# Show that $\det D = a \det D_{n-1}- b^2 \det D_{n-1}$ for $n = 2, 3,\dots$ [closed]

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 $$n = 2, 3,\dots$$

## 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

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• 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. – 5xum Apr 9 at 9:33
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• Are you sure that the formula is right? Should there be a $+$ or a $-$ in there? – Theo Bendit Apr 9 at 9:40
• I asked the question beacause i really have now idea how to solve this problem, I just need some help getting started with it. – SonCOR Apr 9 at 9:44
• It shuld be right now @TheoBendit – SonCOR Apr 9 at 9:45

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