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Mathematics · 2026 · 3 marks
CBSE 2026 · Region 4 · Set 3 · Q27
If A = [2, 0, 1; 2, 1, 3; 1, -1, 0], then compute A^2-5 A +4 I.
If $\displaystyle \mathrm{A}=\left[\begin{array}{rrr}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]$, then compute $\displaystyle \mathrm{A}^{2}-5 \mathrm{~A}+4 \mathrm{I}$.
Marking-scheme solution
$\displaystyle A^{2}=\begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}\begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}=\begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix}$
$\displaystyle A^{2}-5 A+4 I=\begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix}-\begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix}+\begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}$
$\displaystyle A^{2}-5 A+4 I=\begin{bmatrix} -1 & -1 & -3 \\ -1 & -3 & -10 \\ -5 & 4 & 2 \end{bmatrix}$
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CBSE Class 12 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.