If $A$ is matrix from $M_{n-1,n}$ such that sum of all elements in row is zero.
Prove det$(AA^T)=nk^2$, for some $k\in\mathbb{Z}$.
I tried to solve it, but I’m really confused.
I know that $A[1, 1, \dots, 1]^t= 0 * [1, 1,\dots, 1]^t$
So zero is eigenvalue of $A$, so it's singular and $\det A=0$.
So $\det(AA^{T})=\det(A)\det(A^T)=\det(A)\det(A)=\det(A^2)=0$
Is this correct? Is there any other way to solve this? Thank you in advance.