If \(A = \begin{bmatrix} 3 & 2 & 0 1 & 4 & 0 0 & 0 & 5 \end{bmatrix}\), then compute \(A^2 - 7A + 10 I\).
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For block diagonal matrices like this, the multiplication often stays within the blocks, making it easier to track.
Solution and Explanation
Concept:
Matrix multiplication \(A^2=A\times A\).
Scalar multiplication \(kA\) means multiplying every element of \(A\) by \(k\).
\(I\) is the identity matrix of the same order as \(A\).
Step 1: Calculate \(A^2\)
Given: \[ A= \begin{bmatrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix} \] Therefore: \[ A^2= \begin{bmatrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix} \begin{bmatrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix} \] \[ A^2= \begin{bmatrix} 3(3)+2(1) & 3(2)+2(4) & 0 \\ 1(3)+4(1) & 1(2)+4(4) & 0 \\ 0 & 0 & 5(5) \end{bmatrix} \] \[ A^2= \begin{bmatrix} 11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25 \end{bmatrix} \]
Step 2: Calculate \(7A\) and \(10I\)
\[ 7A= 7\begin{bmatrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix} = \begin{bmatrix} 21 & 14 & 0 \\ 7 & 28 & 0 \\ 0 & 0 & 35 \end{bmatrix} \] Also: \[ 10I= \begin{bmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix} \]
Step 3: Compute \(A^2-7A+10I\)
\[ A^2-7A+10I = \begin{bmatrix} 11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25 \end{bmatrix} - \begin{bmatrix} 21 & 14 & 0 \\ 7 & 28 & 0 \\ 0 & 0 & 35 \end{bmatrix} + \begin{bmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix} \] \[ = \begin{bmatrix} 11-21+10 & 14-14 & 0 \\ 7-7 & 18-28+10 & 0 \\ 0 & 0 & 25-35+10 \end{bmatrix} \] \[ = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \]
Final Answer:
Therefore, \[ \boxed{A^2-7A+10I=O} \] Hence, the result is the Zero Matrix.
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