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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

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

2 Answers 2

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

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Note that

$Adj M. M= |M| I$, $|kM|=k^n |M|$ and $Adj M=M^{-1} |M|.$

Let $M=kA$, then $Adj(kA) kA= |kA| I=k^n |A| \implies Adj(kA)A=k^{n-1}|A|$

$\implies Adj(kA)=k^{n-1}|A|A^{-1}=k^{n-1} Adj(A).$

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