Let A = 1 2 - 1 3 , B = 4 0 1 5 , C = 2 0 1 - 2 and a = 4 , b = - 2 . Show that A B T = B T A T .

We have, A = 1 2 - 1 3 , B = 4 0 1 5 , C = 2 0 1 - 2 and a = 4 , b = - 2 . A B = 1 2 - 1 3 4 0 1 5 = 4 + 2 0 + 10 - 4 + 3 0 + 15 = 6 10 - 1 15 ∴ A B T = 6 - 1 10 15 Now, B T A T = 4 1 0 5 1 - 1 2 3 = 4 + 2 − 4 + 3 0 + 10 0 + 15 = 6 − 1 10 15 = A B T Hence, Proved.

Let A = 1 2 - 1 3 , B = 4 0 1 5 , C = 2 0 1 - 2 and a = 4 , b = - 2 . Show that A B T = B T A T .

We have, A = 1 2 - 1 3 , B = 4 0 1 5 , C = 2 0 1 - 2 and a = 4 , b = - 2 . A B = 1 2 - 1 3 4 0 1 5 = 4 + 2 0 + 10 - 4 + 3 0 + 15 = 6 10 - 1 15 ∴ A B T = 6 - 1 10 15 Now, B T A T = 4 1 0 5 1 - 1 2 3 = 4 + 2 − 4 + 3 0 + 10 0 + 15 = 6 − 1 10 15 = A B T Hence, Proved.

Let A = 1 2 - 1 3 , B = 4 0 1 5 , C = 2 0 1 - 2 and a = 4 , b = - 2 . Show that ( A - B ) C = A C - B C .

We have, A = 1 2 - 1 3 , B = 4 0 1 5 , C = 2 0 1 - 2 and a = 4 , b = - 2 . A - B = 1 - 4 2 - 0 - 1 - 1 3 - 5 = - 3 2 - 2 - 2 A - B C = - 3 2 - 2 - 2 2 0 1 - 2 = - 4 - 4 - 6 4 … i Now, A C = 1 2 - 1 3 2 0 1 - 2 = 4 - 4 1 - 6 … ii B C = 4 0 1 5 2 0 1 - 2 = 8 0 7 - 10 … iii A C · B C = 4 - 8 - 4 - 0 1 - 7 - 6 + 10 [ Using Eq. ii and iii ] = - 4 - 4 - 6 4 = ( A - B ) C [ Using Eq. i ] Hence, Proved.

Let A = 1 2 - 1 3 , B = 4 0 1 5 , C = 2 0 1 - 2 and a = 4 , b = - 2 . Show that ( A - B ) T = A T - B T .

We have, A = 1 2 - 1 3 , B = 4 0 1 5 , C = 2 0 1 - 2 and a = 4 , b = - 2 . A - B T = 1 - 4 2 - 0 - 1 - 1 3 - 5 T = - 3 2 - 2 - 2 T = - 3 - 2 2 - 2 And, A T - B T = 1 - 1 2 3 - 4 1 0 5 = - 3 - 2 2 - 2 = A - B T Hence, Proved.

If A = cos θ sin θ - sin θ cos θ , then show that A 2 = cos 2 θ sin 2 θ - sin 2 θ cos 2 θ .

We have, A = cos θ sin θ - sin θ cos θ ∴ A 2 = A · A = cos θ sin θ - sin θ cos θ cos θ sin θ - sin θ cos θ = cos 2 θ - sin 2 θ cos θ sin θ + sin θ cos θ - sin θ cos θ - cos θ sin θ - sin 2 θ + cos 2 θ = cos 2 θ 2 sin θ cos θ - 2 sin θ cos θ cos 2 θ ∵ cos 2 x - sin 2 x = cos 2 x = cos 2 θ sin 2 θ - sin 2 θ cos 2 θ ∵ sin 2 x = 2 sin x cos x ] Hence, proved.

If A = 0 − x x 0 , B = 0 1 1 0 and x 2 = - 1 , then show that ( A + B ) 2 = A 2 + B 2 .

We have A = 0 − x x 0 , B = 0 1 1 0 and x 2 = - 1 . ∴ ( A + B ) = 0 - x + 1 x + 1 0 And ( A + B ) 2 = 0 - x + 1 x + 1 0 0 - x + 1 x + 1 0 = 1 - x 2 0 0 1 - x 2 .......(i) Also A 2 = A A = 0 - x x 0 0 - x x 0 = - x 2 0 0 - x 2 And B 2 = B B = 0 1 1 0 0 1 1 0 = 1 0 0 1 Now A 2 + B 2 = - x 2 + 1 0 0 - x 2 + 1 [Using Eq (i) and (ii)] = 1 - x 2 0 0 1 - x 2 = A + B 2 Hence, proved.

Verify that A 2 = I when A = 0 1 - 1 4 - 3 4 3 - 3 4 .

We have, A = 0 1 - 1 4 - 3 4 3 - 3 4 ∴ A 2 = 0 1 - 1 4 - 3 4 3 - 3 4 0 1 - 1 4 - 3 4 3 - 3 4 ∵ A 2 = A A = 0 + 4 − 3 0 − 3 + 3 0 + 4 − 4 0 − 12 + 12 4 + 9 − 12 − 4 − 12 + 16 0 − 12 + 12 3 + 9 − 12 − 3 − 12 + 16 = 1 0 0 0 1 0 0 0 1 = I Hence, proved.

Prove by Mathematical Induction that A ' n = A n ' , where n ∈ N for any square matrix A .

Let P ( n ) : A ' n = A n ' ∴ P ( 1 ) : ( A ' ) 1 = ( A 1 ) ' ⇒ A ' = A ' ⇒ P ( 1 ) is true. Now let P ( k ) : A ' k = A k ' where k ∈ N . And P ( k + 1 ) : A ' k + 1 = A k + 1 ' We need to prove P k + 1 is true whenever P k is true. ∴ P ( k + 1 ) : A ' k · A ' 1 = A k + 1 ' ⇒ A ' k · A ' 1 = A k ' ( A ) ' = A A k ' = A k + 1 ' ∵ A ' k = A k ' and ( A B ) ' = B ' A ' ∴ A ' k + 1 = A k + 1 ' Hence, proved.

E M B I B E

NCERT Exemplar Mathematics - Class 12>Chapter 3 - Matrices>EXERCISE>Q 32

MEDIUM

12th CBSE

IMPORTANT

Earn 100

Let $A = [\begin{matrix} 1 & 2 \\ - 1 & 3 \end{matrix}], B = [\begin{matrix} 4 & 0 \\ 1 & 5 \end{matrix}], C = [\begin{matrix} 2 & 0 \\ 1 & - 2 \end{matrix}] and a = 4, b = - 2$ . Show that ${(A B)}^{T} = B^{T} A^{T}$ .

Important Questions on Matrices

MEDIUM

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 3 - Matrices>EXERCISE>Q 32

Let $A = [\begin{matrix} 1 & 2 \\ - 1 & 3 \end{matrix}], B = [\begin{matrix} 4 & 0 \\ 1 & 5 \end{matrix}], C = [\begin{matrix} 2 & 0 \\ 1 & - 2 \end{matrix}] and a = 4, b = - 2$ . Show that $(A - B) C = A C - B C$ .

EASY

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 3 - Matrices>EXERCISE>Q 32

Let $A = [\begin{matrix} 1 & 2 \\ - 1 & 3 \end{matrix}], B = [\begin{matrix} 4 & 0 \\ 1 & 5 \end{matrix}], C = [\begin{matrix} 2 & 0 \\ 1 & - 2 \end{matrix}] and a = 4, b = - 2$ . Show that $(A - B)^{T} = A^{T} - B^{T}$ .