Question

If \[A = \begin{bmatrix}3 & - 2 \\ 4 & - 2\end{bmatrix}\], find the value of \[\lambda\]  so that \[A^2 = \lambda A - 2I\]. Hence, find A⁻¹.

Answer 1

\[A = \begin{bmatrix} 3 & - 2 \\4 & - 2 \end{bmatrix}\]
\[ \therefore A^2 = \begin{bmatrix} 1 & - 2\\4 & - 4 \end{bmatrix}\]
Given: 
\[ A^2 = \lambda A - 2I . . . \left( 1 \right)\]
\[ \Rightarrow \begin{bmatrix} 1 & - 2 \\ 4 & - 4 \end{bmatrix} = \lambda\begin{bmatrix} 3 & - 2 \\ 4 & - 2 \end{bmatrix} - 2\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix} 1 & - 2\\4 & - 4 \end{bmatrix} = \begin{bmatrix} 3\lambda & - 2\lambda\\4\lambda & - 2\lambda \end{bmatrix} - \begin{bmatrix} 2 & 0\\0 & 2 \end{bmatrix} \]
\[ \Rightarrow \begin{bmatrix} 1 & - 2\\4 & - 4 \end{bmatrix} = \begin{bmatrix} 3\lambda - 2 & - 2\lambda\\4\lambda & - 2\lambda - 2 \end{bmatrix}\]
On equating corresponding terms, we get
\[ - 2\lambda = - 2\]
\[ \Rightarrow \lambda = 1 \]
\[\text{ On substituting } \lambda = 1\text{ in }\left( 1 \right),\text{ we get}\]
\[ A^2 = A - 2I \]
\[ \Rightarrow A^2 - A = - 2I\]
\[ \Rightarrow A - A^2 = 2I\]
\[ \Rightarrow A^{- 1} \left( A - A^2 \right) = A^{- 1} \times 2I \left(\text{ Pre - multiplying both sides with }A^{- 1} \right)\]
\[ \Rightarrow I - A = 2 A^{- 1} \]
\[2 A^{- 1} = \begin{bmatrix} 1 & 0\\0 & 1 \end{bmatrix} - \begin{bmatrix} 3 & - 2\\4 & - 2 \end{bmatrix} = \begin{bmatrix} 1 - 3 & 0 + 2\\0 - 4 & 1 + 2 \end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{2}\begin{bmatrix} - 2 & 2\\ - 4 & 3 \end{bmatrix}\]