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Question
If A=⎡⎢⎣2−11−12−11−12⎤⎥⎦ find A−1 by the adjoint method.
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Solution
A=⎡⎢⎣2−11−12−11−12⎤⎥⎦A−1=adj(A)|A|Cij=(−1)i+jMij∴C11=(−1)1+1[2−1−12]=3C12=(−1)1+2[−1−112]=(−1)(−1)=1C13=(−1)1+3[−121−1]=−1C21=(−1)2+1[−1112]=(−1)(−3)=3C22=(−1)2+2[2112]=3C23=(−1)2+3[2−11−1]=(−1)(−1)=1C31=(−1)3+1[−112−1]=−1C32=(−1)3+2[21−1−1]=(−1)(−1)=1C33=(−1)3+3[2−1−12]=3C=⎡⎢⎣C11C12C13C21C22C23C31C32C33⎤⎥⎦=⎡⎢⎣31−1331−113⎤⎥⎦=adj(A)=CT=⎡⎢⎣31−1331−113⎤⎥⎦|A|=4A−1=adj(A)|A|=14⎡⎢⎣31−1331−113⎤⎥⎦
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