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

Given matrix $A = \begin{bmatrix} 7 & 3 \\ 5 & -7 \end{bmatrix}$ with the condition $A^{-1} = kA$.

Starting with $A^{-1} = kA$, multiply both sides by $A$ on the left:

$A \cdot A^{-1} = A \cdot kA$

$I = kA^2$

$A^2 = \frac{1}{k}I$

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Calculate $A^2 = A \times A$:

$A^2 = \begin{bmatrix} 7 & 3 \\ 5 & -7 \end{bmatrix} \times \begin{bmatrix} 7 & 3 \\ 5 & -7 \end{bmatrix}$

First row, first column: $7(7) + 3(5) = 49 + 15 = 64$

First row, second column: $7(3) + 3(-7) = 21 - 21 = 0$

Second row, first column: $5(7) + (-7)(5) = 35 - 35 = 0$

Second row, second column: $5(3) + (-7)(-7) = 15 + 49 = 64$

$A^2 = \begin{bmatrix} 64 & 0 \\ 0 & 64 \end{bmatrix}$

$A^2 = 64\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$A^2 = 64I$

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From the equation $A^2 = \frac{1}{k}I$ and $A^2 = 64I$:

$64I = \frac{1}{k}I$

$64 = \frac{1}{k}$

$k = \frac{1}{64}$

Therefore, $k = \frac{1}{64}$.