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$$ \begin{vmatrix} 2 & 2 & ... & 2 & 2 & 1 \\ 2 & 2 & ... & 2 & 2 & 2 \\ 2 & 2 & ... & 3 & 2 & 2 \\ ... & ... & ... & ... & ... & ... \\ 2 & n-1 & ... & 2 & 2 & 2 \\ n & 2 & ... & 2 & 2 & 2 \end{vmatrix} $$ I got this in my linear algebra homework. In the task, it is required to find the determinant of a matrix by the method of representing the sum of determinants. By that I mean this property of determinants: $$ \begin{vmatrix} a & b+e \\ c & d+f \\ \end{vmatrix} = \begin{vmatrix} a & b \\ c & d \\ \end{vmatrix} + \begin{vmatrix} a & e \\ c & f \\ \end{vmatrix} $$

What I have tried:

  1. Add the i-th with the (i-1)-th;
  2. Add the last row to all, getting two determinants, one of which is 0 (because of the (n-1)-st column);
  3. Tried to get n on the diagonal, this is what I got: $$ \begin{vmatrix} 2 & 3 & ... & n-1 & n & n \\ 2 & 3 & ... & n-1 & n & n+1 \\ 2 & 3 & ... & n & n & n+1 \\ ... & ... & ... & ... & ... & ... \\ 2 & n & ... & n-1 & n & n+1 \\ n & 3 & ... & n-1 & n & n+1 \end{vmatrix} $$ The closest one to mine from StackExchange was this one. But I didn't manage to link these two determinants.

No matter how I transform it, nothing worked for me. Any ideas?

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Let's do some elementary operations on rows and columns: $$\begin{align} \begin{vmatrix} 2 & 2 & ... & 2 & 2 & 1 \\ 2 & 2 & ... & 2 & 2 & 2 \\ 2 & 2 & ... & 3 & 2 & 2 \\ ... & ... & ... & ... & ... & ... \\ 2 & n-1 & ... & 2 & 2 & 2 \\ n & 2 & ... & 2 & 2 & 2 \end{vmatrix} &\overset{1}= \begin{vmatrix} 2 & 2 & ... & 2 & \color{red}2\downarrow & 1 \\ 0 & 0 & ... & 0 & 0 & 1 \\ 0 & 0 & ... & 1 & 0 & 1 \\ ... & ... & ... & ... & ... & ... \\ 0 & n-3 & ... & 0 & 0 & 1 \\ n-2 & 0 & ... & 0 & 0 & 1 \end{vmatrix} \\[2mm] &\overset{2}= 2\begin{vmatrix} 0 & 0 & ... & 0 & \leftarrow\color{red}1 \\ 0 & 0 & ... & 1 & 1 \\ ... & ... & ... ... & ... & ... \\ 0 & n-3 & ... & 0 & 1 \\ n-2 & 0 & ... & 0 & 1 \end{vmatrix} \\[2mm] &\overset{3}= (-1)^{n-1}2\begin{vmatrix} 0 & 0 & ... & 1 \\ ... & ... & ... ... & ... \\ 0 & n-3 & ... & 0 \\ n-2 & 0 & ... & 0 \end{vmatrix} \\[2mm] &\overset{4}= (-1)^{n-1 + \frac{(n-2)(n-3)}{2}}2\cdot (n-2)! \end{align}$$ Explanation:
$1)$ Subtract the first row from all others
$2)$ Expand through $(n-1)$-th column
$3)$ Expand through the first row
$4)$ Calculate the anti-diagonal determinant

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  • $\begingroup$ @Christoph Thanks. I hope it is fixed now. $\endgroup$
    – VIVID
    May 4 at 5:54
  • $\begingroup$ Sorry, I was mistaken, the missing factor should have been $(-1)^{\left\lfloor\tfrac{n-2}2\right\rfloor} = (-1)^{\binom{n-2}{2}}$. $\endgroup$
    – Christoph
    May 4 at 5:58
  • $\begingroup$ @Christoph Is my current expression correct? $\endgroup$
    – VIVID
    May 4 at 6:00
  • $\begingroup$ Looks good to me! $\endgroup$
    – Christoph
    May 4 at 6:00
  • $\begingroup$ @Christoph Thank you very much! $\endgroup$
    – VIVID
    May 4 at 6:01

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