Matrix with a parameter determinant $A \in \mathbb R^{n,n}$
$A_n=\begin{bmatrix} 1&2&3&\dots&n\\x&1&2&\dots&n-1\\x&x&1&\dots&n-2\\\vdots&\vdots&\vdots&\ddots&\vdots\\x&x&x&\dots&1\end{bmatrix}$
How to calculate $\det A_n$ for $x\in\mathbb R$?
 A: I denote $A_n$ for the determinant of your matrix of size $n\times n$, rather than the matrix itself:
$$A_n=\begin{vmatrix} 1&2&3&\dots&n\\x&1&2&\dots&n-1\\x&x&1&\dots&n-2\\\vdots&\vdots&\vdots&\ddots&\vdots\\x&x&x&\dots&1\end{vmatrix}$$
And we will prove that
$$A_n=(1-x)^n-(-x)^n$$
First use row operations to write
$$A_n=\begin{vmatrix}
1-x&1&1&\dots&1\\
0&1-x&1&\dots&1\\
0&0&1-x&\dots&1\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
x&x&x&\dots&1\end{vmatrix}$$
To achieve this, you subtract the second row from the first, then the third from the second, etc. Only the last line remains untouched.
From this new determinant, you get immediately by developping along the first column:
$$A_n=(1-x)A_{n-1}+(-1)^{n-1}xB_{n-1}$$
Where $B_n$ is this determinant (of size $n \times n$):
$$B_n=\begin{vmatrix}
1&1&1&\dots&1&1\\
1-x&1&1&\dots&1&1\\
0&1-x&1&\dots&1&1\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\
0&0&0&\dots&1-x&1\end{vmatrix}$$
Now, use column operations to rewrite your determinant$B_n$ as
$$B_n=\begin{vmatrix}
0&0&0&\dots&0&1\\
-x&0&0&\dots&0&1\\
0&-x&0&\dots&0&1\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\
0&0&0&\dots&-x&1\end{vmatrix}$$
To achieve this, subtract the last column from the butlast, and continue likewise to the left.
This is almost the determinant of a companion matrix, you can rewrite (factoring out $(-x)$ from the first $n-1$ columns):
$$B_n=(-x)^{n-1}\begin{vmatrix}
0&0&0&\dots&0&1\\
1&0&0&\dots&0&1\\
0&1&0&\dots&0&1\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\
0&0&0&\dots&1&1\end{vmatrix}$$
The characteristic polynomial of this companion matrix is $(-1)^n(-1-t-\dots+t^n)$, and it's determinant is $(-1)^{n-1}$. You have thus $B_n=(-1)^{n-1}(-x)^{n-1}=x^{n-1}$.
So,
$$A_n=(1-x)A_{n-1}+(-1)^{n-1}xx^{n-2}=(1-x)A_{n-1}+(-x)^{n-1}$$
Now, we can prove by induction our claim, that is
$$A_n=(1-x)^n-(-x)^n$$
If it's true for $n-1$, then
$$A_{n-1}=(1-x)^{n-1}-(-x)^{n-1}$$
Then, using our previous relation,
$$A_n=(1-x)\left((1-x)^{n-1}-(-x)^{n-1}\right) + (-x)^{n-1}$$
$$=(1-x)^n-(1-x)(-x)^{n-1}+(-x)^{n-1}$$
$$=(1-x)^n-(-x)^n$$
Then the property is true for $n$. Since it's trivially true for $n=1$ (then $A_1=1$), it's true for all $n$.
A: The derivation is a bit easier if you know the formula for the determinant of a circulant matrix. First, by applying appropriate row operations, we get
$$
\det(A_n)=\begin{vmatrix}
1-x&1&1&\dots&1&1\\
0&1-x&1&\dots&1&1\\
0&0&1-x&\dots&1&1\\
\vdots&\vdots&\ddots&\ddots&\vdots&\vdots\\
0&0&\cdots&0&1-x&1\\
x&x&x&\dots&x&1\end{vmatrix}.
$$
Then, by applying some column operations, we get
$$
\det(A_n)=\begin{vmatrix}
1-x&x\\
&1-x&x\\
&&\ddots&\ddots\\
&&&1-x&x\\
x&&&&1-x\end{vmatrix}.
$$
This is the determinant of a circulant matrix. If we put $\omega=e^{2\pi i/n}$, we have
$$
\det(A_n)
=\prod_{j=0}^{n-1}(1-x+x\omega^j)
=(-x)^n\prod_{j=0}^{n-1}\left(\frac{1-x}{-x}-\omega^j\right)
=(-x)^n\left[\left(\frac{1-x}{-x}\right)^n-1\right],
$$
i.e. $\det(A_n)=(1-x)^n-(-x)^n$.
