Question

If \[A = \begin{bmatrix}3 & - 5 \\ - 4 & 2\end{bmatrix}\] , find A² − 5A − 14I.

Answer 1

\[Given: A = \begin{bmatrix}3 & - 5 \\ - 4 & 2\end{bmatrix}\]
\[Now, \]
\[ A^2 = AA\]
\[ \Rightarrow A^2 = \begin{bmatrix}3 & - 5 \\ - 4 & 2\end{bmatrix}\begin{bmatrix}3 & - 5 \\ - 4 & 2\end{bmatrix}\]
\[ \Rightarrow A^2 = \begin{bmatrix}9 + 20 & - 15 - 10 \\ - 12 - 8 & 20 + 4\end{bmatrix}\]
\[ \Rightarrow A^2 = \begin{bmatrix}29 & - 25 \\ - 20 & 24\end{bmatrix}\]
\[\]
\[ A^2 - 5A - 14I\]
\[ \Rightarrow A^2 - 5A - 14I = \begin{bmatrix}29 & - 25 \\ - 20 & 24\end{bmatrix} - 5\begin{bmatrix}3 & - 5 \\ - 4 & 2\end{bmatrix} - 14\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\]
\[ \Rightarrow A^2 - 5A - 14I = \begin{bmatrix}29 & - 25 \\ - 20 & 24\end{bmatrix} - \begin{bmatrix}15 & - 25 \\ - 20 & 10\end{bmatrix} - \begin{bmatrix}14 & 0 \\ 0 & 14\end{bmatrix}\]
\[ \Rightarrow A^2 - 5A - 14I = \begin{bmatrix}29 - 15 - 14 & - 25 + 25 + 0 \\ - 20 + 20 + 0 & 24 - 10 - 14\end{bmatrix}\]
\[ \Rightarrow A^2 - 5A - 14I = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}\]