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.]