Question
If \[ A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} \] find \[ A^2 – 5A + 4I \] and hence find matrix \(X\) such that \[ A^2 – 5A + 4I + X = 0. \]
Solution
Step 1: Compute \(A^2\)
\[ A^2 = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} \]Step 2: Compute \(A^2 – 5A\)
\[ 5A = \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix} \] \[ A^2 – 5A = \begin{bmatrix} -5 & -1 & -3 \\ -1 & -7 & -10 \\ -5 & 4 & -2 \end{bmatrix} \]Step 3: Add \(4I\)
\[ 4I = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix} \] \[ A^2 – 5A + 4I = \begin{bmatrix} -1 & -1 & -3 \\ -1 & -3 & -10 \\ -5 & 4 & 2 \end{bmatrix} \]Step 4: Find \(X\)
\[ A^2 – 5A + 4I + X = 0 \Rightarrow X = -(A^2 – 5A + 4I) \] \[ X = \begin{bmatrix} 1 & 1 & 3 \\ 1 & 3 & 10 \\ 5 & -4 & -2 \end{bmatrix} \]Final Answers
\[
A^2 – 5A + 4I =
\begin{bmatrix}
-1 & -1 & -3 \\
-1 & -3 & -10 \\
-5 & 4 & 2
\end{bmatrix}
\]
\[
X =
\begin{bmatrix}
1 & 1 & 3 \\
1 & 3 & 10 \\
5 & -4 & -2
\end{bmatrix}
\]