2

Let $A$ be a square matrix such that $AA^T = I$. If $\det{A} < 0$, find $\det{A}$ and $\det{(I + A)}$.

I managed to get $\det{A} = -1$ but had difficulties when trying to calculate $\det{(I + A)}$.

Chappers
  • 67,606

1 Answers1

3

$$1=\det(I)=\det(AA^t)=\det(A)^2\Rightarrow \det(A)=-1.$$

$$\det(I+A)=\det(AA^t+A)=\det(A)\det(I+A^t)=\det(A)\det((I+A)^t)=-\det(I+A)\Rightarrow \det(I+A)=0.$$

Sonner
  • 3,108