If A=left[begin{array}{ccc}1 & -1 & 1 2 & 1 | vedclass

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(C) Given,$A=\left[\begin{array}{ccc}1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 1 & 1\end{array}ight]$ and $10 B=\left[\begin{array}{ccc}4 & 2 & 2 \ -5 & 0 &

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$5$

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(C) Given,$A=\left[\begin{array}{ccc}1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 1 & 1\end{array}ight]$ and $10 B=\left[\begin{array}{ccc}4 & 2 & 2 \ -5 & 0 & \alpha \ 1 & -2 & 3\end{array}ight]$.Since $B$ is the inverse of $A$,we have $B = A^{-1}$.Thus,$10 A^{-1} = \left[\begin{array}{ccc}4 & 2 & 2 \ -5 & 0 & \alpha \ 1 & -2 & 3\end{array}ight]$.Multiplying both sides by $A$ on the right,we get $10 A^{-1} A = \left[\begin{array}{ccc}4 & 2 & 2 \ -5 & 0 & \alpha \ 1 & -2 & 3\end{array}ight] A$.Since $A^{-1} A = I$,we have $10 I = \left[\begin{array}{ccc}4 & 2 & 2 \ -5 & 0 & \alpha \ 1 & -2 & 3\end{array}ight] \left[\begin{array}{ccc}1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 1 & 1\end{array}ight]$.Calculating the product on the right side:$\left[\begin{array}{ccc}4(1)+2(2)+2(1) & 4(-1)+2(1)+2(1) & 4(1)+2(-3)+2(1) \ -5(1)+0(2)+\alpha(1) & -5(-1)+0(1)+\alpha(1) & -5(1)+0(-3)+\alpha(1) \ 1(1)-2(2)+3(1) & 1(-1)-2(1)+3(1) & 1(1)-2(-3)+3(1)\end{array}ight] = \left[\begin{array}{ccc}10 & 0 & 0 \ -5+\alpha & \alpha+5 & -5+\alpha \ 0 & 0 & 10\end{array}ight]$.Equating this to $10 I = \left[\begin{array}{ccc}10 & 0 & 0 \ 0 & 10 & 0 \ 0 & 0 & 10\end{array}ight]$,we compare the elements.From the element at $(2, 1)$,we get $-5 + \alpha = 0$,which implies $\alpha = 5$.

If $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]$,$10 B=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right]$ and $B$ is the inverse of $A$,then the value of $\alpha$ is

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