Question

For the matrix show that A 3 − 6 A 2 + 5 A + 11 I = O. Hence, find A −1.

Answer 1

Given matrix is,

A=[ 1 1 1 1 2 −3 2 −1 3 ]

The value of A 2 is,

A 2 =[ 1 1 1 1 2 −3 2 −1 3 ][ 1 1 1 1 2 −3 2 −1 3 ] =[ 1+1+2 1+2−1 1−3+3 1+2−6 1+4+3 1−6−9 2−1+6 2−2−3 2+3+9 ] =[ 4 2 1 −3 8 −14 7 −3 14 ]

The value of A 3 is,

A 3 = A 2 A =[ 4 2 1 −3 8 −14 7 −3 14 ][ 1 1 1 1 2 −3 2 −1 3 ] =[ 4+2+2 4+4−1 4−6+3 −3+8−8 −3+16−4 −3−24−42 7−3+28 7−6−14 7+9+42 ] =[ 8 7 1 −23 27 −69 32 −13 58 ]

Substitute the values of A 3 , A 2 and A in A 3 −6 A 2 +5A+11I.

A 3 −6 A 2 +5A+11I=[ 8 7 1 −23 27 −69 32 −13 58 ]−6[ 4 2 1 −3 8 −14 7 −3 14 ]+5[ 1 1 1 1 2 −3 2 −1 3 ]+11[ 1 0 0 0 1 1 0 0 1 ] =[ 8−24+5+11 7−12+5+0 1−6+5+0 −23+18+5+0 27−48+10+11 −69+84−15+0 32−42+10+0 −13+18−5+0 58+84+15+11 ] =[ 0 0 0 0 0 0 0 0 0 ]

Hence, it is verified that A 3 −6 A 2 +5A+11I=0.

Multiply both sides of A 3 −6 A 2 +5A+11I=0 by A −1 we get,

A −1 A 3 − A −1 ( 6 A 2 )+ A −1 ( 5A )+ A −1 ( 11I )=0 A −1 A( A 2 )−6 A −1 A( A )+5 A −1 A+ 11A −1 I= A −1 0 I( A 2 )−6I( A )+5I+ 11A −1 =0 A 2 −6A+5I+ 11A −1 =0

Further simplify the above equation,

A −1 = 6A− A 2 −5I 11 = 1 11 ( 6[ 1 1 1 1 2 −3 2 −1 3 ]−[ 4 2 1 −3 8 −14 7 −3 14 ]−5[ 1 0 0 0 1 0 0 0 1 ] ) = 1 11 ( [ 6−4−5 6−2−0 6−1−0 6+3−5 12−8−5 −18+14−0 12−7−0 −6+3−0 18−14−5 ] ) = 1 11 [ −3 4 5 9 −1 −4 5 −3 −1 ]