Question

If verify that A 3 − 6 A 2 + 9 A − 4 I = O and hence find A −1

Answer 1

Given matrix is,

A=[ 2 −1 1 −1 2 −1 1 −1 2 ]

The value of A 2 is,

A 2 =[ 2 −1 1 −1 2 −1 1 −1 2 ][ 2 −1 1 −1 2 −1 1 −1 2 ] =[ 6 −5 5 −5 6 −5 5 −5 6 ]

And, the value of A 3 is,

A 3 = A 2 A =[ 6 −5 5 −5 6 −5 5 −5 6 ][ 2 −1 1 −1 2 −1 1 −1 2 ] =[ 22 −21 21 −21 22 −21 21 −21 22 ]

Now substitute the value of A 2 and A 3 in A 3 −6 A 2 +9A−4I, we get,

A 3 −6 A 2 +9A−4I=[ 22 −21 21 −21 22 −21 21 −21 22 ]−6[ 6 −5 5 −5 6 −5 5 −5 6 ]+9[ 2 −1 1 −1 2 −1 1 −1 2 ]−4[ 1 0 0 0 1 0 0 0 1 ] =[ 22−36+18−4 −21−( −30 )−9+0 21−30+9+0 −21+30−9+0 22−36+18−4 −22+30−8+0 21−30+9 −21+30−9 22−36+18−4 ] =[ 0 0 0 0 0 0 0 0 0 ]

Hence, it is verified that A 3 −6 A 2 +9A−4I=0.

Now, multiply both sides of A 3 −6 A 2 +9A−4I=0 by A −1 we get,

A −1 A 3 − A −1 ( 6 A 2 )+ A −1 ( 9A )− A −1 ( 4I )=0 A −1 A( A 2 )−6 A −1 A( A )+9 A −1 A− 4A −1 I= A −1 0 I( A 2 )−6I( A )+9I− 4A −1 =0 A 2 −6A+9I− 4A −1 =0

Further simplify the above equation,

A −1 = A 2 −6A+9I 4 = 1 4 ( [ 6 −5 5 −5 6 −5 5 −5 6 ]−6[ 2 −1 1 −1 2 −1 1 −1 2 ]+9[ 1 0 0 0 1 0 0 0 1 ] ) = 1 4 [ 6−12+9 −5+6+0 5−6+0 −5+6+0 6−12+9 −5+6+0 5−6+0 −5+6+0 6−12+9 ] = 1 4 [ 3 1 −1 1 3 1 −1 1 3 ]