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(A) Let $A=\left[\begin{array}{ccc}2 & 0 & -1 \ 5 & 1 & 0 \ 0 & 1 & 3\end{array}ight]$.We know that $A=IA$.$	herefore \left[\begin{array}{ccc}2 &

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$A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \ -15 & 6 & -5 \ 5 & -2 & 2\end{array}ight]$

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(A) Let $A=\left[\begin{array}{ccc}2 & 0 & -1 \ 5 & 1 & 0 \ 0 & 1 & 3\end{array}ight]$.We know that $A=IA$.$	herefore \left[\begin{array}{ccc}2 & 0 & -1 \ 5 & 1 & 0 \ 0 & 1 & 3\end{array}ight]=\left[\begin{array}{ccc}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}ight] A$.Applying $R_{1} ightarrow \frac{1}{2} R_{1}$,we have:$\left[\begin{array}{ccc}1 & 0 & -\frac{1}{2} \ 5 & 1 & 0 \ 0 & 1 & 3\end{array}ight]=\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}ight] A$.Applying $R_{2} ightarrow R_{2}-5 R_{1}$,we have:$\left[\begin{array}{ccc}1 & 0 & -\frac{1}{2} \ 0 & 1 & \frac{5}{2} \ 0 & 1 & 3\end{array}ight]=\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \ -\frac{5}{2} & 1 & 0 \ 0 & 0 & 1\end{array}ight] A$.Applying $R_{3} ightarrow R_{3}-R_{2}$,we have:$\left[\begin{array}{ccc}1 & 0 & -\frac{1}{2} \ 0 & 1 & \frac{5}{2} \ 0 & 0 & \frac{1}{2}\end{array}ight]=\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \ -\frac{5}{2} & 1 & 0 \ \frac{5}{2} & -1 & 1\end{array}ight] A$.Applying $R_{3} ightarrow 2 R_{3}$,we have:$\left[\begin{array}{ccc}1 & 0 & -\frac{1}{2} \ 0 & 1 & \frac{5}{2} \ 0 & 0 & 1\end{array}ight]=\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \ -\frac{5}{2} & 1 & 0 \ 5 & -2 & 2\end{array}ight] A$.Applying $R_{1} ightarrow R_{1}+\frac{1}{2} R_{3}$ and $R_{2} ightarrow R_{2}-\frac{5}{2} R_{3}$,we have:$\left[\begin{array}{ccc}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}ight]=\left[\begin{array}{ccc}3 & -1 & 1 \ -15 & 6 & -5 \ 5 & -2 & 2\end{array}ight] A$.$	herefore A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \ -15 & 6 & -5 \ 5 & -2 & 2\end{array}ight]$.

Find the inverse of the matrix,if it exists: $\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]$

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