I have the following proof that I would like to be walked through because I'm not intuitively seeing what to do:
If $A$ is $n\times n$, prove $\det\left(\operatorname{adj}(A)\right) = \det(A)^{n-1}$.
I know the property of $A\operatorname{adj}(A) = \det(A)I$ is important but I don't know how to apply it to get an answer. Any help is much appreciated.
Thanks,
$\endgroup$ 104 Answers
$\begingroup$A(adj A) = |A|(I)
|A(adj A)| = |(|A| I)|
|A| |adj A| = $|A|^n * |I|$
|A| |adj A| = $|A|^n $
case 1: if |A|$\neq0$
Then we get ,|adj A| = $|A|^{n-1} $
case 2: if |A|$=0$
Then,|adj A|$=0$
And, we again get |adj A| = $|A|^{n-1} $
$\endgroup$ 7 $\begingroup$We have the relation$$A \operatorname{adj}(A)=\det(A)I $$Now take determinant on both the sides we get,$$\det(A)\det(\operatorname{adj}(A))=\det(\det(A)I) \tag1$$use the relation that, for a matrix $A$ that is $n \times n$,
$$\det(kA)=k^n (\det(A))$$where k is a numerical constant. Hence we have
$$\det(\det(A)I)=\det(A)^n \det (I)=\det(A)^n \tag2$$and from equations $(1)$ and $(2)$ it follows that$$\det(A)\det(\operatorname{adj}(A))=\det(A)^n ,$$which implies$$\det(\operatorname{adj}(A)) = \det(A)^{n-1} .$$
$\endgroup$ 2 $\begingroup$We know that $A\cdot adj(A) = detA\cdot I$
Now, this implies $$\det(A\cdot adjA) = \det(\det A\cdot I)$$
The matrix on the right is a diagonal matrix with each diagonal entry equal to $detA$
Thus, its determinant will simply be the product of the diagonal entries,$(\det A)^n$
Also, using the multiplicity of determinant function, we get $\det(A\cdot adjA) = \det A\cdot \det(adjA)$
Case $1$: $\det A \neq 0$
This directly gives us $\det(adjA) = (detA)^{n-1}$
Case $2$: $\det A = 0$
Case $2.1$ : $A = 0$
This simply means that $adjA = 0$ thus satisfying the result obtained in case 1.
Case $2.2$ : $A\neq 0$
This means that $adjA$ isn't invertible. If it would have been, then right multiplication with $(adjA)^{-1}$ would have given $A = 0$, Thus a contradiction.
Now, as $adjA$ isn't invertible, this implies $det(adjA) = 0$ , thus completing the proof.
$\endgroup$ $\begingroup$[Case I] det (Adj(A))=0
[Case II] det (Adj(A)) = nonzero, so Adj(A) is invertible.
Let (Adj(A))^{-1} =B. From A Adj(A)=det(A)I, A Adj(A) B= det(A)I B.
So A = B det(A)I.
Suppose that det(A)=0. Then A = 0. So Adj A =0 implies det (Adj A)=0, a contradiction.
Therefore det (A) = nonzero. Now we have det (Adj(A))=det (A)^{n-1}.
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