linear algebra - Matrix with zeros on diagonal and ones in other places is invertible

Hi guys I am working with this and I am trying to prove to myself that n by n matrices of the type zero on the diagonal and 1 everywhere else are invertible.

I ran some cases and looked at the determinant and came to the conclusion that we can easily find the determinant by using the following $\det(A)=(-1)^{n+1}(n-1)$. To prove this I do induction

n=2 we have the $A=\begin{bmatrix}
0 & 1\\
1 & 0
\end{bmatrix}$ $\det(A)=-1$ and my formula gives me the same thing (-1)(2-1)=-1

Now assume if for $n \times n$ and $\det(A)=(-1)^{n+1}(n-1)$

Now to show for a matrix B of size $n+1 \times n+1$. I am not sure I was thinking to take the determinant of the $n \times n$ minors but I am maybe someone can help me. Also is there an easier way to see this is invertible other than the determinant? I am curious.

Answer

This is easy to calculate by row reduction:

Add all rows to first:
$$\det(A) =\det \begin{bmatrix}
0 & 1 & 1 &...&1 \\
1 & 0 & 1 &...&1 \\
1 & 1 & 0 &...&1 \\
... & ... & ... &...&... \\
1 & 1 & 1 &...&0 \\

\end{bmatrix}=\det \begin{bmatrix}
n-1 & n-1 & n-1 &...&n-1 \\
1 & 0 & 1 &...&1 \\
1 & 1 & 0 &...&1 \\
... & ... & ... &...&... \\
1 & 1 & 1 &...&0 \\
\end{bmatrix} \\
=(n-1)\det \begin{bmatrix}
1 & 1 & 1 &...&1 \\
1 & 0 & 1 &...&1 \\

1 & 1 & 0 &...&1 \\
... & ... & ... &...&... \\
1 & 1 & 1 &...&0 \\
\end{bmatrix}=(n-1)\det \begin{bmatrix}
1 & 1 & 1 &...&1 \\
0 & -1 & 0 &...&0 \\
0 & 0 & -1 &...&0 \\
... & ... & ... &...&... \\
0 & 0 & 0 &...&-1 \\
\end{bmatrix}$$

where in the last row operation I subtracted the first row from each other row.

This shows
$$\det(A)=(n-1)(-1)^{n-1}$$