3
$\begingroup$

$A^2=I$

Take determinant on both sides: $$|A^2|= |I| $$ $$|A|^2= 1$$ $$|A| = +1 \text{ or } -1$$

Is this proof correct?

$\endgroup$
  • $\begingroup$ It means that the matrix is it's own inverse. $\endgroup$ – SalmonKiller May 29 '15 at 18:35
  • $\begingroup$ the determinant of a product is the product of determinants. Can you take it from there? $\endgroup$ – QTHalfTau May 29 '15 at 18:36
  • $\begingroup$ I know that. Wait i'll just edit my answer. To SalmonKiller that is. $\endgroup$ – kay May 29 '15 at 18:36
  • 1
    $\begingroup$ Yes, your proof is correct. $\endgroup$ – Ben Grossmann May 29 '15 at 18:55
6
$\begingroup$

$$\det{A^2}=\det{AA}=\det{A}\det{A}=(\det{A})^2=\det{I}=1$$ $$\therefore\det{A}=\pm 1$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Yes thank you. :) $\endgroup$ – kay May 29 '15 at 18:40
  • $\begingroup$ @kay Glad I could help :) If you're satisfied with my answer, you can click the "accept answer" button next to it, so that the question is marked as solved. $\endgroup$ – Demosthene May 29 '15 at 18:41
3
$\begingroup$

You have $1 = \text{det}I = \text{det}(A^2) = \text{det}A\text{det}A $

It follows that $\text{det}A = \pm 1$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

yes. the determinant have the properties that $$det(AB) = det(A)det(B), det(I) = 1.$$ therefore using the two properties we have, $$det(A^2) = \left(det(A)\right)^2 = 1 \implies det(A) = \pm 1.$$

| cite | improve this answer | |
$\endgroup$

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.