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

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  • $\begingroup$ It looks like you're mixing up $a_1-a_2$ and $a_2-a_1$. $\endgroup$
    – B. Goddard
    Jan 8, 2018 at 12:04
  • $\begingroup$ 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). $\endgroup$
    – Martin R
    Jan 8, 2018 at 12:09
  • $\begingroup$ @B.Goddard I did $C1-C2$ so $a_1-a_2$ $\endgroup$
    – Stu
    Jan 8, 2018 at 12:09
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    $\begingroup$ 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 $\endgroup$
    – Martin R
    Jan 8, 2018 at 12:13
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    $\begingroup$ Focus: $(a_1-a_2)^{n-1} = (-1)^{n-1}(a_2-a_1)$. $\endgroup$
    – B. Goddard
    Jan 8, 2018 at 12:13

1 Answer 1

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

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