If \(\mathrm{A} = \begin{bmatrix}3&1\\-1&2\end{bmatrix}\) and \(\mathrm{I} = \begin{bmatrix}1&0\\0&1\end{bmatrix}\), then the value of \(\mathrm{A}^2 - 5\mathrm{A} + 6\mathrm{I}\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\begin{bmatrix}-1&0\\0&-1\end{bmatrix}\)
By matrix multiplication:
\(\displaystyle A^2 = \begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix} = \begin{bmatrix}9 - 1&3 + 2\\-3 - 2&-1 + 4\end{bmatrix}\)
\(\displaystyle = \begin{bmatrix}8&5\\-5&3\end{bmatrix}\)
Then
\(\displaystyle A^2 - 5A + 6I = \begin{bmatrix}8 - 15 + 6&5 - 5 + 0\\-5 + 5 + 0&3 - 10 + 6\end{bmatrix}\)
\(\displaystyle = \begin{bmatrix}-1&0\\0&-1\end{bmatrix}\)
Shortcut: By the Cayley-Hamilton theorem, a 2 × 2 matrix satisfies \(A^2 - (\mathrm{tr}\,A)A + |A|I = O\). Here tr A = 5 and |A| = 6 + 1 = 7, so
\(A^2 - 5A + 7I = O\)
\(A^2 - 5A + 6I = -I\)