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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\

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 $R_3 	o \frac{1}{2}R_3$: $\left[\begin{array}{ccc}1 & 0 & -1 \ 0 & 1 & 2 \ 0 & 0 & 1\end{array}ight] = \left[\begin{array}{ccc}-2 & 1 & 0 \ 1 & 0 & 0 \ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2}\end{array}ight] A$ Applying $R_1 	o R_1 + R_3$ and $R_2 	o R_2 - 2R_3$: $\left[\begin{array}{lll}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}ight] = \left[\begin{array}{ccc}\frac{1}{2} & \frac{-1}{2} & \frac{1}{2} \ -4 & 3 & -1 \ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2}\end{array}ight] A$ Thus,$A^{-1} = \left[\begin{array}{ccc}\frac{1}{2} & \frac{-1}{2} & \frac{1}{2} \ -4 & 3 & -1 \ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2}\end{array}ight]$.

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

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