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

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.