Answer:
Given, \[A=\left[ \begin{matrix} 2 & 3 \\ -1 & 2 \\ \end{matrix} \right]\] Now, \[{{A}^{2}}=\left[ \begin{matrix} 2 & 3 \\ -1 & 2 \\ \end{matrix} \right]\,\,\,\,\left[ \begin{matrix} 2 & 3 \\ -1 & 2 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 4-3 & 6+6 \\ -\,2-2 & -\,3+4 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 12 \\ -\,4 & 1 \\ \end{matrix} \right]\] \[\therefore \] \[{{A}^{2}}-4A+l\,\left[ \begin{matrix} 1 & 12 \\ -\,4 & 1 \\ \end{matrix} \right]-4\left[ \begin{matrix} 2 & 3 \\ -1 & 2 \\ \end{matrix} \right]+\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 1 & 12 \\ -\,4 & 1 \\ \end{matrix} \right]-\left[ \begin{matrix} 8 & 12 \\ -\,4 & 8 \\ \end{matrix} \right]+\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 1-8+1 & 12-12+0 \\ -\,4+4+0 & 1-8+1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} -\,6 & 0 \\ 0 & -\,6 \\ \end{matrix} \right]=-\,6l\]
You need to login to perform this action.
You will be redirected in
3 sec
