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The matrix $A_n\in\mathbb{R}^{n\times n}$ is given by

$$\left[a_{i,j}\right] = \left\lbrace\begin{array}{cc} 1 & i=j \\ -j & i = j+1\\ i & i = j-1 \\ 0 & \text{other cases} \end{array} \right.$$

Question: How to calculate $\det{A}$ formally correct?


My attempt: Take a look at $A_n$ $$A_n=\left(\begin{array}{cc|cc} && 0 & 0 \\ & A_{n-2}&n-2&0 \\ \hline\\ 0&-n+2&1&n-1 \\ 0&0&-n+1&1 \end{array}\right)$$

By Laplace for the determinat now follows:

$$\det{A_n}=\det{\left(\begin{array}{cc|cc} && 0 & 0 \\ & A_{n-2}&n-2&0 \\ \hline\\ 0&-n+2&1&n-1 \\ 0&0&-n+1&1 \end{array}\right)} =\\= \det{\left(\begin{array}{cc|c} && 0 \\ & A_{n-2}&n-2 \\ \hline\\ 0&-n+2&1 \end{array}\right)}+{\left(n-1\right)\cdot \det{\left(\begin{array}{cc|cc} && 0 \\ & A_{n-2}&0 \\ \hline\\ 0&-n+2&n-1 \\ \end{array}\right)}}=\\= \det{A_{n-1}}+\left(n-1\right)^2\cdot\det{A_{n-2}} $$

This should be the correct result, isn't it? However, but the main question is: Is this proof mathematical ok, i.e. is it formally correct? Or what alternative do I have to calculate the determinant?

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  • $\begingroup$ This is all correct, but you are not yet finished, you just have a recursion. And I am not sure what you mean with "formally correct". It would be nice to specify that you are applying Laplace to the last row, but this is an issue of legibility not of correctness. $\endgroup$ – Phira Jan 11 '14 at 0:05
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It is correct, but you can simplify a little bit:

$$\det{A_n} = \det{\left(\begin{array}{cc|c} && 0 \\ & A_{n-1}&n-1 \\ \hline\\ 0&-n+1&1 \end{array}\right)}= \det{A_{n-1}}+\left(n-1\right)^2\cdot\det{A_{n-2}} $$

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Looks good to me, but you should compare and contrast with the Wikipedia derivation of the recurrence for the determinant of tridiagonal matrices.

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