If $A =\begin{bmatrix}7&3\\5&-7\end{bmatrix}$ be such that $A^{-1} = kA$, then $k$ equals

$\frac{1}{64}$

The correct answer is Option (2) → $\frac{1}{64}$ **

Given matrix:

$A=\begin{pmatrix}7 & 3 \\ 5 & -7\end{pmatrix}$

Condition: $A^{-1}=kA$

For a 2×2 matrix,

$A^{-1}=\frac{1}{\det(A)}\begin{pmatrix}-7 & -3 \\ -5 & 7\end{pmatrix}$

Determinant:

$\det(A)=7(-7)-3(5)=-49-15=-64$

Thus,

$A^{-1}=\frac{1}{-64}\begin{pmatrix}-7 & -3 \\ -5 & 7\end{pmatrix}$

$=\begin{pmatrix}\frac{7}{64} & \frac{3}{64} \\ \frac{5}{64} & -\frac{7}{64}\end{pmatrix}$

Now compare with $kA$:

$kA=k\begin{pmatrix}7 & 3 \\ 5 & -7\end{pmatrix} =\begin{pmatrix}7k & 3k \\ 5k & -7k\end{pmatrix}$

Equating entries:

$7k=\frac{7}{64}\;\Rightarrow\;k=\frac{1}{64}$

The value of $k$ is $\frac{1}{64}$.