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

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

• You could note that the adjugate is continuous and take the limit of both sides of the equation Jul 6, 2022 at 19:33
• To add to @BenGrossmann's comment, consider the matrix $A+tI$, for some $t$ such that it is nonsingular, and take the limit $t\to 0$. Jul 6, 2022 at 20:07

The elements of the adjugate are $$n-1 \times n-1$$ minors of $$A$$ which obviously scale in the right way.
$$Adj M. M= |M| I$$, $$|kM|=k^n |M|$$ and $$Adj M=M^{-1} |M|.$$
Let $$M=kA$$, then $$Adj(kA) kA= |kA| I=k^n |A| \implies Adj(kA)A=k^{n-1}|A|$$
$$\implies Adj(kA)=k^{n-1}|A|A^{-1}=k^{n-1} Adj(A).$$