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This question already has an answer here:

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.

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marked as duplicate by BCLC, Namaste linear-algebra Oct 25 '18 at 12:01

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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 $\det A = x$

$$x \times x = 1 \iff x^2 = 1 \iff x = \pm 1$$

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