Answer:
We have \[\left[ \begin{matrix} 6 & -\,3 \\ -\,2 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\cdot A\] \[\Rightarrow \] \[\left[ \begin{matrix} 1 & -\,\frac{1}{2} \\ -\,2 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} \frac{1}{6} & 0 \\ 0 & 1 \\ \end{matrix} \right]\cdot A\] \[\left[ {{R}_{1}}\to \frac{1}{6}{{R}_{1}} \right]\] \[\Rightarrow \] \[\left[ \begin{matrix} 1 & -\,\frac{1}{2} \\ 0 & 0 \\ \end{matrix} \right]=\left[ \begin{matrix} \frac{1}{6} & 0 \\ \frac{1}{3} & 1 \\ \end{matrix} \right]\cdot A\] \[[{{R}_{2}}\to {{R}_{2}}+2{{R}_{1}}]\] Thus, we have all zeroes in second row of the left-hand side matrix. Hence, \[{{A}^{-1}}\] does not exist.
You need to login to perform this action.
You will be redirected in
3 sec