Question

Using elementary transformations, find the inverse of matrix $\begin{bmatrix}-1&1&2\\ 1&2&3\\ 3&1&1\end{bmatrix}.$

• A. $\begin{bmatrix}1&-1&1\\ -8&7&-5\\ 5&-4&3\end{bmatrix}$
• B. $\begin{bmatrix}2&-1&1\\ -6&7&-5\\ 5&-4&3\end{bmatrix}$
• C. $\begin{bmatrix}2&-1&1\\ -6&4&-5\\ 5&-4&3\end{bmatrix}$
• D. $\begin{bmatrix}1&-1&1\\ -6&4&-5\\ 5&-4&3\end{bmatrix}$

Answer 1

The Correct Option is A

Given, matrix $A =\begin{bmatrix}-1&1&2\\ 1&2&3\\ 3&1&1\end{bmatrix}$, then $A = IA$ $\Rightarrow\begin{bmatrix}-1&1&2\\ 1&2&3\\ 3&1&1\end{bmatrix}=\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}A$ Applying $R_{2}\rightarrow R_{2}+R_{1}, R3\rightarrow R_{3} + 3R_{1}$, we get $\begin{bmatrix}-1&1&2\\ 0&3&5\\ 0&4&7\end{bmatrix}=\begin{bmatrix}1&0&0\\ 0&1&0\\ 3&0&1\end{bmatrix}A$ Applying $R_{1}\rightarrow \left(-1\right)R_{1}$, we get $\begin{bmatrix}1&-1&-2\\ 0&3&5\\ 0&4&7\end{bmatrix}=\begin{bmatrix}-1&0&0\\ 1&1&0\\ 3&0&1\end{bmatrix}A $ Applying $R_{2}\rightarrow R_{2}-R_{3}$, we get $\begin{bmatrix}1&-1&-2\\ 0&-1&-2\\ 0&4&7\end{bmatrix}=\begin{bmatrix}-1&0&0\\ -2&1&-1\\ 3&0&1\end{bmatrix}A$ Applying $R_{1}\rightarrow R_{1} -R_{2}, and R_{3}\rightarrow R_{3} + 4R_{2 }$,we get $\begin{bmatrix}1&0&0\\ 0&-1&-2\\ 0&0&-1\end{bmatrix}=\begin{bmatrix}1&-1&1\\ -2&1&-1\\ -5&4&-3\end{bmatrix}A$ Applying $R_{2}\rightarrow\left(-1\right)R_{2}$, we get $\begin{bmatrix}1&0&0\\ 0&1&2\\ 0&0&-1\end{bmatrix}=\begin{bmatrix}1&-1&1\\ 2&-1&1\\ 5&4&-3\end{bmatrix}A $ Applying $R_{2}\rightarrow R_{2} +2R_{3}$, we get $\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&-1\end{bmatrix}=\begin{bmatrix}1&-1&1\\ -8&7&-5\\ 5&-4&3\end{bmatrix}A$ $A^{-1}=\begin{bmatrix}1&-1&1\\ -8&7&-5\\ 5&-4&3\end{bmatrix}$