$|\text{det}(A)| = 1$ implies $A$ is orthogonal

Question

I know that $A$ orthogonal $\Rightarrow$ |det($A$)| = 1. Now I need to prove or disprove the reversed statement:

$$ |\det(A)| = 1 \Rightarrow A \,\text{ is orthogonal} $$

This is what I'm currently trying:

$$ |\det(A)| = 1 \Rightarrow \det(A)^2 = 1 \Rightarrow \det(AA^t) = 1 $$

But I'm unsure whether this implies, that $AA^t = E_n$. Any help is welcome at this point. Maybe the statement isn't even true.

Answer 1

It's not true:

$$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$

has determinant $1$ but it's not orthogonal since the columns are not orthonormal.

Furthermore,

$$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\cdot\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}^T = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\cdot\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$$

Answer 2

No. Take any matrix $A$ with determinant $d\not=0$ and divide the elements of the first row by $d$. Then the new matrix has determinant $1$. Now it should be easy to find a counterexample to your implication.