Let A be a 3 × 3 real matrix such that A 1 0 1 = 2 1 0 1 , A − 1 0 1 = 4 − 1 0 1 , A 0 1 0 = 2 0 1 0 . Then, the system A − 3 I x y z = 1 2 3 has

Step-by-step solution

Let, A = x 1 y 1 z 1 x 2 y 2 z 2 x 3 y 3 z 3 It is given that, A 1 0 1 = 2 0 2 ⇒ x 1 + z 1 x 2 + z 2 x 3 + z 3 = 2 0 2 ⇒ x 1 + z 1 = 2 , x 2 + z 2 = 0 , x 3 + z 3 = 0 . . . i Also, A − 1 0 1 = − 4 0 4 ⇒ − x 1 + z 1 − x 2 + z 2 − x 3 + z 3 = - 4 0 4 ⇒ − x 1 + z 1 = − 4 , − x 2 + z 2 = 0 , − x 3 + z 3 = 4 . . . i i Now, A 0 1 0 = 0 2 0 ⇒ y 1 y 2 y 3 = 0 2 0 ⇒ y 1 = 0 , y 2 = 2 , y 3 = 0 . . . i i i Using i and i i , ⇒ x 1 = 3 , x 2 = 0 , x 3 = − 1 ⇒ z 1 = − 1 , z 2 = 0 , z 3 = 3 ∴ A = 3 0 − 1 0 2 0 − 1 0 3 Now, A − 3