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  1. MHB Apprentice

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    #1
    I'm a little stuck here. I should determine the following determinant. I first tried to simplify it a little by using elemntary transformations. And then I did Laplace expansion on the last row.

    $\begin{vmatrix}2 & 2 & \cdots & 2 & 2 & 1 \\ 2 & 2 & \cdots & 2 & 2 & 2 \\ 2 & 2 & \cdots & 3 & 2 & 2 \\ \vdots & \vdots & \quad & \vdots & \vdots & \vdots \\ 2 & n-1 & \cdots & 2 & 2 & 2 \\ n & 2 & \cdots & 2 & 2 & 2\end{vmatrix}=\begin{vmatrix}1 & 1 & \cdots & 1 & 1 & 1 \\ 0 & 0 & \cdots & 0 & 0 & 2 \\ 0 & 0 & \cdots & 1 & 0 & 2 \\ \vdots & \vdots & \quad & \vdots & \vdots & \vdots \\ 0 & n-3 & \cdots & 0 & 0 & 2 \\ n-2 & 0 & \cdots & 0 & 0 & 2\end{vmatrix} \\ =(-1)^{n+1} \cdot (n-2) \cdot \begin{vmatrix}1 & 1 & \cdots & 1 & 1 & 1 \\ 0 & 0 & \cdots & 0 & 0 & 2 \\ 0 & 0 & \cdots & 1 & 0 & 2 \\ \vdots & \vdots & \quad & \vdots & \vdots & \vdots \\ 0 & n-4 & \cdots & 0 & 0 & 2 \\ n-3 & 0 & \cdots & 0 & 0 & 2\end{vmatrix} + (-1)^{2n} \cdot 2 \cdot \begin{vmatrix}1 & 1 & \cdots & 1 & 1 & 1 \\ 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & \cdots & 0 & 1 & 0 \\ \vdots & \vdots & \quad & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & n-3 & \cdots & 0 & 0 & 0\end{vmatrix}$.

    The second determinant is equal to 0. And the first determinant is of the same form as the initial, just one degree smaller.
    We have recursion here. I used labels:

    $D_n=\begin{vmatrix}2 & 2 & \cdots & 2 & 2 & 1 \\ 2 & 2 & \cdots & 2 & 2 & 2 \\ 2 & 2 & \cdots & 3 & 2 & 2 \\ \vdots & \vdots & \quad & \vdots & \vdots & \vdots \\ 2 & n-1 & \cdots & 2 & 2 & 2 \\ n & 2 & \cdots & 2 & 2 & 2\end{vmatrix}=\begin{vmatrix}1 & 1 & \cdots & 1 & 1 & 1 \\ 0 & 0 & \cdots & 0 & 0 & 2 \\ 0 & 0 & \cdots & 1 & 0 & 2 \\ \vdots & \vdots & \quad & \vdots & \vdots & \vdots \\ 0 & n-3 & \cdots & 0 & 0 & 2 \\ n-2 & 0 & \cdots & 0 & 0 & 2\end{vmatrix}$

    I have:

    $D_n=(-1)^{n+1} \cdot (n-2) \cdot D_{n-1}$.

    I wasn't sure how to continue this recursion though. I've got something like:

    $2 \cdot (-1)^? \cdot (n-2)!$

    I am not sure what I should get as exponent of -1. I've tried a few things but everything that I've got was different from the solution that my tutor gave us. Yet he said that the solution might be incorrect.
    Those recursions are just so confusing to me. I would appreciate it if someone could explain me how to determine that exponent.

  2. MHB Apprentice

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    #2
    Quote Originally Posted by karseme View Post
    I'm a little stuck here. I should determine the following determinant. I first tried to simplify it a little by using elemntary transformations. And then I did Laplace expansion on the last row.

    $\begin{vmatrix}2 & 2 & \cdots & 2 & 2 & 1 \\ 2 & 2 & \cdots & 2 & 2 & 2 \\ 2 & 2 & \cdots & 3 & 2 & 2 \\ \vdots & \vdots & \quad & \vdots & \vdots & \vdots \\ 2 & n-1 & \cdots & 2 & 2 & 2 \\ n & 2 & \cdots & 2 & 2 & 2\end{vmatrix}=\begin{vmatrix}1 & 1 & \cdots & 1 & 1 & 1 \\ 0 & 0 & \cdots & 0 & 0 & 2 \\ 0 & 0 & \cdots & 1 & 0 & 2 \\ \vdots & \vdots & \quad & \vdots & \vdots & \vdots \\ 0 & n-3 & \cdots & 0 & 0 & 2 \\ n-2 & 0 & \cdots & 0 & 0 & 2\end{vmatrix} \\ =(-1)^{n+1} \cdot (n-2) \cdot \begin{vmatrix}1 & 1 & \cdots & 1 & 1 & 1 \\ 0 & 0 & \cdots & 0 & 0 & 2 \\ 0 & 0 & \cdots & 1 & 0 & 2 \\ \vdots & \vdots & \quad & \vdots & \vdots & \vdots \\ 0 & n-4 & \cdots & 0 & 0 & 2 \\ n-3 & 0 & \cdots & 0 & 0 & 2\end{vmatrix} + (-1)^{2n} \cdot 2 \cdot \begin{vmatrix}1 & 1 & \cdots & 1 & 1 & 1 \\ 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & \cdots & 0 & 1 & 0 \\ \vdots & \vdots & \quad & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & n-3 & \cdots & 0 & 0 & 0\end{vmatrix}$.

    The second determinant is equal to 0. And the first determinant is of the same form as the initial, just one degree smaller.
    We have recursion here. I used labels:

    $D_n=\begin{vmatrix}2 & 2 & \cdots & 2 & 2 & 1 \\ 2 & 2 & \cdots & 2 & 2 & 2 \\ 2 & 2 & \cdots & 3 & 2 & 2 \\ \vdots & \vdots & \quad & \vdots & \vdots & \vdots \\ 2 & n-1 & \cdots & 2 & 2 & 2 \\ n & 2 & \cdots & 2 & 2 & 2\end{vmatrix}=\begin{vmatrix}1 & 1 & \cdots & 1 & 1 & 1 \\ 0 & 0 & \cdots & 0 & 0 & 2 \\ 0 & 0 & \cdots & 1 & 0 & 2 \\ \vdots & \vdots & \quad & \vdots & \vdots & \vdots \\ 0 & n-3 & \cdots & 0 & 0 & 2 \\ n-2 & 0 & \cdots & 0 & 0 & 2\end{vmatrix}$

    I have:

    $D_n=(-1)^{n+1} \cdot (n-2) \cdot D_{n-1}$.

    I wasn't sure how to continue this recursion though. I've got something like:

    $2 \cdot (-1)^? \cdot (n-2)!$

    I am not sure what I should get as exponent of -1. I've tried a few things but everything that I've got was different from the solution that my tutor gave us. Yet he said that the solution might be incorrect.
    Those recursions are just so confusing to me. I would appreciate it if someone could explain me how to determine that exponent.
    I did this by changing the matrix to the one where the anti diagonal becomes the diagonal. The number of swaps needed to do this is n/2 if n is even and (n-1)/2 if n is odd. Each swap contributes -1 so we want the sign to be + when n or n-1 is a multiple of 4 this means the power of -1 is n(n-1)/2. Hope this helps

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