# determinant computing

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

• It looks like you're mixing up $a_1-a_2$ and $a_2-a_1$. – B. Goddard Jan 8 '18 at 12:04
• You have two contradicting statements: At the very end: "We deduce that $D_n=a_1(a_1-a_2)^{n-1}$" (which is correct), and at the beginning: "But I found that $D_n=a_1(a_2-a_1)^{n-1}$" (which is wrong). – Martin R Jan 8 '18 at 12:09
• @B.Goddard I did $C1-C2$ so $a_1-a_2$ – Stu Jan 8 '18 at 12:09
• Then there is no mistake, $a_1(a_1-a_2)^{n-1}$ and $(-1)^{n-1}a_1(a_2-a_1)^{n-1}$ are identical – Martin R Jan 8 '18 at 12:13
• Focus: $(a_1-a_2)^{n-1} = (-1)^{n-1}(a_2-a_1)$. – B. Goddard Jan 8 '18 at 12:13

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 $$C_{n-1}\leftarrow C_{n-1}-C_{n}$$.