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## Let A=⎡⎢⎣112302103⎤⎥⎦Here, |A|=−1(9−2)=−7 (expanding along the second column)|A|I=−7I .....(1)We know that, adjA=CTSo, we will find out co-factors of each element of A.C11=(−1)1+1∣∣∣0203∣∣∣⇒C11=0C12=(−1)1+2∣∣∣3213∣∣∣⇒C12=−(9−2)=−7C13=(−1)1+3∣∣∣3010∣∣∣⇒C13=0C21=(−1)2+1∣∣∣1203∣∣∣⇒C21=−(3−0)=−3C22=(−1)2+2∣∣∣1213∣∣∣⇒C22=3−2=1C23=(−1)2+3∣∣∣1110∣∣∣⇒C23=−(0−1)=1C31=(−1)3+1∣∣∣1202∣∣∣⇒C31=2−0=2C32=(−1)3+2∣∣∣1232∣∣∣⇒C32=−(2−6)=4C33=(−1)3+3∣∣∣1130∣∣∣⇒C33=0−3=−3So, the cofactor matrix C=⎡⎢⎣0−70−31124−3⎤⎥⎦⇒adjA=CT=⎡⎢⎣0−32−71401−3⎤⎥⎦Consider (adjA)A=⎡⎢⎣0−32−71401−3⎤⎥⎦⎡⎢⎣112302103⎤⎥⎦=⎡⎢⎣0−9+20+0+00−6+6−7+3+4−7+0+0−14+2+120+3−30+0+00+2−9⎤⎥⎦=⎡⎢⎣−7000−7000−7⎤⎥⎦=−7⎡⎢⎣100010001⎤⎥⎦⇒(adjA)A=−7I .....(2)Now, A(adjA)=⎡⎢⎣112302103⎤⎥⎦⎡⎢⎣0−32−71401−3⎤⎥⎦=⎡⎢⎣0−7+0−3+1+22+4−60+0+0−9+0+26+0−60+0+0−3+0+32+0−9⎤⎥⎦=⎡⎢⎣−7000−7000−7⎤⎥⎦=−7⎡⎢⎣100010001⎤⎥⎦⇒A(adjA)=−7I ....(3)From (1), (2) and (3), we getA(adjA)=(adjA)A=|A|I

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