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