For matrix \( \mathrm{A}=\begin{bmatrix}2 & -3\\3 & 4\end{bmatrix} \) the value of \( \mathrm{A^2-6A} \) is equal to:(where I is an identity matrix of order 2)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → −17I
\( \displaystyle A^2=\begin{bmatrix}2 & -3\\3 & 4\end{bmatrix}\begin{bmatrix}2 & -3\\3 & 4\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}4-9 & -6-12\\6+12 & -9+16\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}-5 & -18\\18 & 7\end{bmatrix} \)
\( \displaystyle 6A=\begin{bmatrix}12 & -18\\18 & 24\end{bmatrix} \)
\( \displaystyle A^2-6A=\begin{bmatrix}-17 & 0\\0 & -17\end{bmatrix} \)
\( \displaystyle =-17I \)
Shortcut: For a 2 × 2 matrix, \( A^2-(\mathrm{tr}\,A)A+|A|\,I=O \) (Cayley-Hamilton). Here tr A = 2 + 4 = 6 and |A| = 8 + 9 = 17:
\( \displaystyle A^2-6A+17I=O\ \Rightarrow\ A^2-6A=-17I \)