If \(A = \begin{bmatrix}5&3\\2&4\end{bmatrix}\), then the matrix \(A^2 - 6A + 14 I\) is (where I is an identity matrix of order 2)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\begin{bmatrix}15&9\\6&12\end{bmatrix}\)
\(\displaystyle A^2 = \begin{bmatrix}5&3\\2&4\end{bmatrix}\begin{bmatrix}5&3\\2&4\end{bmatrix} = \begin{bmatrix}25 + 6&15 + 12\\10 + 8&6 + 16\end{bmatrix}\)
\(\displaystyle = \begin{bmatrix}31&27\\18&22\end{bmatrix}\)
\(\displaystyle 6A = \begin{bmatrix}30&18\\12&24\end{bmatrix},\quad 14I = \begin{bmatrix}14&0\\0&14\end{bmatrix}\)
\(\displaystyle A^2 - 6A + 14I = \begin{bmatrix}31 - 30 + 14&27 - 18\\18 - 12&22 - 24 + 14\end{bmatrix}\)
\(\displaystyle = \begin{bmatrix}15&9\\6&12\end{bmatrix}\)
Shortcut: By the Cayley-Hamilton theorem, \(A^2 - (\mathrm{tr}\,A)A + |A|I = O\). Here \(\mathrm{tr}\,A = 9\) and \(|A| = 20 - 6 = 14\), so \(A^2 - 9A + 14I = O\) and
\(\displaystyle A^2 - 6A + 14I = 3A = \begin{bmatrix}15&9\\6&12\end{bmatrix}\)