To Prove : A3−6A2+9A−4I=O
Consider, A2=⎡⎢⎣2−11−12−11−12⎤⎥⎦⎡⎢⎣2−11−12−11−12⎤⎥⎦
=⎡⎢⎣4+1+1−2−2−12+1+2−2−2−11+4+1−1−2−22+1+2−1−2−21+1+4⎤⎥⎦
⇒A2=⎡⎢⎣6−55−56−55−56⎤⎥⎦
Now, A3=A2A
=⎡⎢⎣6−55−56−55−56⎤⎥⎦⎡⎢⎣2−11−12−11−12⎤⎥⎦
=⎡⎢⎣12+5+5−6−10−56+5+10−10−6−55+12+5−5−6−1010+5+6−5−10−65+5+12⎤⎥⎦
A3=⎡⎢⎣22−2121−2122−2121−2122⎤⎥⎦
Consider, A3−6A2+9A−4I
=⎡⎢⎣22−2121−2122−2121−2122⎤⎥⎦−6⎡⎢⎣6−55−56−55−56⎤⎥⎦+9⎡⎢⎣2−11−12−11−12⎤⎥⎦−4⎡⎢⎣100010001⎤⎥⎦
=⎡⎢⎣22−2121−2122−2121−2122⎤⎥⎦−⎡⎢⎣36−3030−3036−3030−3036⎤⎥⎦+⎡⎢⎣18−99−918−99−918⎤⎥⎦−⎡⎢⎣400040004⎤⎥⎦
=⎡⎢⎣22−36+18−4−21+30−921−30+9−21+30−922−36+18−4−21+30−921−30+9−21+30−922−36+18−4⎤⎥⎦
=⎡⎢⎣000000000⎤⎥⎦
⇒A3−6A2+9A−4I=O .....(1)
A−1 :
Multiplying eqn (1) by A−1, we get
A3A−1−6A2A−1+9AA−1−4IA−1=OA−1
⇒A2−6A+9I−4A−1=O
⇒4A−1=A2−6A+9I
Substituting the values in above eqn
⇒4A−1=⎡⎢⎣6−55−56−55−56⎤⎥⎦−6⎡⎢⎣2−11−12−11−12⎤⎥⎦+9⎡⎢⎣100010001⎤⎥⎦
⇒4A−1=⎡⎢⎣6−55−56−55−56⎤⎥⎦−⎡⎢⎣12−66−612−66−612⎤⎥⎦+⎡⎢⎣900090009⎤⎥⎦
⇒4A−1=⎡⎢⎣6−12+9−5+65−6−5+66−12+9−5+65−6−5+66−12+9⎤⎥⎦
⇒A−1=14⎡⎢⎣31−1131−113⎤⎥⎦