If $A$ is an orthogonal matrix with $|A|=-1$, show that $|I-A|=0$

Question

Let $A$ be an $n \times n$ orthogonal matrix where $A$ is of even order with $|A|=-1.$ Show that, $|I-A|=0,$ where $I$ denotes the $n \times n$ identity matrix.

---

My approach

$A \cdot A^{\top}=I$

$|A| \cdot\left|A^{\top}\right|=1 \quad$ or $\quad\left|A^{\top}\right|=-1.....(2)$

let $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$

$L = I-A$

$L=\left[\begin{array}{cc}1-a & -b \\ -c & 1-d\end{array}\right]; \quad 2 a d=2 b c$(from eq2)

$|L|=(1-a)(1-d) - bc$

$=-(a+d)$

What to do next? Am I going wrong?

Answer 1

Since $A$ is orthogonal, rewrite $\det(I-A)$ as follows: $$\begin{align}\det(I-A)&=\det(A^TA-A)\\&=\det(A)\det(A^T-I)\\&=\det(A)\det((A^T-I)^T)\\&=-\det(A-I)\end{align}$$ It follows that $$\det(I-A)=(-1)^{n+1}\det(I-A)$$ As $n$ is even, we get $$\boxed{\det(I-A)=0}$$

as desired.