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 

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 $26 \times 3^5$.
