Question

The number of matrices $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix} \text { (where } a, b, c, d \in R \text { ) }$ such that $A^{-1}=-A$ is :

• A. 0
• B. 1
• C. 2
• D. infinite

Answer 1

Answer: (D)

As $A^{-1}$ exists, $\operatorname{det}(A)=a d-b c \neq 0$.
Also, $\operatorname{det}(-A)=\operatorname{det}\begin{bmatrix}-a & -b \\ -c & -d\end{bmatrix}=a d-b c$
$=\operatorname{det}(A)$
since $\operatorname{det}\left(A^{-1}\right)=\frac{1}{\operatorname{det}(A)}, A^{-1}=-A$ implies
$\frac{1}{a d-b c} =a d-b c $
$\Rightarrow(a d-b c)^2 =1 \Rightarrow a d-b c= \pm 1 $.
If $a d-b c=1$, then $A^{-1}=-A$
gives
$A^{-1}=\begin{bmatrix}d & -b \\-c & a\end{bmatrix}=\begin{bmatrix}-a & -b \\-c & -d\end{bmatrix}$
$\Rightarrow a+d=0 $
$\therefore a(-a)-b c=1 $
$\Rightarrow b c=-\left(1+a^2\right) \Rightarrow b c \neq 0 \text { and }$
$b=-\frac{1+a^2}{c} $
$\therefore A=\begin{bmatrix}a & -\frac{\left(1+a^2\right)}{c} \\c & -a\end{bmatrix} \text {, where } c \neq 0 \text {. } \\$
Thus, there are infinite number of such matrices.