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

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

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:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.2 1

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.
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.2 2

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

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:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.2 3
(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

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

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