• $A^t$: transpose of matrix $A$.

  • $\det A$: determinant of matrix $A$.

  • $I$: the identity matrix.

We know:

$A \in M_{n\times n}(\mathbb{R}), \\ AA^t = I, \ \mathrm{and \ det}(A) < 0$

We want: $\det(A + I)$

From $\det A = \det A^t$, $\det A^{-1} = \frac{1}{\det A}$ and $\det A < 0$ we can deduce: $\det A = -1$. I tried calculating

$$\det(A + I) = \det(A(I + A^t)) = \det(A) \det(I + A^t) = - \det(I + A^t)$$

but that didn't lead to anything. I tried using the fact that $A + A^t$ is symmetric and $A - A^t$ is skew-symmetric but I couldn't achieve anything with that either. I thought about interpreting it via a linear transformation but couldn't come up with anything.

I learned that matrices, where $AA^t = I$, are called orthogonal but that's all I know about orthogonal matrices so I'm hoping to find a solution that doesn't get too deep into orthogonality.

Hints would be appreciated. Thank you in advance!


1 Answer 1


Let $D=\det(A+I)$. You have $$D=\det(A)\det(I+A^{-1})=\det(A)\det(I+A^{T})=\det(A)\det(I+A)=\det(A)D$$

So $D=0$ since $\det(A) \neq 1$.

  • $\begingroup$ Oh yes! Because $(A + B)^t = A^t + B^t$ so $det(A^t + I) = det(A + I)$. Thank you $\endgroup$
    – Zara
    Nov 24, 2020 at 10:58
  • $\begingroup$ @Zara You got it :) $\endgroup$ Nov 24, 2020 at 10:59

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