CBSE 2026 · Region 4 · Set 2 · Q27 · 3 marks
If $\displaystyle \mathrm{A}=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$, then compute $\displaystyle \mathrm{A}^{2}-4 \mathrm{~A}-5 \mathrm{I}$.
Marking-scheme solution
$\displaystyle A^{2}=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}=\begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix}$
$\displaystyle A^{2}-4 A-5 I=\begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix}-\begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix}-\begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}$
$\displaystyle A^{2}-4 A-5 I=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$
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CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.