If A = 2 3 - 1 2 and f x = x 2 - 4 x + 7 , show that f A = O . Use this to find A 3 and A 5 .

Given that, A = 2 3 - 1 2 and f x = x 2 - 4 x + 7 To proof : f A = O To find : A 3 and A 5 Now, f x = x 2 - 4 x + 7 f A = A 2 - 4 A + 7 I = 2 3 - 1 2 2 3 - 1 2 - 4 2 3 - 1 2 + 7 1 0 0 1 = 2 × 2 + 3 × 2 2 × 3 + 3 × 2 - 1 × 2 + 2 × - 1 - 1 × 3 + 2 × 2 - 8 12 - 4 8 + 7 0 0 7 = 1 12 - 4 1 - 8 12 - 4 8 + 7 0 0 7 = 1 - 8 + 7 12 - 12 + 0 - 4 + 4 + 0 1 - 8 + 7 = 0 0 0 0 = O Hence, f x = x 2 - 4 x + 7 = O . Taking, A 3 = A 2 A A 3 = 1 12 - 4 1 2 3 - 1 2 A 3 = 1 × 2 + 12 × - 1 1 × 3 + 12 × 2 - 4 × 2 + 1 × - 1 - 4 × 3 + 1 × 2 = - 10 27 - 9 - 10 A 4 = A 2 A 2 A 4 = 1 12 - 4 1 1 12 - 4 1 = - 47 24 - 8 - 47 Now, A 5 = A 4 × A A 5 = - 47 24 - 8 - 47 2 3 - 1 2 = - 118 - 93 31 - 118 Hence, A 3 = - 10 27 - 9 - 10 and A 5 = - 118 - 93 31 - 118 .

If A = 1 1 1 1 , prove that A n = 2 n - 1 2 n - 1 2 n - 1 2 n - 1 , for all positive integers n .

Given that, A = 1 1 1 1 To proof : A n = 2 n - 1 2 n - 1 2 n - 1 2 n - 1 , for all positive integers n Now, We shall prove the result by the principle mathematical induction on n . When n = 1 , we have A 1 = A = 1 1 1 1 = 2 1 - 1 2 1 - 1 2 1 - 1 2 1 - 1 So, the result is true for n = 1 Now, let result be true for n = m , then A m = 2 m - 1 2 m - 1 2 m - 1 2 m - 1 We shall now prove the result for n = m + 1 A m + 1 = A m · A = 2 m - 1 2 m - 1 2 m - 1 2 m - 1 1 1 1 1 A m + 1 = 2 m - 1 × 1 + 2 m - 1 × 12 m - 1 × 1 + 2 m - 1 × 1 2 m - 1 × 1 + 2 m - 1 × 1 2 m - 1 × 1 + 2 m - 1 × 1 A m + 1 = 2 m - 1 + 2 m - 1 2 m - 1 + 2 m - 1 2 m - 1 + 2 m - 1 2 m - 1 + 2 m - 1 A m + 1 = 2 · 2 m - 1 2 · 2 m - 1 2 · 2 m - 1 2 · 2 m - 1 A m + 1 = 2 m 2 m 2 m 2 m This shows that the result is true for n = m + 1 , whenever it is true for n = m . Hence, by principle of mathematical induction the result is valid for any positive integer n .

Find a 2 × 2 matrix B such that B 1 - 2 1 4 = 6 I 2 .

Given that, B 1 - 2 1 4 = 6 I 2 To find : 2 × 2 matrix B Now, Let B = a b c d a b c d 1 - 2 1 4 = 6 1 0 0 1 a × 1 + b × 1 a × - 2 + b × 4 c × 1 + d × 1 c × - 2 + d × 4 = 6 0 0 6 a + b - 2 a + 4 b c + d - 2 c + 4 d = 6 0 0 6 Equate the corresponding terms... a + b = 6 . . . . . . . . 1 - 2 a + 4 b = 0 ⇒ a = 2 b a = 2 b Put in 1 2 b + b = 6 ⇒ b = 2 So, a = 2 b ⇒ a = 2 2 a = 4 c + d = 0 ⇒ c = - d . . . . . . . . . . . 2 2 c + 4 d = 6 ⇒ c - 2 d = - 3 . . . . . . . . . . . 3 c = - d ⇒ - d - 2 d = - 3 ⇒ d = 1 c = - 1 To check the accuracy, put the value of c and d in 2 i.e. c + d = 0 ⇒ - 1 + 1 = 0 Hence, B = 4 2 - 1 1 .

If A = 3 - 4 - 1 2 , find matrix B such that BA = I .

Given that, A = 3 - 4 - 1 2 To find : matrix B such that BA = I Now, BA = I ⇒ B = A - 1 ∴ A = 3 × 2 - - 1 × - 4 = 6 - 4 = 2 adjA = 2 1 4 3 = 2 4 1 3 ∴ A - 1 = 1 A adj A = 1 2 2 4 1 3 ⇒ A - 1 = 1 2 1 2 3 2 = B Hence, B = 1 2 1 2 3 2 .

Find the matrix A such that 1 1 0 1 A = 3 3 5 1 0 1 .

Given that, 1 1 0 1 A = 3 3 5 1 0 1 To find : matrix A Now, Let A = a b c d e f 1 1 0 1 a b c d e f = 3 3 5 1 0 1 1 × a + 1 × d 1 × b + 1 × e 1 × c + 1 × f 0 × a + 1 × d 0 × b + 1 × e 0 × c + 1 × f = 3 3 5 1 0 1 a + d b + e c + f d e f = 3 3 5 1 0 1 Equating the corresponding terms... a + d = 3 . . . . . . . . . . . . . 1 d = 1 . . . . . . . . . 2 a + 1 = 3 ⇒ a = 2 b + e = 3 . . . . . . . . . 3 e = 0 . . . . . . . . . . . . 4 b + 0 = 3 ⇒ b = 3 c + f = 5 . . . . . . . . . . . . 5 f = 1 . . . . . . . . . . . . . . . 6 c + 1 = 5 ⇒ c = 4 Hence, matrix is A = 2 3 4 1 0 1 .

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

Given that, 2 - 1 1 0 - 3 4 A = - 1 - 8 1 - 2 9 22 To find : matrix A Now, 2 - 1 1 0 - 3 4 3 × 2 A = - 1 - 8 1 - 2 9 22 3 × 2 . . . . . . . . . 1 Since, product of matrices on left hand side is possible If number of rows in A = 2 = number of columns of pre-matrix. Also, product matrix on R . H . S is of order 3 × 2 . ∴ A must be of order 2 × 2 . Let A = a b c d 2 × 2 From equation 1 , we have 2 - 1 1 0 - 3 4 a b c d = - 1 - 8 1 - 2 9 22 2 × a + - 1 × c 2 × b + - 1 × d 1 × a + 0 × c 1 × b + 0 × d - 3 × a + 4 × c - 3 × b + 4 × d = - 1 - 8 1 - 2 9 22 2 a - c 2 b - d a + 0 b + 0 - 3 a + 4 c - 3 b + 4 d = - 1 - 8 1 - 2 9 22 Equating the corresponding elements.... 2 a - c = - 1 . . . . . . . . . . . 1 a = 1 . . . . . . . 2 2 - c = - 1 ⇒ c = 3 Also a = 1 , c = 3 satisfies - 3 a + 4 c = 9 . . . . . . . . . . 3 Taking, 2 b - d = - 8 . . . . . . . 4 b = - 2 . . . . . . . . . . 5 - 4 - d = - 8 ⇒ d = 4 Also b = - 2 and d = 4 satisfies - 3 b + 4 d = 22 Hence, a = 1 , c = 3 , b = - 2 , d = 4 Required matrix is A = 1 - 2 3 4 2 × 2 .

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

Given that, 2 - 1 1 0 - 3 4 A = - 1 - 8 - 10 1 - 2 - 5 9 22 15 To find : A Now, 2 - 1 1 0 - 3 4 3 × 2 A = - 1 - 8 - 10 1 - 2 - 5 9 22 15 3 × 3 Since, product of matrices on L . H . S is possible only if Number of rows in A = 2 = Number of columns in pre-matrix. Further, product matrix on R . H . S is of order 3 × 3 . Therefore, A must be a matrix of order 2 × 3 . Let A = a b c d e f 2 × 3 From given equation , we have 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 Equating the corresponding elements, i.e. 2 a - d = - 1 . . . . . . . . . . . . . 1 a = 1 ∴ from 1 , d = 2 + 1 = 3 Clearly a = 1 , d = 3 satisfies - 3 a + 4 d = 9 Also, 2 b - e = - 8 . . . . . . . . . . . . 2 And b = - 2 ∴ from 2 , e = 4 Thus, b = - 2 and e = 4 satisfies - 3 b + 4 e = 22 Also, 2 c - f = - 10 . . . . . . . . . . . . . . 3 And c = - 5 ∴ from 3 , f = 0 Also, c = - 5 , f = 0 satisfies - 3 c + 4 f = 15 . Hence, required matrix is A = 1 - 2 - 5 3 4 0 .

E M B I B E

Understanding ISC Mathematics Class 12 Volume 1>Chapter 3 - Matrices>EXERCISE 3.3

M L Aggarwal Solutions for Chapter: Matrices, Exercise 3: EXERCISE 3.3

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Matrices, Exercise 3: EXERCISE 3.3

Attempt the practice questions on Chapter 3: Matrices, Exercise 3: EXERCISE 3.3 with hints and solutions to strengthen your understanding. Understanding ISC Mathematics Class 12 Volume 1 solutions are prepared by Experienced Embibe Experts.

Questions from M L Aggarwal Solutions for Chapter: Matrices, Exercise 3: EXERCISE 3.3 with Hints & Solutions

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 3 - Matrices>EXERCISE 3.3

Find $x$ , if

$[\begin{matrix} 1 & 2 & 1 \end{matrix}] [\begin{matrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{matrix}] [\begin{matrix} 0 \\ 2 \\ x \end{matrix}] = O \{\}$ .

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 3 - Matrices>EXERCISE 3.3

If $A = [\begin{matrix} 2 & 3 \\ - 1 & 2 \end{matrix}]$ and $f (x) = x^{2} - 4 x + 7$ , show that $f (A) = O$ . Use this to find $A^{3}$ and $A^{5}$ .

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 3 - Matrices>EXERCISE 3.3

If $A = [\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}]$ , prove that $A^{n} = [\begin{matrix} 2^{n - 1} & 2^{n - 1} \\ 2^{n - 1} & 2^{n - 1} \end{matrix}]$ , for all positive integers $n$ .

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 3 - Matrices>EXERCISE 3.3

Find a $2 \times 2$ matrix $B$ such that $B [\begin{matrix} 1 & - 2 \\ 1 & 4 \end{matrix}] = 6 I_{2}$ .

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 3 - Matrices>EXERCISE 3.3

If $A = [\begin{matrix} 3 & - 4 \\ - 1 & 2 \end{matrix}]$ , find matrix $B$ such that $BA = I$ .

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 3 - Matrices>EXERCISE 3.3

Find the matrix $A$ such that $[\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] A = [\begin{matrix} 3 & 3 & 5 \\ 1 & 0 & 1 \end{matrix}]$ .

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 3 - Matrices>EXERCISE 3.3

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

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 3 - Matrices>EXERCISE 3.3

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

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