Prove A² − 7A + 10I₃ = O

Question

If \[ A = \begin{bmatrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix} \] show that \[ A^2 – 7A + 10I_3 = O. \]


Solution

Step 1: Compute \(A^2\)

\[ A^2 = \begin{bmatrix} 11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25 \end{bmatrix} \]

Step 2: Form Expression

\[ A^2 – 7A + 10I_3 = \begin{bmatrix} 11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25 \end{bmatrix} – \begin{bmatrix} 21 & 14 & 0 \\ 7 & 28 & 0 \\ 0 & 0 & 35 \end{bmatrix} + \begin{bmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix} \]

Step 3: Simplify

\[ = \begin{bmatrix} 11 – 21 + 10 & 14 – 14 + 0 & 0 \\ 7 – 7 + 0 & 18 – 28 + 10 & 0 \\ 0 & 0 & 25 – 35 + 10 \end{bmatrix} \] \[ = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \]

Final Result

\[ A^2 – 7A + 10I_3 = O \]

Hence proved.

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