If $A\in\Bbb R^{3\times3}$ is an invertible matrix such that $A^2=A$, then $\det(2A)=8$

Question

True or false?

"If $\pmb{A\in\Bbb R^{3\times3}}$ is an invertible matrix such that $\pmb{A^2=A}$, then $\pmb{\det(2A)=8}$".

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True.

Proof. We start from $\det(2A)$. Then, $2^3\det(A)$. But we know that $A^2=AA=A$, so $A=AA^{-1}$ because $A$ is an invertible matrix, hence $$8\det(A)=8\det(AA^{-1})=8\det(A)\det(A^{-1})=8\det(A)(\det(A))^{-1}=8\det(A)\frac{1}{\det(A)}=8,$$ which is where we wanted to go. $\square$

Is it correct?

Thanks!!

Answer 1

Good, but long! Here $A=AA^{-1}=I$, so $$\det(2A)=2^3 \det(A)=2^3\det(I)=8$$

Moral: The only matrix which is both idempotent and invertible is the identity matrix !