## MP Board Class 12th Maths Important Questions Chapter 3 Matrices

### Matrices Important Questions

Matrices Objective Type Questions:

Question 1.
If A = $$\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}$$ and A + A’ = I, then the value of α is:
(a) $$\frac { \pi }{ 6 }$$
(b) $$\frac { \pi }{ 3 }$$
(c) π
(d) $$\frac { 3\pi }{ 2 }$$
(b) $$\frac { \pi }{ 3 }$$

Question 2.
If A = $$\left[\begin{array}{lll} {2} & {0} & {0} \\ {0} & {2} & {0} \\ {0} & {0} & {2} \end{array}\right]$$, then A5 is equal to:
(a) 5 A
(b) 10 A
(c) 16 A
(d) 32 A
(c) 16 A

Question 3.
If a matrix is both symmetric and skew – symmetric, then:
(a) A is a diagonal matrix
(b) A is zero matrix
(c) A is a square matrix
(d) None of these
(b) A is zero matrix

Question 4.
If A = $$\begin{bmatrix} \alpha & \beta \\ \lambda & -\alpha \end{bmatrix}$$ is such that A2 = I, then:
(a) 1 + α2 + βλ = 0
(b) 1 – α2 + βλ = 0
(c) 1 – α2 – βλ = 0
(d) 1 + α2 – βλ = 0
(c) 1 – α2 – βλ = 0

Question 5.
If A = $$\begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix}$$, then An = ………………………:
(a) A = $$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$, if n is even natural number
(b) A = $$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$, if n is odd natural number
(c) A = $$\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$$, if n ∈ N
(d) None of these
(a) A = $$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$, if n is even natural number

Question 2.
Fill in the blanks:

1. If A = $$\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$$ and B = $$\begin{bmatrix} 1 & 3 \\ 2 & 5 \end{bmatrix}$$, then AB = …………………………….
2. If A = diag [1, -1, 2] and B = diag [2, 3, -1], then value of 3A + 4B will be ……………………
3. A matrix A is said to be idempotent if ………………………
4. A matrix A is said to be orthogonal if ……………………..
5. If [x, 1] $$\begin{bmatrix} 1 & 0 \\ -2 & 0 \end{bmatrix}$$ = 0, then find the value of x will be …………………………….

1. $$\begin{bmatrix} 10 & 26 \\ 7 & 19 \end{bmatrix}$$
2. diag [11, 9, 2]
3. A2 = A
4. AA’ = A’A = I
5. x = 2

Question 3.
Write True/False:

1. Multiplication of matrix is always commutative?
2. Two matrix are said to be comparable if they have same number of rows and columns?
3. If A is a square matrix, then A. adj A = |A| I?
4. A square matrix A is said to be symmetric if A = – AT?
5. Matrix A and B are inverse of each other if AB = BA?

1. False
2. True
3. True
4. False
5. False

Question 4.
Match the Column:

1. (d)
2. (e)
3. (a)
4. (b)
5. (c)

Question 5.
Write the answer in one word/sentence:

1. If A and B are two square matrix of same order, then what is the value of Adj (AB)?
2. A square matrix A is said to be Involountary matrix if?
3. If A = $$\begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}$$, then find the value of A2?
4. If A = [1, 2, 3], then find the value of AAT?
5. If X + Y = $$\begin{bmatrix} 1 & -2 \\ 3 & 4 \end{bmatrix}$$ and X – Y = $$\begin{bmatrix} 3 & 2 \\ -1 & 0 \end{bmatrix}$$, then find the value of X?

2. A2 = I
3. -I
4. [1, 4]
5. $$\begin{bmatrix} 2 & 0 \\ 1 & 2 \end{bmatrix}$$

Question 1.
If A = $$\begin{bmatrix} a^{ 2 }+b^{ 2 } & b^{ 2 }+c^{ 2 } \\ a^{ 2 }+c^{ 2 } & a^{ 2 }+b^{ 2 } \end{bmatrix}$$ and B = $$\begin{bmatrix} 2ab & 2bc \\ -2ac & -2ab \end{bmatrix}$$, then find the value of A + B? (NCERT)
Solution:
A + B =

Question 2.
If A = $$\begin{bmatrix} cos^{ 2 }x & sin^{ 2 }x \\ sin^{ 2 }x & cos^{ 2 }x \end{bmatrix}$$ and B = $$\begin{bmatrix} sin^{ 2 }x & cos^{ 2 }x \\ cos^{ 2 }x & sin^{ 2 }x \end{bmatrix}$$, then find A + B? (NCERT)
Solution:
A + B = $$\begin{bmatrix} cos^{ 2 }x & sin^{ 2 }x \\ sin^{ 2 }x & cos^{ 2 }x \end{bmatrix}$$ + $$\begin{bmatrix} sin^{ 2 }x & cos^{ 2 }x \\ cos^{ 2 }x & sin^{ 2 }x \end{bmatrix}$$

⇒ A + B = $$\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$$

Question 3.
If A = and B = then find 3A – 5B? (NCERT)
Solution:

Question 4.
Simplify: cos θ $$\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}$$ + sinθ $$\begin{bmatrix} sin\theta & -cos\theta \\ cos\theta & sin\theta \end{bmatrix}$$
Solution:
cos θ $$\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}$$ + sinθ $$\begin{bmatrix} sin\theta & -cos\theta \\ cos\theta & sin\theta \end{bmatrix}$$

Question 5.
From the following equation find the value of x and y?
2 $$\begin{bmatrix} x & 5 \\ 7 & y-3 \end{bmatrix}$$ + $$\begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix}$$ = $$\begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix}$$? (NCERT)
Solution:

From defnition of matrix,
2x + 3 = 7
⇒ 2x = 4 ⇒ x = 2
⇒ 2y – 4 = 14
⇒ 2y = 18 ⇒ y = 9
∴x = 2, y = 9.

Question 6.
Find the value of X and Y if X + Y = $$\begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix}$$ and X – Y = $$\begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}$$? (NCERT)
Solution:
Given X + Y = $$\begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix}$$ ……………….. (1)
and X – Y = $$\begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}$$ ………………….. (2)
2X = $$\begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix}$$ + $$\begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}$$
⇒ 2X = $$\begin{bmatrix} 5+3 & 2+6 \\ 0+0 & 9-1 \end{bmatrix}$$
⇒ 2X = $$\begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix}$$
⇒ X = $$\frac{1}{2}$$ $$\begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix}$$ = $$\begin{bmatrix} 4 & 4 \\ 0 & 4 \end{bmatrix}$$
Substracting eqn. (2) from eqn. (1),
2Y = $$\begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix}$$ – $$\begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}$$
⇒ 2Y = $$\begin{bmatrix} 5-3 & 2-6 \\ 0-0 & 9+1 \end{bmatrix}$$
⇒ Y = $$\frac{1}{2}$$ $$\begin{bmatrix} 2 & -4 \\ 0 & 10 \end{bmatrix}$$ = $$\begin{bmatrix} 1 & -2 \\ 0 & 5 \end{bmatrix}$$

Question 7.
Find the value of x and y
2 $$\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix}$$ + $$\begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix}$$ = $$\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$$ (NCERT)
Solution:
Given:
2 $$\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix}$$ + $$\begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix}$$ = $$\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$$
⇒ $$\begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix}$$ + $$\begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix}$$ = $$\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$$
⇒ $$\begin{bmatrix} 2+y & 6+0 \\ 0+1 & 2x+2 \end{bmatrix}$$ = $$\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$$
⇒ $$\begin{bmatrix} 2+y & 6 \\ 1 & 2x+2 \end{bmatrix}$$ = $$\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$$
By defnition of matrix,
2 + y = 5 ⇒ y = 3
2x + 2 = 8
⇒ x + 1 = 4
⇒ x = 3
∴ x = 3, y = 3.

Question 8.
If $$\left[\begin{array}{c} {x+y+z} \\ {x+z} \\ {y+z} \end{array}\right]$$ = [ $$\begin{matrix} 9 \\ 5 \\ 7 \end{matrix}$$ ] find the value of x, y, and z?
Solution: Given $$\left[\begin{array}{c} {x+y+z} \\ {x+z} \\ {y+z} \end{array}\right]$$ = [ $$\begin{matrix} 9 \\ 5 \\ 7 \end{matrix}$$ ]
By defnition of matrix,
x + y + z = 9
x + z = 5
y + z = 7
From eqns. (1) and (2),
x + y + z = 9
⇒ 5 + y = 9 ⇒ y = 4
From eqns. (1) and (3),
x + (y + z) = 9
⇒ x + 7 = 9
⇒ x = 2
Putting the value of x in eqn. (2),
2 + z = 5
⇒ z = 3
∴ x = 2, y = 4, z = 3.

Question 9.
If $$\begin{bmatrix} x+y & 2 \\ 5+z & xy \end{bmatrix}$$ = $$\begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}$$, then fund the value of x, y and z? (NCERT)
Solution:
Given:
$$\begin{bmatrix} x+y & 2 \\ 5+z & xy \end{bmatrix}$$ = $$\begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}$$
By defnition of matrix,
x + y = 6 ……………….. (1)
xy = 8 ………………. (2)
5 + z = 5
⇒ z = 0
From eqn. (1), y = 6 – x
xy = 8
⇒ 6x – x2 = 8
⇒ x2 – 6x + 8 = 0
⇒ x2 – 4x – 2x + 8 = 0
⇒ x (x – 4) – 2 ( x – 4) = 0
⇒ ( x – 2) (x – 4) = 0
⇒ x = 2, 4
When x = 2 then y = 6 – 2 = 4
When x = 4 then y = 6 – 4 = 2
So x = 2, y = 4, z = 0
x = 4, y = 2, z = 0.

Question 10.
If A = $$\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$$ then prove that A2 – 4A + 5I = 0?
Solution:
Given:
A = $$\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$$
A2 = A.A = $$\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$$ $$\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$$
= $$\begin{bmatrix} 1.1+(-1)2 & 1(-1)+(-1).3 \\ 2.1+3.2 & 2(-1)+3.3 \end{bmatrix}$$
= $$\begin{bmatrix} 1-2 & -1-3 \\ 2+6 & -2+9 \end{bmatrix}$$
= $$\begin{bmatrix} -1 & -4 \\ 8 & 7 \end{bmatrix}$$
4A = 4$$\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$$ = $$\begin{bmatrix} 4 & -4 \\ 8 & 12 \end{bmatrix}$$
5I = 5$$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ = $$\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$$
= $$\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$
= 0. Proved.

Question 11.
If A = $$\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}$$ then prove that A2 – 6A + 17I = 0?
Solution:
A = $$\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}$$
A2 = A.A = $$\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}$$ $$\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}$$
= $$\begin{bmatrix} 4-9 & -6-12 \\ 6+12 & -9+16 \end{bmatrix}$$ = $$\begin{bmatrix} -5 & -18 \\ 18 & 7 \end{bmatrix}$$
∴ L.H.S = A2 – 6A + 17I
= $$\begin{bmatrix} -5 & -18 \\ 18 & 7 \end{bmatrix}$$ – 6 $$\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}$$ + 17 $$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
= $$\begin{bmatrix} -5 & -18 \\ 18 & 7 \end{bmatrix}$$ – $$\begin{bmatrix} 12 & -18 \\ 18 & 24 \end{bmatrix}$$ + $$\begin{bmatrix} 17 & 0 \\ 0 & 17 \end{bmatrix}$$
= $$\begin{bmatrix} -5-12+17 & -18+18+0 \\ 18-18+0 & 7-24+17 \end{bmatrix}$$
= $$\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$ = 0 = R.H.S. proved.

Question 12.
If A = $$\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$$ then prove that A2 – 5A + 7I = 0. (NCERT)
Solution:
A2 = A.A = $$\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$$ × $$\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$$
⇒ A2 = $$\begin{bmatrix} 3\times 3-1\times 1 & 3\times 1+1\times 2 \\ -1\times 3+2\times -1 & -1\times 1+2\times 2 \end{bmatrix}$$
⇒ A2 = $$\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$$
5A = 5$$\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$$
⇒ 5A = $$\begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}$$
7I = 7$$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
⇒ 7I = $$\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$$
∴ A2 – 5A + 7I = $$\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$$ – $$\begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}$$ + $$\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$$
⇒ A2 – 5A + 7I = $$\begin{bmatrix} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{bmatrix}$$
⇒ A2 – 5A + 7I = $$\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$
⇒ A2 – 5A + 7I = 0.

Question 13.
If A = $$\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$$ and I = $$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ then find the value of k if A2 = kA – 2I? (NCERT)
Solution:
Given:
A2 = kA – 2I
⇒ kA = A2 + 2I
⇒ k $$\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$$ = $$\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$$ × $$\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$$ + 2 $$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
⇒ $$\begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix}$$ = $$\begin{bmatrix} 3\times 3-2\times 4 & 3\times -2+2\times 2 \\ 4\times 3-2\times 4 & 4\times -2+2\times 2 \end{bmatrix}$$ + $$\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$$
= $$\begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix}$$ = $$\begin{bmatrix} 1 & -2 \\ 4k & -4 \end{bmatrix}$$ + 2$$\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$$
⇒ $$\begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix}$$ = $$\begin{bmatrix} 1+2 & 2+0 \\ 4+0 & -4+2 \end{bmatrix}$$
⇒ $$\begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix}$$ = $$\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$$
⇒ 2k = 2
⇒ k = 1.

Question 14.
If f(x) = x2 – 2x – 3, then find the value of f(A) if A = $$\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$$
Solution:
f(x) = x2 – 2x – 3
∴ f(A) = A2 – 2A – 3I
⇒ f(A) = $$\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$$ × $$\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$$ – 2 $$\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$$ – 3 $$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

⇒ f(A) = $$\begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix}$$ – $$\begin{bmatrix} 2 & 4 \\ 4 & 2 \end{bmatrix}$$ – $$\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$$
⇒ = $$\begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix}$$ – $$\begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix}$$ = $$\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$
⇒ = 0.

Question 15.
If matrix A = $$\left[\begin{array}{ccc} {0} & {a} & {-3} \\ {2} & {0} & {-1} \\ {b} & {1} & {0} \end{array}\right]$$ skew – symmetric matrix, then find the value of ‘a’ and ‘b’? (CBSE 2018)
Solution:
If A is skew symmmetric matrix then A’ = -A
Given:
A = $$\left[\begin{array}{ccc} {0} & {a} & {-3} \\ {2} & {0} & {-1} \\ {b} & {1} & {0} \end{array}\right]$$
A’ = $$\left[\begin{array}{ccc} {0} & {2} & {b} \\ {a} & {0} & {1} \\ {-3} & {-1} & {0} \end{array}\right]$$
– A = $$\left[\begin{array}{ccc} {0} & {-a} & {3} \\ {-2} & {0} & {1} \\ {-b} & {-1} & {0} \end{array}\right]$$
∵ A’ = – A
$$\left[\begin{array}{ccc} {0} & {2} & {b} \\ {a} & {0} & {1} \\ {-3} & {-1} & {0} \end{array}\right]$$ = $$\left[\begin{array}{ccc} {0} & {-a} & {3} \\ {-2} & {0} & {1} \\ {-b} & {-1} & {0} \end{array}\right]$$
∴ 2 = -a or a = -2
-3 = -b or b = 3.

Question 16.
If A = $$\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}$$, then prove that AA-1 = I?
Solution:
Given:
A = $$\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}$$
A-1 = $$\frac { adjA }{ |A| }$$
|A| = |$$\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}$$|
= cos2α – (-sin2 α)
= cos2α + sin2α = 1 ………………………… (1)
∴|A| = 1
We know adj A = $$\begin{bmatrix} C_{ 11 } & C_{ 21 } \\ C_{ 12 } & C_{ 22 } \end{bmatrix}$$
Where C11 = (-1)2 cos α = cos α
C12 = (-1)3 sin α = -sin α
C21 = (-1)3 (-sin α ) = sin α
C22 = (-1)4cos α = cos α
∴ adj A = $$\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}$$ …………….. (2)
∴ A-1 = $$\frac { adjA }{ |A| }$$ = $$\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}$$
∴ A.A-1 = $$\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}$$ $$\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}$$

= $$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ = I. Proved.

Question 17.
If A = $$\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}$$ then prove that:
A. (Adj A) = |A| I?
Solution:
Given A = $$\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}$$
A11 = cos α, A12 = sin α, A21 = – sin α, A22 = cos α
Adj A = $$\begin{bmatrix} A_{ 11 } & A_{ 21 } \\ A_{ 12 } & A_{ 22 } \end{bmatrix}$$ = $$\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}$$
A.(Adj A) = $$\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}$$ $$\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}$$

∴ A.(Adj A) = I = |A| I. Proved.

Question 18.
Prove that the square matrix A = $$\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}$$ is orthogonal matrix?
Solution:
Given:
A = $$\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}$$
∴ A’ = $$\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}$$
Now A.A’ = $$\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}$$ $$\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}$$

= $$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ = I
Similarly A’.A = I
Then A.A’ = A’A = I
So A is orthogonal matrix.
Proved.

Question 19.
If A = $$\begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix}$$ and B = $$\begin{bmatrix} 6 & 7 \\ 8 & 9 \end{bmatrix}$$, then find the value of (AB)-1?
Solution:
Given:
A = $$\begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix}$$, B = $$\begin{bmatrix} 6 & 7 \\ 8 & 9 \end{bmatrix}$$
= $$\begin{bmatrix} 3.6+2.8 & 3.7+2.9 \\ 7.6+5.8 & 7.7+5.9 \end{bmatrix}$$ = $$\begin{bmatrix} 34 & 39 \\ 82 & 94 \end{bmatrix}$$
|AB| = $$\begin{bmatrix} 34 & 39 \\ 82 & 94 \end{bmatrix}$$
= $$\begin{bmatrix} 34 & 39 \\ 82 & 94 \end{bmatrix}$$ = 3196 – 3198 = -2 ≠ 0
AB11 = 94, AB12 = -82, AB21 = -39, AB22 = 34
adj AB = $$\begin{bmatrix} 94 & -39 \\ -82 & 34 \end{bmatrix}$$
(AB)-1 = $$\frac { adjAB }{ |AB| }$$ = $$\frac{1}{ -2}$$ $$\begin{bmatrix} 94 & -39 \\ -82 & 34 \end{bmatrix}$$
= $$\begin{bmatrix} -47 & \frac { 39 }{ 2 } \\ 41 & -17 \end{bmatrix}$$

Question 20.
If A = $$\begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix}$$ then prove that:
2A-1 = 9I – A? (CBSE 2018)
Solution:
Given:
A = $$\begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix}$$
∴ A11 = 7, A12 = – (- 4), A21 = – (- 3), A22 = 2
∴ adj A = $$\begin{bmatrix} A_{ 11 } & A_{ 21 } \\ A_{ 12 } & A_{ 22 } \end{bmatrix}$$ = $$\begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix}$$
Now |A| = $$\begin{vmatrix} 2 & -3 \\ -4 & 7 \end{vmatrix}$$ = 14 – 12 = 2
∴ A-1 = $$\frac { adjA }{ |A| }$$ = $$\frac{1}{2}$$ $$\begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix}$$
⇒ 2A-1 = $$\begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix}$$
Again, 9I – A = 9 $$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ – $$\begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix}$$
= $$\begin{bmatrix} 9 & 0 \\ 0 & 9 \end{bmatrix}$$ – $$\begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix}$$ = $$\begin{bmatrix} 9-2 & 0+3 \\ 0+4 & 9-7 \end{bmatrix}$$
∴ 9I – A = $$\begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix}$$
From eqns. (1) and (2),
2A-1 = 9I – A. proved.

Question 21.
For matrix A and B prove that (AB)’ = B’A’ where A = [ $$\begin{matrix} 1 \\ -4 \\ 3 \end{matrix}$$ ], B = [-1 2 1]? (NCERT)
Solution:
AB = [ $$\begin{matrix} 1 \\ -4 \\ 3 \end{matrix}$$ ]3×1 [-1 2 1]1×3
AB = $$\left[\begin{array}{rrr} {-1 \times 1} & {1 \times 2} & {1 \times 1} \\ {-4 \times-1} & {-4 \times 2} & {-4 \times 1} \\ {-3 \times 1} & {3 \times 2} & {3 \times 1} \end{array}\right]$$
⇒ AB = [ $$\begin{matrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{matrix}$$ ]
(AB)’ = [ $$\begin{matrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{matrix}$$ ] …………………… (1)
B = [ -1 2 1]’
B’ = [ $$\begin{matrix} -1 \\ 2 \\ 1 \end{matrix}$$ ]
A’ = [ $$\begin{matrix} -1 \\ -4 \\ 3 \end{matrix}$$ ]
A’ = [1 -4 3]1 ×3
B’A’ = [ $$\begin{matrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{matrix}$$ ]
From eqns. (1) and (2),
(AB)’ = B’A’. proved.

Matrices Long Answer Type Questions – II

Question 1.
If A = $$\begin{bmatrix} 1 & 4 \\ 3 & 5 \end{bmatrix}$$ then prove that:
Solution:
Given:
A = $$\begin{bmatrix} 1 & 4 \\ 3 & 5 \end{bmatrix}$$
Then, |A| = $$\begin{bmatrix} 1 & 4 \\ 3 & 5 \end{bmatrix}$$ = 5 – 12 = -7
A11 = 5, A12 = -3, A21 = -4, A22 = 1
∴ adj A = $$\begin{bmatrix} A_{ 11 } & A_{ 21 } \\ A_{ 12 } & A_{ 22 } \end{bmatrix}$$ = $$\begin{bmatrix} 5 & -4 \\ -3 & 1 \end{bmatrix}$$
⇒ A. adj A = $$\begin{bmatrix} 1 & 4 \\ 3 & 5 \end{bmatrix}$$ × $$\begin{bmatrix} 5 & -4 \\ -3 & 1 \end{bmatrix}$$

⇒ A.adj A = $$\begin{bmatrix} -7 & 0 \\ 0 & -7 \end{bmatrix}$$ = -7$$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
and (adj A) . A = $$\begin{bmatrix} 5 & -4 \\ -3 & 1 \end{bmatrix}$$ × $$\begin{bmatrix} 1 & 4 \\ 3 & 5 \end{bmatrix}$$

⇒ (adj A).A = $$\begin{bmatrix} -7 & 0 \\ 0 & -7 \end{bmatrix}$$
⇒ (adj A).A = – 7 $$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
⇒ (adj A).A = |A| I
From eqns. (1) and (2),

Question 2.
If A = $$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$ then prove that:
Solution:
Solve like Q.No. 1.

Question 3.
If matrix A = [ $$\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}$$ ] then prove that: A-1 = A?
Solution:
Given:
A = [ $$\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}$$ ]
By formula A-1 = $$\frac { adjA }{ |A| }$$
Then, |A| = [ $$\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}$$ ] = 1.(0-1) = -1
A11 = (-1)2 (0 – 0) = 0
A12 = (-1)3 (0 – 0) = 0
A13 = (-1)4 ( 0 – 1) = -1, A21 = (-1)3 (0 – 0) = 0
A22 = (-1)4 ( 0 – 1) = -1, A23 = (-1)5 (0 – 0) = 0
A31 = (-1)4 ( 0 – 1) = -1, A32 = (-1)5 (0 – 0) = 0
A33 = (-1)6 ( 0 – 0) = 0
adj A = $$\left[\begin{array}{ccc} {A_{11}} & {A_{21}} & {A_{31}} \\ {A_{12}} & {A_{22}} & {A_{32}} \\ {A_{13}} & {A_{23}} & {A_{33}} \end{array}\right]$$ = [ $$\begin{matrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{matrix}$$ ]
= [- $$\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}$$ ]
∴ image 14 = [ $$\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}$$ ]
∴ A-1 = A.

Question 4.
Matrix A = [ $$\begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix}$$ ], find inverse of matrix A?
Solution:
Given A = [ $$\begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix}$$ ]
∴ |A| = 2 (4 × 2 – 7 × 1) – 3 (3 × 2 – 3 × 1) + 1 (3 × 7 – 3 × 4)
= 2 ( 8 – 7) – 3 ( 6 – 3) + 1 (21 – 12)
= 2(1) – 3(3) + 1(9)
= 2 – 9 + 9 = 2
If |A| ≠ 0 then A-1 will exist.
Now, A11 = + (8 – 7) = 1, A12 = – (6 – 3) = -3
A13 = + (21 – 12), A21 = – (6 – 7) = 1
A22 = +(4 – 3) = 1, A23 = – (14 – 9) = -5
A31 = + (3 – 4) = -1, A32 = – (2 – 3) = 1
A33 = + (8 – 9) = -1
∴ adj A = $$\left[\begin{array}{ccc} {A_{11}} & {A_{21}} & {A_{31}} \\ {A_{12}} & {A_{22}} & {A_{32}} \\ {A_{13}} & {A_{23}} & {A_{33}} \end{array}\right]$$ = [ $$\begin{matrix} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1 \end{matrix}$$ ]
∵ A-1 = $$\frac { adjA }{ |A| }$$
∴A-1 = $$\frac{1}{2}$$ [ $$\begin{matrix} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1 \end{matrix}$$ ]
⇒ A-1 = $$\left[\begin{array}{ccc} {\frac{1}{2}} & {\frac{1}{2}} & {\frac{-1}{2}} \\ {\frac{-3}{2}} & {\frac{1}{2}} & {\frac{1}{2}} \\ {\frac{9}{2}} & {\frac{-5}{2}} & {\frac{-1}{2}} \end{array}\right]$$

Question 5.
If A = [ $$\begin{matrix} 1 & 2 & 3 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}$$ ], then find the value of A-1?
Solution:
Solve like Q.No.4.
[ $$\begin{matrix} 1 & -3 & 2 \\ -3 & 3 & -1 \\ 2 & -1 & 0 \end{matrix}$$ ]

Question 6.
If A = [ $$\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}$$ ], then find the value of A-1?
Solution:
Given:
A = [ $$\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}$$ ]
|A| = [ $$\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}$$ ]
⇒ |A| = 1 (1 – 4) + 2 ( 4 – 2) + 2 ( 4 – 2)
= – 3 + 4 + 4 = 5
If |A| ≠ 0, then A-1 exists
A11 = $$\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix}$$ = 1 – 4 = -3
A12 = – $$\begin{vmatrix} 2 & 2 \\ 2 & 1 \end{vmatrix}$$ = – ( 2 – 4) = 2
A13 = $$\begin{vmatrix} 2 & 1 \\ 2 & 2 \end{vmatrix}$$ = 4 – 2 = 2
A21 = – $$\begin{vmatrix} 2 & 2 \\ 2 & 1 \end{vmatrix}$$ = – ( 2 – 4) = 2
A22 = $$\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix}$$ = 1 – 4 = – 3
A23 = – $$\begin{vmatrix} 1 & 2 \\ 2 & 2 \end{vmatrix}$$ = – ( 2 – 4) = 2
A31 = $$\begin{vmatrix} 2 & 2 \\ 1 & 2 \end{vmatrix}$$ = 4 – 2 = 2
A32 = – $$\begin{vmatrix} 1 & 2 \\ 2 & 2 \end{vmatrix}$$ = – (2 – 4) = 2
A33 = $$\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix}$$ = 1 – 4 = -3
∴adj A = $$\left[\begin{array}{ccc} {A_{11}} & {A_{21}} & {A_{31}} \\ {A_{12}} & {A_{22}} & {A_{32}} \\ {A_{13}} & {A_{23}} & {A_{33}} \end{array}\right]$$
= [ $$\begin{matrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{matrix}$$ ]
∴A-1 = $$\frac { adjA }{ |A| }$$ = $$\frac{1}{5}$$ [ $$\begin{matrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{matrix}$$ ]

Question 7.
If A = $$\begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix}$$ then prove that: A2 – 2A + 3I = 0?
Solution:
Given:
A = $$\begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix}$$
A2 = A.A = $$\begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix}$$ $$\begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix}$$
= $$\begin{bmatrix} 2.2+3(-1) & 2.3+3.0 \\ -1.2+0(-1) & -1.3+0.0 \end{bmatrix}$$
= $$\begin{bmatrix} 1 & 6 \\ -2 & -3 \end{bmatrix}$$
2A = $$\begin{bmatrix} 4 & 6 \\ -2 & 0 \end{bmatrix}$$
3I = $$\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$$
∴ A2 – 2A + 3I = $$\begin{bmatrix} 1 & 6 \\ -2 & -3 \end{bmatrix}$$ – $$\begin{bmatrix} 4 & 6 \\ -2 & 0 \end{bmatrix}$$ + $$\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$$
= $$\begin{bmatrix} 4 & 6 \\ -2 & 0 \end{bmatrix}$$ – $$\begin{bmatrix} 4 & 6 \\ -2 & 0 \end{bmatrix}$$ = $$\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$
= 0. Proved.

Question 8.
If A = $$\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}$$ then prove A2 – 6A + 17I = 0 and find A-1?
Solution:
Given:
A = $$\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}$$
A2 = A.A = $$\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}$$ $$\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}$$
= $$\begin{bmatrix} 4-9 & -6-12 \\ 6+12 & -9+16 \end{bmatrix}$$ = $$\begin{bmatrix} -5 & -18 \\ 18 & 7 \end{bmatrix}$$
∴L.H.S = A2 – 6A + 17I
= $$\begin{bmatrix} -5 & -18 \\ 18 & 7 \end{bmatrix}$$ – 6 $$\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}$$ + 17$$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
= $$\begin{bmatrix} -5 & -18 \\ 18 & 7 \end{bmatrix}$$ – $$\begin{bmatrix} 12 & -18 \\ 18 & 24 \end{bmatrix}$$ + $$\begin{bmatrix} 17 & 0 \\ 0 & 17 \end{bmatrix}$$
= $$\begin{bmatrix} -5-12+17 & -18+18+0 \\ 18-18+0 & 7-24+17 \end{bmatrix}$$ = $$\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$
= 0 = R.H.S. Proved.
Then |A| = $$\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}$$ = 8 + 9 = 17
∴ A11 = (-1)1+1 4 = 4
A12 = (-1)1+2 (3) = -3
A21 = (-1)2+1 (-3) = 3
A22 = (-1)2+2 (2) = 2.
and adj A = $$\begin{bmatrix} A_{ 11 } & A_{ 12 } \\ A_{ 12 } & A_{ 22 } \end{bmatrix}$$ = $$\begin{bmatrix} 4 & 3 \\ -3 & 2 \end{bmatrix}$$
∴A-1 = $$\frac { adjA }{ |A| }$$
= $$\frac{1}{17}$$ $$\begin{bmatrix} 4 & 3 \\ -3 & 2 \end{bmatrix}$$

Question 9.
If A = $$\begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix}$$, then prove A2 + 4A – 42 I = 0 and find A-1?
Solution:
Solve like Q.No. 8.
A-1 = $$\frac{1}{42}$$ $$\begin{bmatrix} -4 & 5 \\ 2 & 0 \end{bmatrix}$$.

Question 10.
(A) Solve the following equations by matrix method:
x + y + z = 3
2x – y + z = 2
x – 2y + 3z = 2.
Solution:
If A = [ $$\begin{matrix} 1 & 1 & 1 \\ 2 & -1 & 1 \\ 1 & -2 & 3 \end{matrix}$$ ], X = [ $$\begin{matrix} x \\ y \\ z \end{matrix}$$ ] and B = [ $$\begin{matrix} 3 \\ 2 \\ 2 \end{matrix}$$ ]
|A| = $$\left|\begin{array}{ccc} {1} & {1} & {1} \\ {2} & {-1} & {1} \\ {1} & {-2} & {3} \end{array}\right|$$
= 1( – 3 + 2) – 2 (3 + 2) + 1(1 + 1)
= -1 – 10 + 2 = -9
A11 = $$\begin{vmatrix} -1 & 1 \\ -2 & 3 \end{vmatrix}$$ = – 3 + 2 = – 1
A12 = – $$\begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix}$$ = – ( 6 – 1) = – 5
A13 = $$\begin{vmatrix} 2 & -1 \\ 1 & -2 \end{vmatrix}$$ = – 4 + 1 = – 3
A21 = – $$\begin{vmatrix} 1 & 1 \\ -2 & 3 \end{vmatrix}$$ = – (3 + 2) = – 5
A22 = $$\begin{vmatrix} 1 & 1 \\ 1 & 3 \end{vmatrix}$$ = 3 -1 = 2
A23 = – $$\begin{vmatrix} 1 & 1 \\ 1 & -2 \end{vmatrix}$$ = – ( -2 -1) = 3
A31 = $$\begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix}$$ = 1 + 1 = 2
A32 = – $$\begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}$$ = – (1 – 2) = 1
A33 = $$\begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix}$$ = -1 -2 = -3
and adj A = [ $$\begin{matrix} -1 & -5 & 2 \\ 5 & 2 & 1 \\ -3 & 3 & -3 \end{matrix}$$ ]
∴ A-1 = $$\frac { adjA }{ |A| }$$
⇒ A-1 = $$\frac{-1}{9}$$ [ $$\begin{matrix} -1 & -5 & 2 \\ 5 & 2 & 1 \\ -3 & 3 & -3 \end{matrix}$$ ]
⇒ A-1 = image 14
X = A-1B

∴ x = 1, y = 1, z = 1.

Question 10.
(B) Solve the following equations by matrix method:
x + y + z = 6
x + 2y = 3z = 14
x + 4y + 9z = 36
Solution:
x + y + z = 6
x + 2y + 3z = 14
x + 4y + 9z = 36.
Where
A = [ $$\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 9 \end{matrix}$$ ], B = [ $$\begin{matrix} 6 \\ 14 \\ 36 \end{matrix}$$ ] , X = [ $$\begin{matrix} x \\ y \\ z \end{matrix}$$ ]
= 6 – 6 + 2 = 2

X = A-1 B
⇒ [ $$\begin{matrix} x \\ y \\ z \end{matrix}$$ ] = [ $$\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}$$ ]
∴ x = 1, y = 2, z = 3.

Question 11.
If A’ = $$\left[\begin{array}{rr} {3} & {4} \\ {-1} & {2} \\ {0} & {1} \end{array}\right]$$ and B = $$\left[\begin{array}{rrr} {-1} & {2} & {1} \\ {1} & {2} & {3} \end{array}\right]$$ then prove the following:
(i) (A + B)’ = A’ + B’
(ii) (A – B)’ = A’ – B’. (NCERT)
Solution:
(i) Given

Question 12.
If A = $$$\left[\begin{array}{rrr} {-1} & {2} & {3} \\ {5} & {7} & {9} \\ {-2} & {1} & {1} \end{array}\right]$$$ and B = $$\left[\begin{array}{rrr} {-4} & {1} & {-5} \\ {1} & {2} & {0} \\ {1} & {3} & {1} \end{array}\right]$$ then prove that:
(i) (A + B)’ = A’ + B’
(ii) (A – B)’ = A’ – B’. (NCERT)
Solution:
solve like Q.No.11.

Question 13.
Express matrix A = $$\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$$ as sum of a symmetric and a skew symmetric matrix? (NCERT)
Solution:

Given:

Eqn. (1) and symmetric matrix and eqn. (2) is skew symmetric matrix.

Question 14.
(A) By using elementary operations, find the inverse of matrix A = $$\begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix}$$
Solution:
Using A = AI

Question 14.
(B) By using elementary operation, find the inverse of matrix A = $$\begin{bmatrix} 3 & 10 \\ 2 & 7 \end{bmatrix}$$?
Solution:
Solve like Q.No. 14 (A).
$$\begin{bmatrix} 7 & -10 \\ -2 & 3 \end{bmatrix}$$

Question 15.
Solve the following system of equations by using matrix method: (NCERT, CBSE 2011)
$$\frac{2}{x}$$ + $$\frac{3}{y}$$ + $$\frac{10}{z}$$ = 4
$$\frac{4}{x}$$ – $$\frac{6}{y}$$ + $$\frac{5}{z}$$ = 1
$$\frac{6}{x}$$ + $$\frac{9}{y}$$ – $$\frac{20}{z}$$ = 2, x, y, z, ≠ 0.
Solution:
Let $$\frac{1}{x}$$ = u,
$$\frac{1}{y}$$ = v
$$\frac{1}{z}$$ = w, then
2u + 3v + 10w = 4
4u – 6v + 5w = 1
6u + 9v – 20 w = 2
Applying formula AX = B
where

⇒ |A| = 2 × (120 – 45) -3 (- 80 – 30) + 10 (36 + 36)
⇒ |A| = 150 + 330 + 720 = 1200
⇒ |A| ≠ 0, hence A-1 exists.
Applying formula
X = A-1B

As $$\frac{1}{x}$$ = u, $$\frac{1}{y}$$ = v, and $$\frac{1}{z}$$ = $$\frac{1}{z}$$ = w
∴$$\frac{1}{x}$$ = $$\frac{1}{2}$$, $$\frac{1}{y}$$ = $$\frac{1}{3}$$ and $$\frac{1}{z}$$ = $$\frac{1}{5}$$
⇒ x = 2, y = 3 and z = 5. is required solution.

Question 16.
Find A-1 where A = $$\left[\begin{array}{ccc} {1} & {2} & {-3} \\ {2} & {3} & {2} \\ {3} & {-3} & {-4} \end{array}\right]$$, corresponding equation is:
x + 2y – 3z = -4
2x + 3y + 2z = 2
3x – 3y – 4z = 11. (CBSE 2008, 10, 12)
Solution:
Given:
A = $$\left[\begin{array}{ccc} {1} & {2} & {-3} \\ {2} & {3} & {2} \\ {3} & {-3} & {-4} \end{array}\right]$$
∴ |A| = $$\left[\begin{array}{ccc} {1} & {2} & {-3} \\ {2} & {3} & {2} \\ {3} & {-3} & {-4} \end{array}\right]$$
⇒ |A| = 1(- 12 + 6) – 2 (- 8 – 6) – 3 (- 6 – 9)
⇒ |A| = 1 (-12 + 6) -2 (-8 -6) -3 (-6 -9)
⇒ |A| ≠ 0
⇒ Hence A-1 exists.
Hence

Equation of above matrix
x + 2y – 3z = – 4
2x + 3y + 2z = 2
3x – 3y – 4z = 11
Solution of above equations
AX = B
Where A = $$\left[\begin{array}{ccc} {1} & {2} & {-3} \\ {2} & {3} & {2} \\ {3} & {-3} & {-4} \end{array}\right]$$, X = [ $$\begin{matrix} x \\ y \\ z \end{matrix}$$ ] , B = [ $$\begin{matrix} -4 \\ 2 \\ 11 \end{matrix}$$ ]
Applying formula
X = A-1B

Question 17.
Find the product of matrix $$\left[\begin{array}{ccc} {-4} & {4} & {4} \\ {-7} & {1} & {3} \\ {5} & {-3} & {-1} \end{array}\right]$$ $$\left[\begin{array}{ccc} {1} & {-1} & {1} \\ {1} & {-2} & {-2} \\ {2} & {1} & {3} \end{array}\right]$$ and with the help of product of matrix solve the equations? (CBSE 2012)
x – y + z = 4
x – 2y – 2z = 9
2x + y + 3z = 1.
Solution:
Let

Writing above equation in matrix form
AX = C
Where

⇒ x = 3, y = -2, z = -1, (by equating of two matrix)

Question 18.
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs. 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs. 90. The cost of 6 kg onion, 2 kg wheat and 3 kg rice is Rs. 70. Find the cost of each item per kg by matrix method?
Solution:
Let the cost of 1 kg onion = Rs. x
1 kg wheat = Rs. y
and 1 kg rice = Rs. z
According to equation
4x + 3y + 2z = 60
2x + 4y + 6z = 90
6x + 2y + 3z = 70
Matrix form will be
AX = B
Where

⇒ Hence A-1 exist.
Applying formula
X = A-1 B

∴ x = Rs. 5, y = Rs. 8, z = Rs. 8.