determinant $s=n+1$ Need help
$A=\begin{vmatrix}
s&s&s &\cdots & s&s\\
s&1&s &\cdots & s&s\\
s&s&2 &\cdots & s&s\\\vdots & \vdots&\vdots&\ddots&\vdots&\vdots&\\
s&s&s &\cdots & n-1&s\\
s&s&s &\cdots & s&n\\
\end{vmatrix}$
a) calculate $\det A$ when $s =n+1$
Prove that in that case is invertible
b) It may happened that $s=4n$ and $\det A=26\times 3^5$
in a) $\det A = -1$ right? how to prove the invertile?
b) plz some hint 
 A: Hint: One way of finding the determinant is to do row operations and track how this changes the determinant.  Start by subtracting the first row from all the others.  You should be able to come up with a formula for the determinant in terms of $s$ and $n$.
Also for part (a) remember the following theorem:


*

*A square matrix is invertible if and only if it's determinant is invertible.

A: If $\det A \ne 0$ the A is invertible, you can construct the inverse by Cramer formulae.
A: Adding a multiple of one row to another row in a square matrix does not change the determinant (link).  Hence, by subtracting the first row from all the others, we find
$$A=\begin{pmatrix}
s&s&s &\cdots & s&s\\
s&1&s &\cdots & s&s\\
s&s&2 &\cdots & s&s\\
\vdots & \vdots&\vdots&\ddots&\vdots&\vdots&\\
s&s&s &\cdots & n-1&s\\
s&s&s &\cdots & s&n\\
\end{pmatrix}$$
and
$$B=\begin{pmatrix}
s&s&s &\cdots & s&s\\
0 & 1-s & 0 & \cdots & 0 & 0 \\
0 & 0 & 2-s & \cdots & 0 & 0 \\
\vdots & \vdots&\vdots&\ddots&\vdots&\vdots&\\
0 & 0 & 0 & \cdots & n-1-s & 0 \\
0 & 0 & 0 & \cdots & 0 & n-s \\
\end{pmatrix}$$
have the same determinant.  The determinant of $B$ can be "read off" using the Leibniz Formula; the only non-zero contribution to the sum comes from the identity permutation.  Hence $$\det(A)=\det(B)=s\prod_{i=1}^n (i-s).$$  In the case when $s=n+1$, we have
\begin{align*}
\det(A) &= (n+1) \times -n \times -(n-1) \times \cdots \times -1 \\
 &= (-1)^{n-1} (n+1)!.
\end{align*}
Since $\det(A) \neq 0$ the matrix $A$ is invertible.
In the case when $s=4n$, we have
\begin{align*}
\det(A) &= 4n \times (1-4n) \times (2-4n) \times \cdots \times (n-4n).
\end{align*}
For no value of $n$ does this equal $26 \times 3^5$.
