InterviewSolution
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InterviewSolution
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If `A=[(k,l),(m,n)]` and `kn!=lm,` show that `A^(2)-(k+n)A+(kn-lm)l=O.` Hence, find `A^(-1)` |
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Answer» We, have, `A[(k,l),(m,n)]`, then `|A|=|(k,l),(m,n)|` `=kn-ml!=0` `therefore" " A^(-1) exists.` Now, `A^(2)=A.A=[(k,l),(m,n)][(k,l)(m,n)=[(k^(2)+lm,kl+ln),(mk+nm,ml+n^(2))]` `therefore " " A^(2)-(k+n)A+(kn-lm)I` `= [(k^(2)+lm,kl+ln),(mk+nm,ml+n^(2))]-[(k-n)[(k,l),(m,n)]+(kn-lm)[(1,0),(0,1)]` `= [(k^(2)+lm,kl+ln),(mk+nm,ml+n^(2))][(k^(2)+nk,kl+nl),(km+nm,kn+n^(2))] +[(kn-lm,0),(0,kn-lm)]` `[(k^(2)+lm-K^(2)-nk+kn-lm,kl+ln-kl-ln),(mk+nm-km-nm,ml+n^(2)-kn-n^(2)+kn-lm)]` `[(0,0),(0,0)]=O` `AsA^(2)-(k+n)A+(kn-lm)I=O` `rArr" " (kn-lm)I=(k+n)A-A^(2)` `rArr" " (kn-lm)IA^(-1)=(k+n)A-A^(2))A^(-1)` `rArr" " (kn-lm)A^(-1)=(k+n)A A^(-1)-A(A A^(-1))` `=(k+n)I-AI` `=(k+n)I-A` `=(k+n)[(1,0),(0,1)]-[(k,l),(m,n)]` `=[(k+n,0),(0,k+n)]-[(k,l),(m,n)]` ` rArr " " (kn-lm)A^(-1)=[(n,-1),(-m,k)]` Hence `A^(-1)=(1)/((kn-lm))[(n,-1),(-m,k)]` |
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