Question

If $A$ is a $2\times 2$ matrix such that $A\begin{bmatrix} \, \, \, 1 \\ -1 \end{bmatrix}=\begin{bmatrix} -1 \\ \, \, \, 2 \end{bmatrix}$ and $A^{2}\begin{bmatrix} \, \, \, 1 \\ -1 \end{bmatrix}=\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ , then trace of $A$ is (where, the trace of the matrix is the sum of all principal diagonal elements of the matrix)

• A. $1$
• B. $0$
• C. $2$
• D. $5$

Answer 1

Answer: (A)

Let, $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$
$A\begin{bmatrix} 1 \\ -1 \end{bmatrix}=\begin{bmatrix} -1 \\ 2 \end{bmatrix}\Rightarrow \begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} 1 \\ -1 \end{bmatrix}=\begin{bmatrix} -1 \\ 2 \end{bmatrix}$
On solving, we get,
$a-b=-1, \, c-d=2$
Also,
$A^{2}\begin{bmatrix} 1 \\ -1 \end{bmatrix}=A\begin{bmatrix} A \begin{bmatrix} 1 \\ -1 \end{bmatrix} \end{bmatrix}=A\begin{bmatrix} -1 \\ 2 \end{bmatrix}=\begin{bmatrix} 1 \\ 0 \end{bmatrix}$
This gives, $-a+2b=1, \, -c+2d=0$
$\Rightarrow b=0, \, a=-1, \, d=2, \, c=4$