0
$\begingroup$

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.

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

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 $n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.