Let A be an invertible $n \times n$ matrix whose inverse is itself. Prove that $\det(A)$ is either $1$ or $-1$.
I'm really lost in class. I don't even know where to start. Please help.
2026-03-25 17:40:16.1774460416
Proving an invertible matrix which is its own inverse has determinant $1$ or $-1$
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Since $A = A^{-1},\;$ we know that $A^2 = AA = I.\;$ The determinant of $\;I = 1$
$$\det(AB) = \det (A) \det (B) \implies \det(AA) = \det(I) = 1 = \det(A) \det(A)$$
What must we conclude about $\det A$? Recall, the determinant is a scalar; we may as well call $\det A = x$
$$x \times x = 1 \iff x^2 = 1 \iff x = \pm 1$$