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Byju's Answer
Standard XII
Mathematics
Inverse of a Matrix
If A = begin ...
Question

If A=⎡⎢⎣2−11−12−11−12⎤⎥⎦ verify that A3−6A2+9A−4I=0 and hence, find A−1

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Solution

Given, A=⎡⎢⎣2−11−12−11−12⎤⎥⎦
∴ A2=AA−⎡⎢⎣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⎤⎥⎦=⎡⎢⎣6−55−56−55−56⎤⎥⎦

and 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⎤⎥⎦=⎡⎢⎣22−2121−2122−2121−2122⎤⎥⎦
∴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+30−9−021−30+9−0−21+30−9−022−36+18−4−21+30−9−021−30+9−0−21+30−9−022−36+18−4⎤⎥⎦=⎡⎢⎣000000000⎤⎥⎦

A3−6A2+9A−4I=0⇒(AAA)A−1−6(AA)A−1+9AA−1−4IA−1=0(Pre−multiplying by A−1as|A|≠0)⎡⎢⎣∵|A|=∣∣ ∣∣2−11−12−11−12∣∣ ∣∣=2(4−1)+1(−2+1)+1(1−2)=6−1−1=4≠0⎤⎥⎦
⇒AA(AA−1)−6A(AA−1)+9(AA−1)−4(IA−1)=0
⇒AAI−6AI=9I−4A−1=0 (using AA−1−I and IA−1=A−1)
⇒A2−6A+9I=4A−1 (using A2I=A2 and AI=A)
⇒A−1=14(A2−6A+9I)

=14⎧⎪⎨⎪⎩⎡⎢⎣6−55−56−55−56⎤⎥⎦⎫⎪⎬⎪⎭−6⎡⎢⎣2−11−12−11−12⎤⎥⎦+9⎡⎢⎣100010001⎤⎥⎦

=14⎧⎪⎨⎪⎩⎡⎢⎣6−55−56−55−56⎤⎥⎦⎫⎪⎬⎪⎭−⎡⎢⎣12−66−612−66−612⎤⎥⎦+⎡⎢⎣900090009⎤⎥⎦

=14⎡⎢⎣6−12+9−5+6+05−6+0−5+6+06−12+9−5+6+05−6+0−5+6+06−12+9⎤⎥⎦=14⎡⎢⎣31−1131−113⎤⎥⎦


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