# Determinant of a matrix's second power equal to 1, so what will be its first power's value?

If it is given that $$B^2=0$$. Then can I directly take the determinant on both sides and prove that determinant of $$B=0$$?

Given:$$B2=0$$

Taking determinant both sides

$$|B^2|=|B|^2$$

$$|B|^2=|0|$$

$$|B|=0$$

This was correct.

But I now have a doubt when $$A^2=I$$, why can't $$A=I$$ the same way we did for $$B=0$$

If $$A^2=I$$

$$|A|^2=|I|$$

$$|A|=1$$

Why is this wrong?

• $|A|=\pm 1$. $|A|$ is not always positive. Commented Jan 16, 2022 at 11:40
• What are the solutions of $x^2=1$ ? Commented Jan 16, 2022 at 11:43
• $|A|=±1$ Is this the correct answer? Commented Jan 16, 2022 at 11:45
• The determinant can of course also be negative, so $\det(A)=-1$ is possible as well (in fact for odd size of $A$ , $A=-I$ is a concrete solution with determinant $-1$). Hence $|A|=\pm 1$ is correct. For $0$, we have only one solution because $-0$ and $+0$ fall together. Commented Jan 16, 2022 at 11:51
• Perhaps you considered $|A|$ to be something like the absolute value, which is why I would prefer the notation "det" Commented Jan 16, 2022 at 11:56