For a matrix A = 2 − 1 1 3 , show that A 2 − 5 A + 7 I 2 = O .

Given, A = 2 - 1 1 3 Now, A 2 = 2 - 1 1 3 2 - 1 1 3 = 4 - 1 - 2 - 3 2 + 3 - 1 + 9 = 3 - 5 5 8 5 A = 5 2 - 1 1 3 = 10 - 5 5 15 7 I 2 = 7 1 0 0 1 = 7 0 0 7 Now, A 2 - 5 A + 7 I 2 = 3 - 5 5 8 - 10 - 5 5 15 + 7 0 0 7 = 3 - 10 + 7 - 5 + 5 + 0 5 - 5 + 0 8 - 15 + 7 = 0 0 0 0 = O Hence, proved.

For a matrix A = 2 − 1 1 3 , show that A 2 − 5 A + 7 I 2 = O .

Given, A = 2 - 1 1 3 Now, A 2 = 2 - 1 1 3 2 - 1 1 3 = 4 - 1 - 2 - 3 2 + 3 - 1 + 9 = 3 - 5 5 8 5 A = 5 2 - 1 1 3 = 10 - 5 5 15 7 I 2 = 7 1 0 0 1 = 7 0 0 7 Now, A 2 - 5 A + 7 I 2 = 3 - 5 5 8 - 10 - 5 5 15 + 7 0 0 7 = 3 - 10 + 7 - 5 + 5 + 0 5 - 5 + 0 8 - 15 + 7 = 0 0 0 0 = O Hence, proved.

If A = 2 5 3 1 , find the value of A 2 − 3 A − 13 I 2 .

Given, A = 2 5 3 1 Now, A 2 = 2 5 3 1 2 5 3 1 = 4 + 15 10 + 5 6 + 3 15 + 1 = 19 15 9 16 3 A = 3 2 5 3 1 = 6 15 9 3 13 I 2 = 13 1 0 0 1 = 13 0 0 13 Now, A 2 - 3 A - 13 I 2 = 19 15 9 16 - 6 15 9 3 - 13 0 0 13 = 19 - 6 - 13 15 - 15 - 0 9 - 9 - 0 16 - 3 - 13 = 0 0 0 0

If A = 1 2 2 2 1 2 2 2 1 , then show that A satisfies the matrix equation X 2 − 4 X − 5 I 3 = O .

Given A = 1 2 2 2 1 2 2 2 1 Given matrix equation is X 2 - 4 X - 5 I 3 = O Substitute A in place of X in order to prove that A satisfies the above condition. ⇒ A 2 - 4 A - 5 I 3 = O Now, A 2 = 1 2 2 2 1 2 2 2 1 1 2 2 2 1 2 2 2 1 = 1 + 4 + 4 2 + 2 + 4 2 + 4 + 2 2 + 2 + 4 4 + 1 + 4 4 + 2 + 2 2 + 4 + 2 4 + 2 + 2 4 + 4 + 1 = 9 8 8 8 9 8 8 8 9 4 A = 4 1 2 2 2 1 2 2 2 1 = 4 8 8 8 4 8 8 8 4 5 I 3 = 5 1 0 0 0 1 0 0 0 1 = 5 0 0 0 5 0 0 0 5 Now, A 2 - 4 A - 5 I 3 = 9 8 8 8 9 8 8 8 9 - 4 8 8 8 4 8 8 8 4 - 5 0 0 0 5 0 0 0 5 = 9 - 4 - 5 8 - 8 - 0 8 - 8 - 0 8 - 8 - 0 9 - 4 - 5 8 - 8 - 0 8 - 8 - 0 8 - 8 - 0 9 - 4 - 5 = 0 0 0 0 0 0 0 0 0 = O Hence, proved.

If A = 1 4 2 3 , show that A 2 − 4 A − 5 I 2 = O .

Given, A = 1 4 2 3 Now, A 2 = 1 4 2 3 1 4 2 3 = 1 + 8 4 + 12 2 + 6 8 + 9 = 9 16 8 17 4 A = 4 1 4 2 3 = 4 16 8 12 5 I 2 = 5 1 0 0 1 = 5 0 0 5 Now, A 2 - 4 A - 5 I 2 = 9 16 8 17 - 4 16 8 12 - 5 0 0 5 = 9 - 4 - 5 16 - 16 - 0 8 - 8 - 0 17 - 12 - 5 = 0 0 0 0 = O Hence, proved.

A matrix A is called idempotent, if A 2 = A . prove that A = 2 − 2 1 − 1 is idempotent.

Given, A = 2 - 2 1 - 1 A matrix A is called idempotent, if A 2 = A A 2 = 2 - 2 1 - 1 2 - 2 1 - 1 = 2 × 2 + - 2 1 2 - 2 + - 2 - 1 1 × 2 + - 1 1 1 - 2 + - 1 - 1 = 4 - 2 - 4 + 2 2 - 1 - 2 + 1 = 2 - 2 1 - 1 = A ∴ A 2 = A Hence, proved.

Show that 2 − 2 − 4 − 1 3 4 1 − 2 − 3 is idempotent.

Let A = 2 − 2 − 4 − 1 3 4 1 − 2 − 3 A matrix A is said to idempotent if A 2 = A Now, A 2 = A A = 2 - 2 - 4 - 1 3 4 1 - 2 - 3 2 − 2 − 4 − 1 3 4 1 − 2 − 3 = 4 + 2 - 4 - 4 - 6 + 8 - 8 - 8 + 12 - 2 - 3 + 4 2 + 9 - 8 4 + 12 - 12 2 + 2 - 3 - 2 - 6 + 6 - 4 - 8 + 9 = 2 − 2 − 4 − 1 3 4 1 − 2 − 3 = A Hence, A 2 = A .