question_answer

If \[A=\left[ \begin{matrix}    6 & -\,3  \\    -\,2 & 1  \\ \end{matrix} \right],\] show that \[{{A}^{-1}}\] does not exist.

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.