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(A) To find the inverse of matrix $A$ using elementary row operations, we write $A = IA$: $\left[\begin{array}{lll}0 & 1 & 2 \ 1 & 2 & 3 \ 3 & 1 & 1

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$A^{-1} = \left[\begin{array}{ccc}1/2 & -1/2 & 1/2 \ -4 & 3 & -1 \ 5/2 & -3/2 & 1/2\end{array}ight]$

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(A) To find the inverse of matrix $A$ using elementary row operations, we write $A = IA$: $\left[\begin{array}{lll}0 & 1 & 2 \ 1 & 2 & 3 \ 3 & 1 & 1\end{array}ight] = \left[\begin{array}{lll}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}ight] A$Applying $R_1 \leftrightarrow R_2$: $\left[\begin{array}{lll}1 & 2 & 3 \ 0 & 1 & 2 \ 3 & 1 & 1\end{array}ight] = \left[\begin{array}{lll}0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1\end{array}ight] A$Applying $R_3 	o R_3 - 3R_1$: $\left[\begin{array}{rrr}1 & 2 & 3 \ 0 & 1 & 2 \ 0 & -5 & -8\end{array}ight] = \left[\begin{array}{rrr}0 & 1 & 0 \ 1 & 0 & 0 \ 0 & -3 & 1\end{array}ight] A$Applying $R_1 	o R_1 - 2R_2$: $\left[\begin{array}{rrr}1 & 0 & -1 \ 0 & 1 & 2 \ 0 & -5 & -8\end{array}ight] = \left[\begin{array}{rrr}-2 & 1 & 0 \ 1 & 0 & 0 \ 0 & -3 & 1\end{array}ight] A$Applying $R_3 	o R_3 + 5R_2$: $\left[\begin{array}{ccc}1 & 0 & -1 \ 0 & 1 & 2 \ 0 & 0 & 2\end{array}ight] = \left[\begin{array}{ccc}-2 & 1 & 0 \ 1 & 0 & 0 \ 5 & -3 & 1\end{array}ight] A$Applying $R_3 	o \frac{1}{2}R_3$: $\left[\begin{array}{ccc}1 & 0 & -1 \ 0 & 1 & 2 \ 0 & 0 & 1\end{array}ight] = \left[\begin{array}{ccc}-2 & 1 & 0 \ 1 & 0 & 0 \ 5/2 & -3/2 & 1/2\end{array}ight] A$Applying $R_1 	o R_1 + R_3$ and $R_2 	o R_2 - 2R_3$: $\left[\begin{array}{lll}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}ight] = \left[\begin{array}{ccc}1/2 & -1/2 & 1/2 \ -4 & 3 & -1 \ 5/2 & -3/2 & 1/2\end{array}ight] A$Thus, $A^{-1} = \left[\begin{array}{ccc}1/2 & -1/2 & 1/2 \ -4 & 3 & -1 \ 5/2 & -3/2 & 1/2\end{array}ight]$.

Obtain the inverse of the following matrix using elementary row operations: $A = \left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$

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