- 2, 3), T (1, – 4)
Solution:
i. Here, A(x₁, y₁) ≡ A(5, 8), B(x₂, y₂) = B(5, 0), C(x₃, y₃) = C(1,0)
Area of a triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
A(ΔABC) = \(\frac{1}{2}\left|\begin{array}{lll}
5 & 8 & 1 \\
5 & 0 & 1 \\
1 & 0 & 1
\end{array}\right|\)
= \(\frac{1}{2}\)[5(0 – 0) – 8(5 – 1) + 1(0 – 0)]
= \(\frac{1}{2}\)[0 – 8(4) + 0]
= \(\frac{1}{2}\)(-32)
= -16
Since area cannot be negative,
A(ΔABC) = 16 sq. units

ii. Here, P(x₁, y₁) ≡ P(3/2, 1), Q(x₂, y₂) ≡ Q(4, 2), R(x₃, y₃) ≡ R(4,-\(\frac{1}{2}\) )


Since area cannot be negative
A(ΔPQR) = 25/8 sq. units

iii. Here, M(x₁, y₁) ≡ M(0, 5), N(x₂, y₂) ≡ N(-2, 3)
T(x₃, y₃) ≡ T(1, -4)
Area of a triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
∴ A(ΔMNT) = \(\frac{1}{2}\left|\begin{array}{ccc}
0 & 5 & 1 \\
-2 & 3 & 1 \\
1 & -4 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [ 0 – 5 (-2 -1) + 1 (8 – 3)]
= \(\frac{1}{2}\)[-5 (-3) + 1(5)]
= \(\frac{1}{2}\) (15 + 5)
= \(\frac{1}{2}\) (20)
= 10 sq. units

Question 6.
Find the area of quadrilateral whose vertices are A (- 3,1), B (- 2, – 2), C (1,4), D (3, – 1).
Solution:
A(-3, 1), B(-2, -2), C(l, 4), D(3, -1)

A(□ ABDC) = A(ΔABD) + A(ΔADC)
Area of triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
A(ΔABD) = \(\frac{1}{2}\left|\begin{array}{ccc}
-3 & 1 & 1 \\
-2 & -2 & 1 \\
3 & -1 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [-3 (-2 + 1) – 1(-2 – 3) + 1(2 + 6)
= \(\frac{1}{2}\) [-3(-1) – 1(-5) + 1(8)]
= \(\frac{1}{2}\) (3 + 5 + 8)
= \(\frac{1}{2}\) (16)
∴ A(ΔABD) = 8 sq. units
A(ΔADC) = \(\frac{1}{2}\left|\begin{array}{ccc}
-3 & 1 & 1 \\
3 & -1 & 1 \\
1 & 4 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [-3(-1-4) – 1(3 – 1) + 1(12 + 1)]
= \(\frac{1}{2}\) [-3(-5) – 1(2) + 1(13)]
= \(\frac{1}{2}\) [15 – 2 + 13]
= \(\frac{1}{2}\) (26)
∴ A(ΔADC) = 13 sq. units
∴ A(□ ABDC) = A(ΔABD) + A(ΔADC)
= 8 + 13
= 21 sq. units

Question 7.
Find the value of k, if the area of triangle whose vertices are P (k, 0), Q (2,2), R (4,3) is \(\frac{3}{2}\) sq. units.
Solution:
Here, P(x₁, y₁) ≡ P(k, 0), Q(x₂, y₂) ≡ Q(2, 2), R(x₃, y₃) ≡ R(4,3)
∴ A(ΔPQR) = \(\frac{3}{2}\) sq. units
Area if triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
∴ \(\pm \frac{3}{2}=\frac{1}{2}\left|\begin{array}{lll}
k & 0 & 1 \\
2 & 2 & 1 \\
4 & 3 & 1
\end{array}\right|\)
∴ ± \( [k(2 – 3) – 0 + 1 (6 – 8)]
∴ ± [latex\frac{3}{2}=\frac{1}{2}\) (-k -2)
∴ ± 3 = -k – 2
∴ 3 = -k – 2 or -3 = -k – 2
∴ k = -5 or k = 1

Question 8.
Examine the collinearity of the following set of points:
i. A (3, – 1), B (0, – 3), C (12, 5)
ii. P (3, – 5), Q (6,1), R (4, 2)
iii. L(0,1/2), M(2,-1), N(-4, 7/2)
Solution:
i. Here, A(x₁, y₁) ≡ A(3, -1), B(x₂, y₂) ≡ B(0, -3), C(x₃, y₃) ≡ C(12, 5)
If A(∆ABC) = 0, then the points A, B, C are collinear.
∴ A(∆ABC) = \(\frac{1}{2}\left|\begin{array}{ccc}
3 & -1 & 1 \\
0 & -3 & 1 \\
12 & 5 & 1
\end{array}\right|\)
= \(\frac{1}{2}\)[3(-3 – 5) – (-1) (0 – 12) + 1(0 + 36)]
= \(\frac{1}{2}\)[3(-8)+ 1(-12)+ 1(36)]
= \(\frac{1}{2}\)(-24 – 12 + 36)
= 0
∴ The points A, B, C are collinear.

ii. Here, P(x₁, y₁) ≡ P(3, -5), Q(x₂, y₂) ≡ Q(6, 1), R(x₃, y₃) ≡ R(4,2)
∴ If A(∆PQR) = 0, then the points P,Q, R are collinear
∴ A(∆PQR) = \(\frac{1}{2}\left|\begin{array}{ccc}
3 & -5 & 1 \\
6 & 1 & 1 \\
4 & 2 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [3(1-2) – (-5)(6 – 4) + 1(12 – 4)]
= \(\frac{1}{2}\) [3(-1) + 5(2) + 1(8)]
= \(\frac{1}{2}\)(-3 + 10 + 8)= \(\frac{15}{2}\) ≠ 0
∴ The points P, Q, R are non-collinear.

iii. Here, L(x₁, y₁) ≡ L(0,1/2), M(x₂, y₂) ≡ M(2, -1), N(x₃, y₃) ≡ N(-4, 7/2)
If A(∆LMN) = 0, then the points L, M, N are collinear.
∴ A(∆LMN) = \(\frac{1}{2}\left|\begin{array}{ccc}
0 & \frac{1}{2} & 1 \\
2 & -1 & 1 \\
-4 & \frac{7}{2} & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [0 – ]\(\frac{1}{2}\) (2 + 4) + 1(7 – 4)]
= \(\frac{1}{2}\)[ –\(\frac{1}{2}\) (6) + 1(3)]
= \(\frac{1}{2}\) (-3 + 3) = 0
∴ The points L, M, N are collinear.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.2

Question 1.
Without expanding, evaluate the following determinants.
i. \(\left|\begin{array}{ccc}
1 & a & b+c \\
1 & b & c+a \\
1 & c & a+b
\end{array}\right|\)
ii. \(\left|\begin{array}{ccc}
2 & 3 & 4 \\
5 & 6 & 8 \\
6 x & 9 x & 12 x
\end{array}\right|\)
iii. \(\left|\begin{array}{lll}
2 & 7 & 65 \\
3 & 8 & 75 \\
5 & 9 & 86
\end{array}\right|\)
Solution:
i. Let D = \(\left|\begin{array}{ccc}
1 & a & b+c \\
1 & b & c+a \\
1 & c & a+b
\end{array}\right|\)
Applying C₃ → C₃ + C₂, we get .
D = \(\left|\begin{array}{lll}
1 & a & a+b+c \\
1 & b & a+b+c \\
1 & c & a+b+c
\end{array}\right|\)
Taking (a + b + c) common from C₃, we get
D = (a + b + c) \(\left|\begin{array}{lll}
1 & a & 1 \\
1 & b & 1 \\
1 & c & 1
\end{array}\right|\)
= (a + b + c)(0)
… [∵ C₁ and C₃ are identical]
= 0

ii. \(\left|\begin{array}{ccc}
2 & 3 & 4 \\
5 & 6 & 8 \\
6 x & 9 x & 12 x
\end{array}\right|\)
Taking (3x) common from R₃, we get
D = 3x \(\left|\begin{array}{lll}
2 & 3 & 4 \\
5 & 6 & 8 \\
2 & 3 & 4
\end{array}\right|\)
= (3x)(0) = 0
… [∵ R₁ and R₃ are identical]
= 0

iii. Let D = \(\left|\begin{array}{lll}
2 & 7 & 65 \\
3 & 8 & 75 \\
5 & 9 & 86
\end{array}\right|\)
Applying Cx₃ → C₃ – 9C₂, we get
D = \(\left|\begin{array}{lll}
2 & 7 & 2 \\
3 & 8 & 3 \\
5 & 9 & 5
\end{array}\right|\)
= 0 …[∵ C₁and C₃ are identical]

Question 2.
Prove that \(\left|\begin{array}{lll}
{x}+y & y+\mathbf{z} & \mathbf{z}+{x} \\
\mathbf{z}+{x} & {x}+y & y+\mathbf{z} \\
{y}+\mathbf{z} & \mathbf{z}+{x} & {x}+{y}
\end{array}\right|=2\left|\begin{array}{ccc}
{x} & y & \mathbf{z} \\
\mathbf{z} & {x} & y \\
y & \mathbf{z} & {x}
\end{array}\right|\)
Solution:

Question 3.
Using properties of determinant, show that
i. \(\left|\begin{array}{ccc}
a+b & a & b \\
a & a+c & c \\
b & c & b+c
\end{array}\right|=4 a b c\)
ii. \(\left|\begin{array}{ccc}
1 & \log _{x} y & \log _{x} z \\
\log _{y} x & 1 & \log _{y} z \\
\log _{2} x & \log _{x} y & 1
\end{array}\right|=0\)
Solution:
i. L.H.S. = \(\)
Applying C₁ → C₁ – (C₂ + C₃), we get
L.H.S. = \(\left|\begin{array}{ccc}
0 & a & b \\
-2 c & a+c & c \\
-2 c & c & b+c
\end{array}\right|\)
Taking (-2) common from C₁, we get
L.H.S. = -2\(\left|\begin{array}{ccc}
0 & a & b \\
c & a+c & c \\
c & c & b+c
\end{array}\right|\)
Applying C₂ → C₂ – C₁ and C₃ → C₃ – C₁, we get
L.H.S. = -2\(\left|\begin{array}{lll}
0 & a & b \\
c & a & 0 \\
c & 0 & b
\end{array}\right|\)
= -2[0(ab – 0) – a(bc – 0) + b(0 – ac)]
= -2(0 – abc – abc)
= -2(-2abc)
= 4abc = R.H.S.

ii.

Question 4.
Solve the following equations.
i. \(\left|\begin{array}{lll}
x+2 & x+6 & x-1 \\
x+6 & x-1 & x+2 \\
x-1 & x+2 & x+6
\end{array}\right|=0\)
ii. \(
Solution:
i. [latex\left|\begin{array}{lll}
x+2 & x+6 & x-1 \\
x+6 & x-1 & x+2 \\
x-1 & x+2 & x+6
\end{array}\right|=0\)
Applying R₂ → R₂ – R₁ and R₃ → R₃ – R₁, we get
\(\left|\begin{array}{ccc}
x+2 & x+6 & x-1 \\
4 & -7 & 3 \\
-3 & -4 & 7
\end{array}\right|\) =0
∴ (x + 2)(- 49 + 12) – (x + 6)(28 + 9) + (x- 1)(- 16 – 21) = 0
∴ (x + 2) (-37) – (x + 6) (37) + (x – 1) (-37) = 0
∴ -37(x + 2+ x + 6 + x – 1) = 0
∴ 3x + 7 = 0
∴ x = \(\frac{-7}{3}\)

ii. \(\left|\begin{array}{ccc}
x-1 & x & x-2 \\
0 & x-2 & x-3 \\
0 & 0 & x-3
\end{array}\right|=0\)
Applying R₂ → R₂ – R₃, we get
\(\left|\begin{array}{ccc}
x-1 & x & x-2 \\
0 & x-2 & 0 \\
0 & 0 & x-3
\end{array}\right|=0\)
∴ (x – 1)(x – 2)(x – 3) – 0] – x(0 – 0) + (x – 2)(0 – 0) =
∴ (x – 1)(x – 2)(x – 3) = 0
∴ x — 1 = 0 or x-2 = 0 or x-3 = 0
∴ x = 1 or x = 2 or x = 3

Question 5.
If \(\left|\begin{array}{lll}
4+x & 4-x & 4-x \\
4-x & 4+x & 4-x \\
4-x & 4-x & 4+x
\end{array}\right|\) = 0, then find the values of x.
Solution:

(12 -x)[1(4x² – 0) – (4 – x)(0 – 0) + (4 – x)(0 – 0)] = 0
∴ (12 – x)(4x²) = 0
∴ x²(12 – x) = 0
∴ x = 0 or 12 – x = 0
∴ x = 0 or x = 12

Question 6.
Without expanding determinant, show that
\(\left|\begin{array}{lll}
1 & 3 & 6 \\
6 & 1 & 4 \\
3 & 7 & 12
\end{array}\right|+4\left|\begin{array}{lll}
2 & 3 & 3 \\
2 & 1 & 2 \\
1 & 7 & 6
\end{array}\right|=10\left|\begin{array}{lll}
1 & 2 & 1 \\
3 & 1 & 7 \\
3 & 2 & 6
\end{array}\right|\)
Solution:
L.H.S. = \(\left|\begin{array}{ccc}
1 & 3 & 6 \\
6 & 1 & 4 \\
3 & 7 & 12
\end{array}\right|+4\left|\begin{array}{lll}
2 & 3 & 3 \\
2 & 1 & 2 \\
1 & 7 & 6
\end{array}\right|\)
In 1st determinant, taking 2 common from C₃ we get

Interchanging rows and columns, we get
L.H.S. = \(\left|\begin{array}{ccc}
10 & 20 & 10 \\
3 & 1 & 7 \\
3 & 2 & 6
\end{array}\right|\)
Taking 10 common from R₁, we get
L.H.S = 10\(\left|\begin{array}{lll}
1 & 2 & 1 \\
3 & 1 & 7 \\
3 & 2 & 6
\end{array}\right|\) = R.H.S.