If \( A=\begin{bmatrix}1 & 0\\ -1 & 5\end{bmatrix} \) and I is identity matrix of order 2, then which of following statements are TRUE?
A. \( A^{2}-6A-mI=0 \) if \( m=-5 \).
B. \( A^{2}-6A+mI=0 \) if \( m=-5 \).
C. \( A \) is a non-singular matrix.
D. \( A^{2}=A \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A and C only
Squaring A,
\( \displaystyle A^{2}=\begin{bmatrix}1&0\\-1&5\end{bmatrix}\begin{bmatrix}1&0\\-1&5\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}1&0\\-6&25\end{bmatrix} \)
Then
\( \displaystyle A^{2}-6A=\begin{bmatrix}1&0\\-6&25\end{bmatrix}-\begin{bmatrix}6&0\\-6&30\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}-5&0\\0&-5\end{bmatrix} \)
\( \displaystyle =-5I \)
Also
\( \displaystyle |A|=1\cdot5-0\cdot(-1)=5 \)
So \( A^{2}-6A+5I=O \) (the 0 in the statements is the zero matrix O), \( |A|=5 \) and \( A^{2}=\begin{bmatrix}1&0\\-6&25\end{bmatrix} \).
Checking the statements
(A) \( A^{2}-6A-mI=0 \) if \( m=-5 \) — Correct. With m = −5 it reads \( A^{2}-6A+5I=O \).
(B) \( A^{2}-6A+mI=0 \) if \( m=-5 \) — Incorrect. It reads \( A^{2}-6A-5I=-10I\ne O \).
(C) A is a non-singular matrix — Correct. \( |A|=5\ne0 \).
(D) \( A^{2}=A \) — Incorrect. The (2, 1) entry of \( A^{2} \) is −6, but that of A is −1.
Shortcut: By the Cayley-Hamilton theorem,
\( \displaystyle A^{2}-(\mathrm{tr}A)A+|A|I=O \).
With tr A = 6 and |A| = 5 this gives
\( \displaystyle A^{2}-6A+5I=O \) directly.