Question

If , show that . Hence find .

Answer 1

Given matrix is,

A=[ 3 1 −1 2 ]

The value of A 2 is,

A 2 =[ 3 1 −1 2 ][ 3 1 −1 2 ] =[ 9−1 3+2 −3−2 −1+4 ] =[ 8 5 −5 3 ]

Now,

A 2 −5A+7I=[ 8 5 −5 3 ]−5[ 3 1 −1 2 ]+7[ 1 0 0 1 ] =[ 8 5 −5 3 ]−[ 15 5 −5 10 ]+[ 7 0 0 7 ] =[ 8−15+7 5−5+0 −5+5+0 3−10+7 ] =[ 0 0 0 0 ]

Hence, it is verified that A 2 −5A+7I=0.

Since A 2 −5A+7I=0

Multiply both sides by A −1 ,

A −1 ( AA )− A −1 ( 5A )+ A −1 ( 7I )= A −1 0 ( A −1 A )A−5( A −1 A )+7 A −1 =0 IA−5I+7 A −1 =0 A−5I+7 A −1 =0

Further simplify the above equation,

A -1 = 5I−A 7

Substitute the values of A and I in the above equation,

A −1 = 1 7 ( 5[ 1 0 0 1 ]−[ 3 1 −1 2 ] ) = 1 7 ( [ 5 0 0 5 ]−[ 3 1 −1 2 ] ) = 1 7 [ 2 −1 1 3 ]