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My problem

Consider the matrix $$ D_n = \underbrace{\begin{bmatrix} a & b & 0 & \cdots & 0 \\ b & a & b & \ddots & \vdots \\ 0 & b & a & \ddots & 0 \\ \vdots & \ddots & \ddots & \ddots & b\\ 0 & \cdots & 0 & b & a \\ \end{bmatrix}}_{n \text{ columns}} $$ Show that $\det D_{n} = a \det D_{n-1} - b^2 \det D_{n-2}$ for $n = 2, 3,\dots$

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closed as off-topic by GNUSupporter 8964民主女神 地下教會, Saad, callculus, Adrian Keister, Alexander Gruber Apr 9 at 16:19

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – GNUSupporter 8964民主女神 地下教會, Saad, callculus, Adrian Keister, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for. $\endgroup$ – 5xum Apr 9 at 9:33
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    $\begingroup$ Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote. $\endgroup$ – 5xum Apr 9 at 9:33
  • $\begingroup$ Are you sure that the formula is right? Should there be a $+$ or a $-$ in there? $\endgroup$ – Theo Bendit Apr 9 at 9:40
  • $\begingroup$ I asked the question beacause i really have now idea how to solve this problem, I just need some help getting started with it. $\endgroup$ – SonCOR Apr 9 at 9:44
  • $\begingroup$ It shuld be right now @TheoBendit $\endgroup$ – SonCOR Apr 9 at 9:45
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Write $D_n$ in two equivalent ways: $$D_n = \begin{bmatrix} a & \begin{matrix}b & 0 & 0 & \cdots & 0\end{matrix} \\ \begin{matrix}b \\ 0 \\ 0 \\ \vdots \\ 0 \end{matrix} & D_{n-1} \end{bmatrix} = \begin{bmatrix} a & b & \begin{matrix}0 & 0 & \cdots & 0\end{matrix} \\ b & a & \begin{matrix}b & 0 & \cdots & 0\end{matrix} \\ \begin{matrix} 0 \\ 0 \\ \vdots \\ 0 \end{matrix} & \begin{matrix} b \\ 0 \\ \vdots \\ 0 \end{matrix} & D_{n-2} \end{bmatrix}.$$

If we expand the cofactors down the first column, from the first of the two matrices, our first term will be $aD_{n-1}$. For the second term, we appeal to the second representation; we get $$-b \det \begin{bmatrix} b & \begin{matrix}0 & 0 & 0 & \cdots & 0\end{matrix} \\ \begin{matrix}b \\ 0 \\ 0 \\ \vdots \\ 0 \end{matrix} & D_{n-2} \end{bmatrix},$$ which we can expand along the top row, to get $$-b^2 \det D_{n-2}.$$ The rest of the column is $0$, so our total determinant is $$\det D_n = a \det D_{n-1} - b^2 \det D_{n - 2}.$$

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The recursion should be $\text{det }D_n = a\text{det }D_{n-1}-b^2\text{det}D_{n-2}$. Try expanding the determinant of $D_n$ over the first row. The answer follows!

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  • $\begingroup$ $$ det D_{n-2} : b * det \underbrace{\begin{bmatrix} a & b & 0 & \cdots \\ b & a & b & \ddots \\ 0 & b & a & \ddots \\ 0 & \cdots & 0 & b \\ \end{bmatrix}}_{n \text{ columns}} $$ $\endgroup$ – SonCOR Apr 9 at 10:04
  • $\begingroup$ One more hint: $D_n = \begin{bmatrix}a & b & 0 \ldots & 0\\ b &\\ 0 \\ \vdots& & D_{n-1}\\ 0\end{bmatrix}$. $\endgroup$ – Geethu Joseph Apr 9 at 10:05
  • $\begingroup$ Then:. $$ b * b * det \underbrace{\begin{bmatrix} a & b & 0 \\ b & a & b \\ 0 & b & a \\ 0 & 0 & 0 \\ \end{bmatrix}}_{n \text{ columns}} $$ = b^{2}det D_{n-2} $\endgroup$ – SonCOR Apr 9 at 10:07

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