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I have found this exercise in a book, and having troubles solving it:

How to calculate this determinant?

$$\det\begin{pmatrix} 5 & 6 & 0 & 0 & 0 & \cdots & 0 \\ 4 & 5 & 2 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 3 & 2 & 0 & \cdots & 0 \\ 0 & 0 & 1 & 3 & 2 & \cdots & \vdots \\ 0 & 0 & 0 & 1 & \ddots & \ddots & 0 \\ & & & & \ddots & 3 & 2 \\ 0 & 0 & 0 & \cdots & 0 & 1 & 3 \\ \end{pmatrix} _{n\times n}$$

Thanks!!

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do you know that a determinant can be solved by multiplying the diagonal entries only? In this case its just 5*5*3^N-2 –  Raul Dec 14 '13 at 17:27
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This is for triangular matrices, Raul... –  Jeremy Daniel Dec 14 '13 at 17:29
    
@JeremyDaniel i admit i wrote it wrong, I meant by multiplying the diagonals and adding them together, + for the diagonals taken from the left and - from the right. –  Raul Dec 14 '13 at 17:46
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4 Answers

Develop the determinant on the last line. You should obtain a linear recurrence of order $2$.

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You can do row operations to eliminate all nonzero elements below the diagonal. Then the determinant is the product of the remaining elements on the diagonal.

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HINT. This is a tridiagonal matrix. Try writing doing a few small cases, $3 \times 3$, $4 \times 4, \cdots$, and see if you can't notice a pattern in how you do one determinant to the next size determinant. Then apply that pattern to the matrix at hand (using cofactors will be most helpful!).

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In general, when calculating a determinant of tridiagonal matrix of this form $D_n= \begin{vmatrix} c& b & 0 & 0 & \ldots & 0 \\ a& c & b & 0 & \ldots & 0 \\ 0& a & c & b & \ldots & 0 \\ \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & \ldots & 0 & a & c & b \\ 0 & \ldots & 0 & 0 & a & c \end{vmatrix}$
you can use Laplace expansion twice to obtain recurrent relation.

$D_n= \begin{vmatrix} c& b & 0 & 0 & \ldots & 0 \\ a& c & b & 0 & \ldots & 0 \\ 0& a & c & b & \ldots & 0 \\ \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & \ldots & 0 & a & c & ab \\ 0 & \ldots & 0 & 0 & a & c \end{vmatrix} \overset{(1)}= cD_{n-1} - b \begin{vmatrix} a & 0 & 0 & \ldots & 0 \\ a & c & b & \ldots & 0 \\ \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & \ldots & a & c & ab \\ 0 & \ldots & 0 & a & c \end{vmatrix} \overset{(2)}= cD_{n-1} - ab D_{n-2}$

The equality (1) is Laplace for the first row and the equality (2) is Laplace for the first column.


This is not exactly the situation from the original post, but we can modify it to a very similar situation using reflection w.r.t vertical and horizontal axis. This can be done by reverting the order of rows and then the order of columns. Since this is the same as doing even number of exchanges of rows/columns, this does not change the determinant. So to solve the original question you simply have to solve the following recurrence:

$D_n= \begin{vmatrix} 3& 1 & 0 & 0 & \ldots & 0 \\ 2& 3 & 1 & 0 & \ldots & 0 \\ 0& 2 & 3 & 1 & \ldots & 0 \\ \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & \ldots & 0 & 2 & 5 & 4 \\ 0 & \ldots & 0 & 0 & 6 & 5 \end{vmatrix}$

$D_3= \begin{bmatrix} 3 & 1 & 0 \\ 2 & 5 & 4 \\ 0 & 6 & 5 \end{bmatrix} $

$D_4= \begin{bmatrix} 3 & 1 & 0 & 0\\ 2 & 3 & 1 & 0 \\ 0 & 2 & 5 & 4 \\ 0 & 0 & 6 & 5 \end{bmatrix} $

$D_n=3D_{n-1}-2D_{n-2}$

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