determinant computing

Question

Find the determinant of $A=\begin{pmatrix}a_1&a_2&\cdots&a_n\\a_1&a_1&\ddots&\vdots\\\vdots&\ddots&\ddots&a_2\\a_1&\cdots&a_1&a_1\end{pmatrix}$

The correction gives me $D_n=\color{red}{(-1)^{n-1}}a_1(a_2-a_1)^{n-1}$

But I found $D_n=a_1(a_1-a_2)^{n-1}$ So where is y mistake?

My attempt : $D_n=\begin{array}{|cccc|}a_1&a_2&\cdots&a_n\\a_1&a_1&\ddots&\vdots\\\vdots&\ddots&\ddots&a_2\\a_1&\cdots&a_1&a_1\end{array}=\begin{array}{|cccc|}a_1-a_2&a_2&\cdots&a_n\\0&a_1&\ddots&\vdots\\\vdots&\ddots&\ddots&a_2\\0&\cdots&a_1&a_1\end{array}=(a_1-a_2)\cdot\begin{array}{|cccc|}1&a_2&\cdots&a_n\\0&a_1&\ddots&\vdots\\\vdots&\ddots&\ddots&a_2\\0&\cdots&a_1&a_1\end{array}$

$D_n=(a_1-a_2)\cdot (-1)^2\cdot D_{n-1}=(a_1-a_2)D_{n-1}$. Since $D_1=a_1$

We deduce $D_n=a_1(a_1-a_2)^{n-1}$

Answer 1

As pointed out in the comments, there is no mistake since $$ (a_1-a_2)^{n-1}=\left((-1)(a_2-a_1)\right)^{n-1}=(-1)^{n-1}(a_2-a_1) ^{n-1}. $$ An other way to derive the result is to do the substitutions $C_i\leftarrow C_i-C_{i+1}$, starting from $C_1\leftarrow C_1-C_{2}$ and finishing by $C_{n-1}\leftarrow C_{n-1}-C_{n}$.