If x y 4 z + 6 x + y = 8 w 0 6 , then find values of x , y , z and w .

We have x y 4 z + 6 x + y = 8 w 0 6 By equality of matrix x + y = 6 and x y = 8 ⇒ x = 6 - y and ( 6 - y ) y = 8 ⇒ y 2 - 6 y + 8 = 0 ⇒ y 2 - 4 y - 2 y + 8 = 0 ⇒ ( y - 2 ) ( y - 4 ) = 0 ⇒ y = 2 or y = 4 ∴ x = 6 - 4 = 2 or x = 6 - 2 = 4 [ ∵ x = 6 - y ] Also z + 6 = 0 ⇒ z = - 6 and w = 4 . ∴ ( x = 2 , y = 4 ) or ( x = 4 , y = 2 ) , z = - 6 and w = 4 .

If A = 1 5 7 12 and B = 9 1 7 8 , find a matrix C such that 3 A + 5 B + 2 C is a null matrix.

We have A = 1 5 7 12 and B = 9 1 7 8 Let C = a b c d ∴ 3 A + 5 B + 2 C = O ⇒ 3 15 21 36 + 45 5 35 40 + 2 a 2 b 2 c 2 d = 0 0 0 0 ⇒ 48 + 2 a 20 + 2 b 56 + 2 c 76 + 2 d = 0 0 0 0 By equality of matrices, we have ⇒ 2 a + 48 = 0 ⇒ a = - 24 Also, 20 + 2 b = 0 ⇒ b = - 10 And, 56 + 2 c = 0 ⇒ c = - 28 And, 76 + 2 d = 0 ⇒ d = - 38 Thus, the required matrix is C = - 24 - 10 - 28 - 38 .

If A = 3 - 5 - 4 2 , then find A 2 - 5 A - 14 I . Hence, obtain A 3 .

We have, A = 3 - 5 - 4 2 . . . . ( i ) ∴ A 2 = A A = 3 - 5 - 4 2 3 - 5 - 4 2 = 29 - 25 - 20 24 . . . . ( i i ) ∴ A 2 - 5 A - 14 I = 29 - 25 - 20 24 - 15 - 25 - 20 10 - 14 0 0 14 = 0 0 0 0 Now, A 2 - 5 A - 14 I = 0 ⇒ A A 2 - 5 A A - 14 A I = 0 [Pre-multiply both sides by A ] ⇒ A 3 - 5 A 2 - 14 A = 0 [ ∵ A I = A ] ⇒ A 3 = 5 A 2 + 14 A ⇒ A 3 = 5 29 - 25 - 20 24 + 14 3 - 5 - 4 2 [Using ( i ) and i i ] ⇒ A 3 = 145 - 125 - 100 120 + 42 - 70 - 56 28 ⇒ A 3 = 187 - 195 - 156 148 .

Find the values of a , b , c and d if 3 a b c d = a 6 - 1 2 d + 4 a + b c + d 3

We have 3 a b c d = a 6 - 1 2 d + 4 a + b c + d 3 ⇒ 3 a 3 b 3 c 3 d = a + 4 6 + a + b c + d - 1 3 + 2 d By equality of matrices, we have: 3 a = a + 4 ⇒ 2 a = 4 ⇒ a = 2 Also, 3 b = 6 + a + b ⇒ 3 b - b = 6 + 2 ∵ a = 2 ⇒ 2 b = 8 ⇒ b = 4 Also, 3 d = 3 + 2 d ⇒ d = 3 And, 3 c = c + d - 1 ⇒ 2 c = 3 - 1 ∵ d = 3 ⇒ 2 c = 2 ⇒ c = 1 ∴ a = 2 , b = 4 , c = 1 and d = 3 .

Find the matrix A such that 2 - 1 1 0 - 3 4 A = - 1 - 8 - 10 1 - 2 - 5 9 22 15 .

We have 2 - 1 1 0 - 3 4 3 × 2 A = - 1 - 8 - 10 1 - 2 - 5 9 22 15 3 × 3 From the given equation, it is clear that order of A should be 2 × 3 . Let A = a b c d e f ∴ 2 - 1 1 0 - 3 4 a b c d e f = - 1 - 8 - 10 1 - 2 - 5 9 22 15 ⇒ 2 a - d 2 b - e 2 c - f a + 0 b + 0 c + 0 - 3 a + 4 d - 3 b + 4 e - 3 c + 4 f = - 1 - 8 - 10 1 - 2 - 5 9 22 15 ⇒ 2 a - d 2 b - e 2 c - f a b c - 3 a + 4 d - 3 b + 4 e - 3 c + 4 f = - 1 - 8 - 10 1 - 2 - 5 9 22 15 By equality of matrices, we get a = 1 , b = - 2 , c = - 5 . And, 2 a - d = - 1 ⇒ 2 1 + 1 = d ∵ a = 1 ⇒ d = 3 And, 2 b - e = - 8 ⇒ e = 2 ( - 2 ) + 8 ∵ b = - 2 ⇒ e = - 4 + 8 ⇒ e = 4 And, 2 c - f = - 10 ⇒ f = 2 c + 10 ⇒ f = 2 - 5 + 10 ⇒ f = 0 ∴ A = 1 − 2 − 5 3 4 0 .

If A = 1 2 4 1 , find A 2 + 2 A + 7 I .

We have A = 1 2 4 1 Now, A 2 = A · A ⇒ A 2 = 1 2 4 1 1 2 4 1 ⇒ A 2 = 1 + 8 2 + 2 4 + 4 8 + 1 ⇒ A 2 = 9 4 8 9 So, A 2 + 2 A + 7 = 9 4 8 9 + 2 4 8 2 + 7 0 0 7 ∴ A 2 + 2 A + 7 = 18 8 16 18 .

If the matrix 0 a 3 2 b - 1 c 1 0 is a skew symmetric matrix, find the values of a , b and c .

Let A = 0 a 3 2 b - 1 c 1 0 Since, A is skew-symmetric matrix. ∴ A ' = - A ⇒ 0 2 c a b 1 3 - 1 0 = - 0 a 3 2 b - 1 c 1 0 ⇒ 0 2 c a b 1 3 - 1 0 = 0 - a - 3 - 2 - b + 1 - c - 1 0 By equality of matrices, we get a = - 2 , c = - 3 and b = - b ⇒ b = 0 ∴ a = - 2 , b = 0 and c = - 3 .

E M B I B E

NCERT Exemplar Mathematics - Class 12>Chapter 3 - Matrices>EXERCISE>Q 37

EASY

12th CBSE

IMPORTANT

Earn 100

Find inverse, by elementary row operations (if possible), of the matrix $[\begin{matrix} 1 & - 3 \\ - 2 & 6 \end{matrix}]$ .

Important Questions on Matrices

MEDIUM

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 3 - Matrices>EXERCISE>Q 38

If

[\begin{matrix} x y & 4 \\ z + 6 & x + y \end{matrix}] = [\begin{array}{l} 8 & w \\ 0 & 6 \end{array}]

, then find values of

x, y, z

and

w

.

MEDIUM

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 3 - Matrices>EXERCISE>Q 39

If

A = [\begin{array}{l} 1 & 5 \\ 7 & 12 \end{array}]

and

B = [\begin{array}{l} 9 & 1 \\ 7 & 8 \end{array}]

, find a matrix

C

such that

3 A + 5 B + 2 C

is a null matrix.

MEDIUM

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 3 - Matrices>EXERCISE>Q 40

If

A = [\begin{matrix} 3 & - 5 \\ - 4 & 2 \end{matrix}]

, then find

A^{2} - 5 A - 14 I

. Hence, obtain

A^{3}

.

MEDIUM

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 3 - Matrices>EXERCISE>Q 41

Find the values of

a, b, c

and

d

if

3 [\begin{array}{l} a & b \\ c & d \end{array}] = [\begin{matrix} a & 6 \\ - 1 & 2 d \end{matrix}] + [\begin{matrix} 4 & a + b \\ c + d & 3 \end{matrix}]

HARD

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 3 - Matrices>EXERCISE>Q 42

Find the matrix $A$ such that $[\begin{matrix} 2 & - 1 \\ 1 & 0 \\ - 3 & 4 \end{matrix}] A = [\begin{matrix} - 1 & - 8 & - 10 \\ 1 & - 2 & - 5 \\ 9 & 22 & 15 \end{matrix}]$ .