Prove that $\operatorname{adj} (kA) = k ^ {n – 1} \operatorname{adj} A$

Question

Prove that $\operatorname{adj} (kA) = k ^ {n – 1} \operatorname{adj} A$ where $A$ is $n \times n$ matrix.

Hello, I know the proof for this when $\det A \neq 0$:

For $k = 0$, this holds.

Now, for $k \neq 0$

$$(kA) \operatorname{adj} (kA) = k ^ n \det (A) I_n$$
$$A \operatorname{adj} (kA) = k ^ {n – 1}$$
Pre-multiply both sides by $\operatorname{adj} A$ (this is possible)
$$\underbrace{\operatorname{adj} (A) A}\operatorname{adj}(kA) = \det (A)k^{n – 1}\operatorname{adj} A$$
$$\det A \cdot I_n \operatorname{adj}(kA) = \det (A)k^{n – 1} \operatorname{adj} A$$
Cancel $\det A$:
$$\operatorname{adj} (kA) = k^{n – 1} \operatorname{adj} A$$

But, what if $\det A = 0$?

Thanks

Answer 1

The elements of the adjugate are $n-1 \times n-1$ minors of $A$ which obviously scale in the right way.