Determinant of matrix $A\in\mathbb{R}^{n,n}$

Question

How to prove that for matrix $A\in\mathbb{R}^{n,n}$ we have $$\det A = \det \begin{vmatrix} x & x& x& ...&x&x\\ 1-x& 1&1 & ...&1& 1\\ 0& 1-x& 1& ...&1&1\\ 0& 0&1-x& ...&1&1\\ &&\ldots&& \\ 0& 0& 0& ...&1-x&1\\ \end{vmatrix} =x^n $$

Answer 1

Let the determinant of $A\in\mathbb R^{n\times n}$ be a polynomial in $x$ denoted as $P_n(x); P_1(x)=\det[x]=x$. Expand along the first column,

$\det A=P_n(x)= \begin{vmatrix} x& x & x & \ldots & x& x \\ 1-x & 1 & 1&\ldots &1& 1 \\ 0 & 1-x & 1 & \ldots &1& 1 \\ 0 & 0 & 1-x & \ldots &1& 1 \\ \vdots&\vdots&\vdots&&\vdots&\vdots\\ 0 & 0 & 0 & \ldots&1-x&1 \notag \end{vmatrix}_{n}\\=x\begin{vmatrix} 1& 1 &\ldots&1 \\ 1-x & 1&\ldots &1\\ 0 & 1-x& \ldots &1\\ 0 & 0& \ldots &1\\ \vdots&\vdots&&\vdots\\ 0 & 0& \ldots&1-x\notag \end{vmatrix}_{n-1}+(x-1)\begin{vmatrix} x& x&\ldots & x\\ 1-x & 1&\ldots &1\\ 0 & 1-x& \ldots &1\\ 0 & 0& \ldots &1\\ \vdots&\vdots&&\vdots\\ 0 & 0 & \ldots&1-x\notag \end{vmatrix}_{n-1}$

Note that the first determinant is just $\frac {P_{n-1}(x)}x$ and the second one $P_{n-1}(x)$.

$\implies P_n(x)=x\cdot\frac{P_{n-1}(x)}x+(x-1)P_{n-1}(x)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =xP_{n-1}(x)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =x^2P_{n-2}(x)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =x^3P_{n-3}(x)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =x^{n-1}P_{n-(n-1)}(x)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =x^n$

Answer 2

Subtract every column other than the last by the last column. Then move the last column to the first (this gives you a factor of $(-1)^{n-1}$ in the determinant). The result is a lower triangular matrix whose main diagonal is $(x,-x,-x,\ldots,-x)$. Thus the determinant of the original matrix is $(-1)^{n-1}x(-x)^{n-1}=x^n$.