If \(A=\begin{bmatrix}1 & 3\\ -1 & 3\end{bmatrix}\) and \(A^2 - 4A + kI = O\), where I is the identity matrix and O is the zero matrix of order 2, then the value of k is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 6
Multiplying A by itself, row by column:
\( \displaystyle A^2 = \begin{bmatrix}1 & 3\\ -1 & 3\end{bmatrix}\begin{bmatrix}1 & 3\\ -1 & 3\end{bmatrix} \)
\( \displaystyle = \begin{bmatrix}1-3 & 3+9\\ -1-3 & -3+9\end{bmatrix} \)
\( \displaystyle = \begin{bmatrix}-2 & 12\\ -4 & 6\end{bmatrix} \)
Subtracting 4A:
\( \displaystyle A^2 - 4A = \begin{bmatrix}-2 & 12\\ -4 & 6\end{bmatrix} - \begin{bmatrix}4 & 12\\ -4 & 12\end{bmatrix} \)
\( \displaystyle = \begin{bmatrix}-6 & 0\\ 0 & -6\end{bmatrix} = -6I \)
So the equation becomes
\(-6I + kI = O\)
\(k = 6\)
Shortcut: By the Cayley-Hamilton theorem, every 2 × 2 matrix satisfies \(A^2 - (\text{tr}\,A)A + |A|\,I = O\). Here
\(\text{tr}\,A = 1 + 3 = 4\)
\(|A| = 3 - (3)(-1) = 6\)
so \(A^2 - 4A + 6I = O\) and k = 6 at once.