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Question
If A=⎡⎢⎣1233−21421⎤⎥⎦, then show that A3−23A−40I=O
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Solution
Calculating A3
Given: A=⎡⎢⎣1233−21421⎤⎥⎦
A2=A⋅A=⎡⎢⎣1233−21421⎤⎥⎦⎡⎢⎣1233−21421⎤⎥⎦
⇒A2=⎡⎢⎣1+6+122−4+63+2+33−6+46+4+29−2+14+6+48−4+212+2+1⎤⎥⎦
⇒A2=⎡⎢⎣1948112814615⎤⎥⎦
A3=A2A=⎡⎢⎣1948112814616⎤⎥⎦⎡⎢⎣1233−21421⎤⎥⎦
A3=⎡⎢⎣19+12+3238−8+1657+4+81+36+322−24+163+12+814+18+6028−12+3042+6+15⎤⎥⎦
A3=⎡⎢⎣63466969−623924663⎤⎥⎦
Calculating A3−23A−40I:
A3−23A−40I
=⎡⎢⎣63466969−623924663⎤⎥⎦−23⎡⎢⎣1233−21421⎤⎥⎦−40⎡⎢⎣100010001⎤⎥⎦
⎡⎢⎣63466969−623924663⎤⎥⎦−⎡⎢⎣23×123×223×323×323×(−2)23×123×423×(2)23×1⎤⎥⎦−⎡⎢⎣400004000040⎤⎥⎦
=⎡⎢⎣63466969−623924663⎤⎥⎦−⎡⎢⎣23466969−4623924623⎤⎥⎦−⎡⎢⎣400004000040⎤⎥⎦
=⎡⎢⎣63−23−4046−46−069−69−069−69−0−6+46−4023−23−092−92−046−46−063−23−40⎤⎥⎦=⎡⎢⎣000000000⎤⎥⎦=0
∴A3−23A−40I=O
Hence proved.
Given: A=⎡⎢⎣1233−21421⎤⎥⎦
A2=A⋅A=⎡⎢⎣1233−21421⎤⎥⎦⎡⎢⎣1233−21421⎤⎥⎦
⇒A2=⎡⎢⎣1+6+122−4+63+2+33−6+46+4+29−2+14+6+48−4+212+2+1⎤⎥⎦
⇒A2=⎡⎢⎣1948112814615⎤⎥⎦
A3=A2A=⎡⎢⎣1948112814616⎤⎥⎦⎡⎢⎣1233−21421⎤⎥⎦
A3=⎡⎢⎣19+12+3238−8+1657+4+81+36+322−24+163+12+814+18+6028−12+3042+6+15⎤⎥⎦
A3=⎡⎢⎣63466969−623924663⎤⎥⎦
Calculating A3−23A−40I:
A3−23A−40I
=⎡⎢⎣63466969−623924663⎤⎥⎦−23⎡⎢⎣1233−21421⎤⎥⎦−40⎡⎢⎣100010001⎤⎥⎦
⎡⎢⎣63466969−623924663⎤⎥⎦−⎡⎢⎣23×123×223×323×323×(−2)23×123×423×(2)23×1⎤⎥⎦−⎡⎢⎣400004000040⎤⎥⎦
=⎡⎢⎣63466969−623924663⎤⎥⎦−⎡⎢⎣23466969−4623924623⎤⎥⎦−⎡⎢⎣400004000040⎤⎥⎦
=⎡⎢⎣63−23−4046−46−069−69−069−69−0−6+46−4023−23−092−92−046−46−063−23−40⎤⎥⎦=⎡⎢⎣000000000⎤⎥⎦=0
∴A3−23A−40I=O
Hence proved.
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