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Question
If A=[4521] ,show that A−1=16(A−5I)
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Solution
|A|=∣∣∣4521∣∣∣=4−10=−6≠0
∴A−1 exists.
Consider AA−1
∴[4521]A−1=[1001]
By (14)R1 weget
⎡⎣15421⎤⎦A−1=⎡⎣14001⎤⎦
By R2−2R1 we get
⎡⎢
⎢⎣1540−32⎤⎥
⎥⎦A−1=⎡⎢
⎢⎣140−121⎤⎥
⎥⎦
By (−23)R2 we get
⎡⎣15401⎤⎦A−1=⎡⎢
⎢⎣14013−23⎤⎥
⎥⎦
By R1−54R2 we get
[1001]A−1=⎡⎢
⎢⎣−165613−23⎤⎥
⎥⎦
∴A−1=16[−152−4] ................(1)
16(A−5I)=16{[4521]−5[1001]}
=16{[4521]−[5005]}
=16[−152−4] ..........(2)
From (1) and (2) , A−1=16(A−5I)
∴A−1 exists.
Consider AA−1
∴[4521]A−1=[1001]
By (14)R1 weget
⎡⎣15421⎤⎦A−1=⎡⎣14001⎤⎦
By R2−2R1 we get
⎡⎢ ⎢⎣1540−32⎤⎥ ⎥⎦A−1=⎡⎢ ⎢⎣140−121⎤⎥ ⎥⎦
By (−23)R2 we get
⎡⎣15401⎤⎦A−1=⎡⎢ ⎢⎣14013−23⎤⎥ ⎥⎦
By R1−54R2 we get
[1001]A−1=⎡⎢ ⎢⎣−165613−23⎤⎥ ⎥⎦
∴A−1=16[−152−4] ................(1)
16(A−5I)=16{[4521]−5[1001]}
=16{[4521]−[5005]}
=16[−152−4] ..........(2)
From (1) and (2) , A−1=16(A−5I)
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