If $A$ is a $3 \times 3$ matrix such that $|adj A| = 9$ and $|kA^{-1}| = 9$, then the value of $k$ are:

Given $A$ is a $3 \times 3$ matrix with $|adj A| = 9$ and $|kA^{-1}| = 9$.

For any $n \times n$ matrix, $|adj A| = |A|^{n-1}$.

Since $A$ is a $3 \times 3$ matrix:

$|adj A| = |A|^{3-1}$

$|adj A| = |A|^2$

Substituting the given value:

$|A|^2 = 9$

$|A| = \pm 3$

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For a scalar $k$ and $n \times n$ matrix $B$, $|kB| = k^n|B|$.

Since $A$ is a $3 \times 3$ matrix:

$|kA^{-1}| = k^3|A^{-1}|$

For inverse matrices, $|A^{-1}| = \frac{1}{|A|}$.

Therefore:

$|kA^{-1}| = k^3 \cdot \frac{1}{|A|}$

Substituting the given value:

$k^3 \cdot \frac{1}{|A|} = 9$

$k^3 = 9|A|$

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When $|A| = 3$:

$k^3 = 9 \times 3$

$k^3 = 27$

$k = 3$

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When $|A| = -3$:

$k^3 = 9 \times (-3)$

$k^3 = -27$

$k = -3$

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Therefore, the value of $k = \pm 3$.