# If $A$ is a matrix, and $A^2=I$, then can I say that $|A|= \pm1$?

$A^2=I$

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

Is this proof correct?

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

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