# If A is 3 × 3 invertible matrix, then show that for any scalar k (non-zero),kA is invertible and (kA)1 = 1/k.A-1.

We know that,

${\left(kA\right)}^{-1}=\frac{1}{\left|kA\right|}×adj\left(kA\right)$ -----------(1)

[since $\left|kA\right|={k}^{n}\left|A\right|$]

$⇒{\left(kA\right)}^{-1}=\frac{1}{k}×{A}^{-1}$

Hence Proved.

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