Question

If  \[A = \begin{bmatrix}4 & 3 \\ 2 & 5\end{bmatrix}\], find x and y such that 

\[A^2 = xA + yI = O\] . Hence, evaluate A⁻¹.

Answer 1

\[A = \begin{bmatrix}4 & 3 \\ 2 & 5\end{bmatrix}\]
\[ \therefore A^2 = \begin{bmatrix}22 & 27 \\ 18 & 31\end{bmatrix}\]
Now, 
\[ A^2 - xA + yI = O\]
\[ \Rightarrow \begin{bmatrix}22 & 27 \\ 18 & 31\end{bmatrix} - \begin{bmatrix}4x & 3x \\ 2x & 5x\end{bmatrix} + \begin{bmatrix}y & 0 \\ 0 & y\end{bmatrix} = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}22 - 4x - y & 27 - 3x \\ 18 - 2x & 31 - 5x - y\end{bmatrix} = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}\]
Thus, we have
\[22 - 4x + y = 0, 27 - 3x = 0, 18 - 2x = 0\text{ and }31 - 5x + y = 0\]
\[ \Rightarrow - 3x = - 27\]
\[ \Rightarrow x = 9\]
\[\text{ On putting }x = 9\text{ in }22 - 4x + y = 0,\text{ we get }\]
\[22 - 36 + y = 0\]
\[ \Rightarrow - 14 = - y\]
\[ \Rightarrow y = 14\]
Now, 
\[ A^2 - 9A + 14I = 0\]
\[ \Rightarrow A^2 - 9A = - 14I\]
\[ \Rightarrow A^{- 1} A^2 - 9A A^{- 1} = - 14I A^{- 1} \left[\text{ Pre - multiplying both sides by }A^{- 1} \right]\]
\[ \Rightarrow A - 9I = - 14 A^{- 1} \]
\[ \Rightarrow A^{- 1} = - \frac{1}{14}\left( A - 9I \right)\]
\[ \Rightarrow A^{- 1} = - \frac{1}{14}\left\{ \begin{bmatrix}4 & 3 \\ 2 & 5\end{bmatrix} - \begin{bmatrix}9 & 0 \\ 0 & 9\end{bmatrix} \right\} = \frac{1}{14}\begin{bmatrix}5 & - 3 \\ - 2 & 4\end{bmatrix}\]