Question

Show that the matrix  \[A = \begin{bmatrix}2 & 3 \\ 1 & 2\end{bmatrix}\]satisfies the equation A³ − 4A² + A = O

Answer 1

\[\text{We have} , A = \begin{bmatrix}2 & 3 \\ 1 & 2\end{bmatrix}\]
\[ \therefore A^2 = AA\]
\[ \Rightarrow A^2 = \begin{bmatrix}2 & 3 \\ 1 & 2\end{bmatrix}\begin{bmatrix}2 & 3 \\ 1 & 2\end{bmatrix}\]
\[ \Rightarrow A^2 = \begin{bmatrix}4 + 3 & 6 + 6 \\ 2 + 2 & 3 + 4\end{bmatrix}\]
\[ \Rightarrow A^2 = \begin{bmatrix}7 & 12 \\ 4 & 7\end{bmatrix}\]
\[ \Rightarrow A^3 = A^2 A\]
\[ \Rightarrow A^3 = \begin{bmatrix}7 & 12 \\ 4 & 7\end{bmatrix}\begin{bmatrix}2 & 3 \\ 1 & 2\end{bmatrix}\]
\[ \Rightarrow A^3 = \begin{bmatrix}14 + 12 & 21 + 24 \\ 8 + 7 & 12 + 14\end{bmatrix}\]
\[ \Rightarrow A^3 = \begin{bmatrix}26 & 45 \\ 15 & 26\end{bmatrix}\]
\[Now, A^3 - 4 A^2 + A\]
\[ \Rightarrow A^3 - 4 A^2 + A = \begin{bmatrix}26 & 45 \\ 15 & 26\end{bmatrix} - 4\begin{bmatrix}7 & 12 \\ 4 & 7\end{bmatrix} + \begin{bmatrix}2 & 3 \\ 1 & 2\end{bmatrix}\]
\[ \Rightarrow A^3 - 4 A^2 + A = \begin{bmatrix}26 & 45 \\ 15 & 26\end{bmatrix} - \begin{bmatrix}28 & 48 \\ 16 & 28\end{bmatrix} + \begin{bmatrix}2 & 3 \\ 1 & 2\end{bmatrix}\]
\[ \Rightarrow A^3 - 4 A^2 + A = \begin{bmatrix}26 - 28 + 2 & 45 - 48 + 3 \\ 15 - 16 + 1 & 26 - 28 + 2\end{bmatrix}\]
\[ \Rightarrow A^3 - 4 A^2 + A = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix} = 0\]
Hence proved .