# prove $\det(adj{A})=\det(A)$ for any $A\in{M}_{n\times{n}}$

I'm using the book "linear algebra" by Hoffman and Kunze as I'm preparing to a test on the course "linear algebra" in university. I couldn't find answers to the exercises presented in the book. one of the exercises is:

prove: $\det(\operatorname{adj}{A})=\det(A)$ for any $A\in{M}_{n\times{n}}$

I have never heard of this claim (which is odd as this one might be very useful).

I don't want to prove this using the following method : let $A=a_{ij}$, as I'd want to find an expression for $\det(A)$ using $a_{ij}$ and the minors of $A$, and to do the same for $\operatorname{adj}A$. this is probably the worst way try to prove such question, without even me knowing it will work.

this is a problem as I would like to have more intuition about time-wasting methods (and might even just Wrong) to make proofs in a test.

• What is $\text{adj } A$ in this book? If it's what I think of as $\text{adj } A$ then this is not true. – Lord Shark the Unknown Jun 28 '18 at 8:54
• Are you sure it is "adjacency matrix" and not "adjugate matrix"? (It is still not true for the latter as the previous comment states except $2\times 2$ case). – A.Γ. Jun 28 '18 at 8:56
• your'e right, fixed it – Jneven Jun 28 '18 at 9:03
• if $A=\left[\begin{array}{cc} a_{11} & a_{12}\\ a_{21} & a_{22} \end{array}\right],$ $adjA=\left[\begin{array}{cc} a_{22} & -a_{12}\\ -a_{21} & a_{11} \end{array}\right]$ – Jneven Jun 28 '18 at 9:10
• Then $\det(\operatorname{adj}{A})=\det(A)^{\color{red}{n-1}}$ (see Properties in the link above). – A.Γ. Jun 28 '18 at 9:12

I think the problem in Section 5.2 of the book is for the case $n=2$. The result is not valid for other values of $n$. For $n=2$ we have $A (adj A)=det(A) I$ Take determinants and use the fact that $det (AB)=det (A) det(B)$ to get $det (A) det (adj (A))=(det(A)) ^{2}$. This gives $det (adj (A))=det (A)$ at least when $det (A) \neq 0$. When $det (A)=0$ you can approximate $A$ by $A+c_nI$ with $c_n \to 0$ and $A+c_nI$ invertible for all $n$.