If A = 1 1 3 5 2 6 - 2 - 1 - 3 , then find A 3 .

We have, A = 1 1 3 5 2 6 - 2 - 1 - 3 Now, A 2 = A · A = 1 1 3 5 2 6 − 2 − 1 − 3 1 1 3 5 2 6 − 2 − 1 − 3 = 1 + 5 − 6 1 + 2 − 3 3 + 6 − 9 5 + 10 − 12 5 + 4 − 6 15 + 12 − 18 − 2 − 5 + 6 − 2 − 2 + 3 − 6 − 6 + 9 = 0 0 0 3 3 9 − 1 − 1 − 3 Now, A 3 = A 2 · A = 0 0 0 3 3 9 − 1 − 1 − 3 1 1 3 5 2 6 − 2 − 1 − 3 = 0 + 0 + 0 0 + 0 + 0 0 + 0 + 0 3 + 15 − 18 3 + 6 − 9 9 + 18 − 27 − 1 − 5 + 6 − 1 − 2 + 3 − 3 − 6 + 9 = 0 0 0 0 0 0 0 0 0

If A = 1 − 2 1 0 1 − 1 3 − 1 1 , then find A 3 − 3 A 2 − A − 3 I , where I is unit matrix of order 3 .

Given, A = 1 − 2 1 0 1 − 1 3 − 1 1 Now, A 2 = A ⋅ A = 1 − 2 1 0 1 − 1 3 − 1 1 1 − 2 1 0 1 − 1 3 − 1 1 = 1 + 0 + 3 − 2 − 2 − 1 1 + 2 + 1 0 + 0 − 3 0 + 1 + 1 0 − 1 − 1 3 − 0 + 3 − 6 − 1 − 1 3 + 1 + 1 = 4 − 5 4 − 3 2 − 2 6 − 8 5 And, A 3 = A 2 · A = 4 − 5 4 − 3 2 − 2 6 − 8 5 1 − 2 1 0 1 − 1 3 − 1 1 = 4 + 0 + 12 − 8 − 5 − 4 4 + 5 + 4 − 3 + 0 − 6 6 + 2 + 2 − 3 − 2 − 2 6 + 0 + 15 − 12 − 8 − 5 6 + 8 + 5 = 16 − 17 13 − 9 10 − 7 21 − 25 19 Now, A 3 − 3 A 2 − A − 3 I = 16 − 17 13 − 9 10 − 7 21 − 25 19 − 3 4 − 5 4 − 3 2 − 2 6 − 8 5 − 1 − 2 1 0 1 − 1 3 − 1 1 − 3 1 0 0 0 1 0 0 0 1 = 16 − 17 13 − 9 10 − 7 21 − 25 19 − 12 − 15 12 − 9 6 − 6 18 − 24 20 − 1 − 2 1 0 1 − 1 3 − 1 1 − 3 0 0 0 3 0 0 0 3 = 16 − 12 − 1 − 3 − 17 + 15 + 2 + 0 13 − 12 − 1 − 0 − 9 + 9 + 0 − 0 10 − 6 − 1 − 3 − 7 + 6 + 1 + 0 21 − 18 − 3 + 0 − 25 + 24 + 1 + 0 19 − 15 − 1 − 3 = 0 0 0 0 0 0 0 0 0 = O 3 × 3 ∴ A 3 − 3 A 2 − A − 3 I = 0

If I = 1 0 0 1 and E = 0 1 0 0 , then show that ( a I + b E ) 3 = a 3 I + 3 a 2 b E , where I is unit matrix of order 2 .

Given, I = 1 0 0 1 and E = 0 1 0 0 . Consider, a I + b E = a 1 0 0 1 + b 0 1 0 0 = a 0 0 a + 0 b 0 0 = a b 0 a And, ( a I + b E ) 2 = a b 0 a a b 0 a = a 2 2 a b 0 a 2 And, ( a I + b E ) 3 = a 2 2 a b 0 a 2 a b 0 a = a 3 3 a 2 b 0 a 3 = a 3 0 0 a 3 + 0 3 a 2 b 0 0 = a 3 1 0 0 1 + 3 a 2 b 0 1 0 0 = a 3 I + 3 a 2 b E . Hence proved.

If A = diag ⁡ a 1 , a 2 , a 3 , then for any integer n ≥ 1 show that A n = diag ⁡ a 1 n , a 2 n , a 3 n .

We shall prove the result by Principle of mathematical induction on n . Step 1 : If n = 1 , by definition of integral power of a matrix, we have A 1 = a 1 1 0 0 0 a 2 1 0 0 0 a 3 1 = a 1 0 0 0 a 2 0 0 0 a 3 = A So, the result is true for n = 1 . Step 2 : Let the result be true for n = m . Then, A m = a 1 m 0 0 0 a 2 m 0 0 0 a 3 m . . . 1 Now, we shall check if result is true for n = m + 1 . Here, A m + 1 = a 1 m + 1 0 0 0 a 2 m + 1 0 0 0 a 3 m + 1 By definition of integral power of a matrix, we have A m + 1 = A m A ⇒ A m + 1 = a 1 m 0 0 0 a 2 m 0 0 0 a 3 m a 1 0 0 0 a 2 0 0 0 a 3 from 1 ⇒ A m + 1 = a 1 a 1 m + 0 + 0 0 + 0 + 0 0 + 0 + 0 0 + 0 + 0 0 + a 2 a 2 m + 0 0 + 0 + 0 0 + 0 + 0 0 + 0 + 0 0 + 0 + a 3 a 3 m ⇒ A m + 1 = a 1 m + 1 0 0 0 a 2 m + 1 0 0 0 a 3 m + 1 This shows that when result is true for n = m , it is also true for n = m + 1 . Hence, by principle of mathematical induction, result is valid for any positive integer n .

If θ − ϕ = π 2 , then show that cos 2 ⁡ θ cos ⁡ θ sin ⁡ θ cos ⁡ θ sin ⁡ θ sin 2 ⁡ θ cos 2 ⁡ ϕ cos ⁡ ϕ sin ⁡ ϕ cos ⁡ ϕ sin ⁡ ϕ sin 2 ⁡ ϕ = O .

Consider, cos 2 ⁡ θ cos ⁡ θ sin ⁡ θ cos ⁡ θ sin ⁡ θ sin 2 ⁡ θ cos 2 ⁡ ϕ cos ⁡ ϕ sin ⁡ ϕ cos ⁡ ϕ sin ⁡ ϕ sin 2 ⁡ ϕ = cos 2 ⁡ θ cos 2 ⁡ ϕ + cos ⁡ θ cos ⁡ ϕ sin ⁡ θ sin ⁡ ϕ cos 2 ⁡ ϕ cos ⁡ θ sin ⁡ θ + sin 2 ⁡ θ cos ⁡ ϕ sin ⁡ ϕ cos 2 ⁡ θ cos ⁡ ϕ sin ⁡ ϕ + cos ⁡ θ sin ⁡ θ sin 2 ⁡ ϕ cos ⁡ θ sin ⁡ θ cos ⁡ ϕ sin ⁡ ϕ + sin 2 ⁡ θ sin 2 ⁡ ϕ = cos ⁡ θ cos ⁡ ϕ cos ⁡ θ cos ⁡ ϕ + sin ⁡ θ sin ⁡ ϕ sin θ cos ϕ cos ⁡ ϕ cos ⁡ θ + sin ⁡ θ sin ⁡ ϕ cos ⁡ θ sin ⁡ ϕ cos ⁡ θ cos ϕ + sin ⁡ θ sin ⁡ ϕ sin θ sin ϕ cos ⁡ θ cos ⁡ ϕ + sin ⁡ θ sin ⁡ ϕ = cos ⁡ θ cos ⁡ ϕ cos ⁡ ( θ − ϕ ) cos ⁡ θ sin ⁡ ϕ cos ⁡ ( θ − ϕ ) sin ⁡ θ cos ⁡ ϕ cos ⁡ ( θ − ϕ ) sin ⁡ θ sin ⁡ ϕ cos ⁡ ( θ − ϕ ) ∵ cos A - B = cos A cos B + sin A sin B = 0 0 0 0 ∵ θ − ϕ = ( 2 n + 1 ) π 2 , n ∈ Z ∴ cos ⁡ ( θ − ϕ ) = cos ⁡ ( 2 n + 1 ) π 2 = 0 = O Hence proved.

If A = 3 − 4 1 − 1 then show that A n = 1 + 2 n − 4 n n 1 − 2 n for any integer n ≥ 1 , by using mathematical induction.

We shall prove the result by mathematical induction on n . Step 1 : When n = 1 , by the definition of integral powers of a matrix, we have A 1 = A = 3 − 4 1 − 1 = 1 + 2 ( 1 ) − 4 ( 1 ) 1 1 − 2 ( 1 ) So, the result is true for n = 1 . Step 2 : Let the result be true for n = m . Then, A m = 1 + 2 m − 4 m m 1 − 2 m . . . 1 Now we shall show that result is true for n = m + 1 i.e. A m + 1 = 1 + 2 ( m + 1 ) − 4 ( m + 1 ) m + 1 1 − 2 ( m + 1 ) By definition of integral powers of a square matrix, we have A m + 1 = A m ⋅ A ⇒ A m + 1 = 1 + 2 m − 4 m m 1 − 2 m 3 − 4 1 − 1 from 1 ⇒ A m + 1 = 3 + 6 m − 4 m − 4 − 8 m + 4 m 3 m + 1 − 2 m − 4 m − 1 + 2 m ⇒ A m + 1 = 3 + 2 m − 4 − 4 m m + 1 − 1 − 2 m ⇒ A m + 1 = 1 + 2 ( m + 1 ) − 4 ( m + 1 ) m + 1 1 − 2 ( m + 1 ) This shows that result is true for n = m + 1 , whenever it is true for n = m Hence, by principle of mathematical induction result is valid for any positive integer n .

E M B I B E

Intermediate First Year Mathematics paper - 1A>Algebra, Vector Algebra, Trigonometry>Chapter 3 - Matrices>Exercise 3 (b)

Telangana Board Solutions for Exercise 2: Exercise 3 (b)

Author:Telangana Board

Telangana Board Mathematics Solutions for Exercise - Telangana Board Solutions for Exercise 2: Exercise 3 (b)

Attempt the practice questions from Exercise 2: Exercise 3 (b) with hints and solutions to strengthen your understanding. Intermediate First Year Mathematics paper - 1A solutions are prepared by Experienced Embibe Experts.

Questions from Telangana Board Solutions for Exercise 2: Exercise 3 (b) with Hints & Solutions

EASY

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 3 - Matrices>Exercise 3 (b)>Q I

If $A = [\begin{array}{cc} 2 & 4 \\ - 1 & k \end{array}]$ and $A^{2} = O$ , then find the value of $k$ .

MEDIUM

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 3 - Matrices>Exercise 3 (b)>Q II

If $A = [\begin{matrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ - 2 & - 1 & - 3 \end{matrix}]$ , then find $A^{3}$ .

HARD

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 3 - Matrices>Exercise 3 (b)>Q II

If $A = [\begin{array}{ccc} 1 & - 2 & 1 \\ 0 & 1 & - 1 \\ 3 & - 1 & 1 \end{array}]$ , then find $A^{3} - 3 A^{2} - A - 3 I$ , where $I$ is unit matrix of order $3$ .

MEDIUM

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 3 - Matrices>Exercise 3 (b)>Q II

If $I = [\begin{array}{ll} 1 0 \\ 0 1 \end{array}]$ and $E = [\begin{array}{ll} 0 1 \\ 0 0 \end{array}]$ , then show that $(a I + b E)^{3} = a^{3} I + 3 a^{2} b E$ , where $I$ is unit matrix of order $2$ .

MEDIUM

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 3 - Matrices>Exercise 3 (b)>Q III

If $A = diag [a_{1}, a_{2}, a_{3}]$ , then for any integer $n \geq 1$ show that $A^{n} = diag [{a_{1}}^{n}, {a_{2}}^{n}, {a_{3}}^{n}]$ .

MEDIUM

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 3 - Matrices>Exercise 3 (b)>Q III

If $θ - ϕ = \frac{π}{2}$ , then show that $[\begin{array}{lr} \cos^{2} θ \cos θ \sin θ \\ \cos θ \sin θ \sin^{2} θ \end{array}] [\begin{array}{lr} \cos^{2} ϕ \cos ϕ \sin ϕ \\ \cos ϕ \sin ϕ \sin^{2} ϕ \end{array}] = O$ .

HARD

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 3 - Matrices>Exercise 3 (b)>Q III

If $A = [\begin{array}{ll} 3 - 4 \\ 1 - 1 \end{array}]$ then show that $A^{n} = [\begin{array}{cc} 1 + 2 n & - 4 n \\ n & 1 - 2 n \end{array}]$ for any integer $n \geq 1$ , by using mathematical induction.