# What is $\det(A + I)$ when $AA^t = I$ and $\det(A) < 0$?

Notation:

• $$A^t$$: 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!

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 $$\det(A) \neq 1$$.
• Oh yes! Because $(A + B)^t = A^t + B^t$ so $det(A^t + I) = det(A + I)$. Thank you