Question

If \[A = \begin{bmatrix}2 & - 1 \\ - 1 & 2\end{bmatrix}\] and I is the identity matrix of order 2, then show that A²= 4 A − 3 I. Hence find A⁻¹.

Answer 1

\[A = \begin{bmatrix}2 & - 1 \\ - 1 & 2\end{bmatrix}\]

\[ A^2 = A \cdot A = \begin{bmatrix}2 & - 1 \\ - 1 & 2\end{bmatrix}\begin{bmatrix}2 & - 1 \\ - 1 & 2\end{bmatrix}\]

\[ = \begin{bmatrix}5 & - 4 \\ - 4 & 5\end{bmatrix}\]

Also,

\[4A = 4\begin{bmatrix}2 & - 1 \\ - 1 & 2\end{bmatrix} = \begin{bmatrix}8 & - 4 \\ - 4 & 8\end{bmatrix}\]

\[3I = 3\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}3 & 0 \\ 0 & 3\end{bmatrix}\]

\[\therefore 4A - 3I = \begin{bmatrix}8 & - 4 \\ - 4 & 8\end{bmatrix} - \begin{bmatrix}3 & 0 \\ 0 & 3\end{bmatrix}\]

\[ = \begin{bmatrix}5 & - 4 \\ - 4 & 5\end{bmatrix}\]

\[ = A^2\]

\[\therefore 4A - 3I = \begin{bmatrix}8 & - 4 \\ - 4 & 8\end{bmatrix} - \begin{bmatrix}3 & 0 \\ 0 & 3\end{bmatrix}\]

\[ = \begin{bmatrix}5 & - 4 \\ - 4 & 5\end{bmatrix}\]

\[ = A^2\]

Hence,

\[4A - 3I = A^2\]

Now,

\[3I = 4A - A^2\]

Pre-multiply both sides by

\[A^{- 1}\] , we get

\[3 A^{- 1} I = 4 A^{- 1} A - A^{- 1} A \cdot A\]

\[ \Rightarrow 3 A^{- 1} = 4I - I \cdot A \left( A^{- 1} A = I \right)\]

\[ \Rightarrow 3 A^{- 1} = 4I - A \left( I \cdot A = A \right)\]

\[ \Rightarrow 3 A^{- 1} = 4\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} - \begin{bmatrix}2 & - 1 \\ - 1 & 2\end{bmatrix}\]

\[ \Rightarrow 3 A^{- 1} = \begin{bmatrix}4 & 0 \\ 0 & 4\end{bmatrix} - \begin{bmatrix}2 & - 1 \\ - 1 & 2\end{bmatrix} = \begin{bmatrix}2 & 1 \\ 1 & 2\end{bmatrix}\]

\[ \Rightarrow A^{- 1} = \frac{1}{3}\begin{bmatrix}2 & 1 \\ 1 & 2\end{bmatrix} = \begin{bmatrix}\frac{2}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{2}{3}\end{bmatrix}\]