Question

If \( A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} \), then compute \( A^{2} - 5A + 4I \).

Hint:

Find A squared by row-by-column multiplication, subtract 5A, and add 4 on the diagonal.

Answer 1

Step 1: Understanding the Concept:
We find \(A^2\) by matrix multiplication, find \(5A\) by scaling every entry, and add \(4I\) where \(I\) is the \(3\times 3\) identity matrix.

Step 2: Compute \(A^2\):
Multiply each row of \(A\) by the columns of \(A\).
Row 1: \((2\cdot2+0\cdot2+1\cdot1,\ 2\cdot0+0\cdot1+1\cdot(-1),\ 2\cdot1+0\cdot3+1\cdot0) = (5,-1,2)\).
Row 2: \((4+2+3,\ 0+1-3,\ 2+3+0) = (9,-2,5)\).
Row 3: \((2-2+0,\ 0-1+0,\ 1-3+0) = (0,-1,-2)\).
\[ A^{2} = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} \]

Step 3: Compute \(5A\) and \(4I\):
\[ 5A = \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix},\qquad 4I = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix} \]

Step 4: Combine entry by entry:
\[ A^{2} - 5A + 4I = \begin{bmatrix} 5-10+4 & -1-0+0 & 2-5+0 \\ 9-10+0 & -2-5+4 & 5-15+0 \\ 0-5+0 & -1+5+0 & -2-0+4 \end{bmatrix} \]
\[ = \begin{bmatrix} -1 & -1 & -3 \\ -1 & -3 & -10 \\ -5 & 4 & 2 \end{bmatrix} \]

Final Answer:
\[ \boxed{A^{2}-5A+4I = \begin{bmatrix} -1 & -1 & -3 \\ -1 & -3 & -10 \\ -5 & 4 & 2 \end{bmatrix}} \]