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Question
Find A2−5A+6I, if A=⎡⎢⎣22101−1130⎤⎥⎦
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Solution
We have A2=A×A
=⎡⎢⎣22101−1130⎤⎥⎦⎡⎢⎣22101−1130⎤⎥⎦
=⎡⎢⎣2(2)+0(2)+1(1)2(2)+1(2)+3(1)1(2)+(−1)(1)+0(1)2(0)+0(1)+1(−1)2(0)+1(1)+3(−1)1(0)+(−1)(1)+0(−1)2(1)+0(3)+1(0)2(1)+1(3)+3(0)1(1)+(−1)(3)+0(0)⎤⎥⎦
=⎡⎢⎣4+0+14+2+32−2+00+0+10+1−30−1+02+0+02+3+01−3+0⎤⎥⎦
=⎡⎢⎣590−1−2−125−2⎤⎥⎦
∴A2−5A+6I
=⎡⎢⎣590−1−2−125−2⎤⎥⎦−5⎡⎢⎣22101−1130⎤⎥⎦+6⎡⎢⎣100010001⎤⎥⎦
=⎡⎢⎣590−1−2−125−2⎤⎥⎦−⎡⎢⎣1010505−55150⎤⎥⎦+⎡⎢⎣600060006⎤⎥⎦
=⎡⎢⎣5−109−100−5−1−0−2−5−1+52−55−15−2−0⎤⎥⎦+⎡⎢⎣600060006⎤⎥⎦
=⎡⎢⎣−5−1−5−1−74−3−10−2⎤⎥⎦+⎡⎢⎣600060006⎤⎥⎦
=⎡⎢⎣−5+6−1+0−5+0−1+0−7+64+0−3+0−10+0−2+6⎤⎥⎦
=⎡⎢⎣1−1−5−1−14−3−104⎤⎥⎦
=⎡⎢⎣22101−1130⎤⎥⎦⎡⎢⎣22101−1130⎤⎥⎦
=⎡⎢⎣2(2)+0(2)+1(1)2(2)+1(2)+3(1)1(2)+(−1)(1)+0(1)2(0)+0(1)+1(−1)2(0)+1(1)+3(−1)1(0)+(−1)(1)+0(−1)2(1)+0(3)+1(0)2(1)+1(3)+3(0)1(1)+(−1)(3)+0(0)⎤⎥⎦
=⎡⎢⎣4+0+14+2+32−2+00+0+10+1−30−1+02+0+02+3+01−3+0⎤⎥⎦
=⎡⎢⎣590−1−2−125−2⎤⎥⎦
∴A2−5A+6I
=⎡⎢⎣590−1−2−125−2⎤⎥⎦−5⎡⎢⎣22101−1130⎤⎥⎦+6⎡⎢⎣100010001⎤⎥⎦
=⎡⎢⎣590−1−2−125−2⎤⎥⎦−⎡⎢⎣1010505−55150⎤⎥⎦+⎡⎢⎣600060006⎤⎥⎦
=⎡⎢⎣5−109−100−5−1−0−2−5−1+52−55−15−2−0⎤⎥⎦+⎡⎢⎣600060006⎤⎥⎦
=⎡⎢⎣−5−1−5−1−74−3−10−2⎤⎥⎦+⎡⎢⎣600060006⎤⎥⎦
=⎡⎢⎣−5+6−1+0−5+0−1+0−7+64+0−3+0−10+0−2+6⎤⎥⎦
=⎡⎢⎣1−1−5−1−14−3−104⎤⎥⎦
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