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Question
Given A=⎡⎢⎣111241231⎤⎥⎦,B=[2334]. Find P such that BPA=[101010].
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Solution
A=⎡⎢⎣111241231⎤⎥⎦B=[2334]BPA=[101010]PA=B−1×[101010]B−1=1|B|[4−3−32]T=−1[4−3−32]PA=[−433−2]×[101010]PA=[−43−43−23]P=[−43−43−23]×A−1A−1=1|A|⎡⎢⎣10−22−1−2−312⎤⎥⎦T|A|=1(4−3)−1(2−2)+1(6−8)|A|=1−2=−1A−1=⎡⎢⎣−1−2301−121−2⎤⎥⎦P=[−43−43−23]×⎡⎢⎣−1−2301−121−2⎤⎥⎦P=[4−88+3−4−12−3+8−3+6−6−2+39+2−6]P=[−47−73−55]
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