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

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.