If matrix \( \mathrm{A}=\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix} \), then \( \mathrm{A}^2+\mathrm{I} \) is equal to: (Where I is an identity matrix of order 2)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4A
\( \displaystyle \mathrm{A}^2=\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}4+3 & 6+6\\2+2 & 3+4\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}7 & 12\\4 & 7\end{bmatrix} \)
\( \displaystyle \mathrm{A}^2+\mathrm{I}=\begin{bmatrix}8 & 12\\4 & 8\end{bmatrix} \)
\( \displaystyle =4\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}=4\mathrm{A} \)
Shortcut: By the Cayley-Hamilton theorem for a 2 × 2 matrix,
\( \displaystyle \mathrm{A}^2-(\mathrm{tr\,A})\mathrm{A}+|\mathrm{A}|\,\mathrm{I}=\mathrm{O} \)
Here tr A = 4 and |A| = 4 − 3 = 1, so
\( \displaystyle \mathrm{A}^2-4\mathrm{A}+\mathrm{I}=\mathrm{O} \)
that is, \( \mathrm{A}^2+\mathrm{I}=4\mathrm{A} \).