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How to show that the determinant of the following $(n\times n)$ matrix

$$\begin{pmatrix} 5 & 2 & 0 & 0 & 0 & \cdots & 0 \\ 2 & 5 & 2 & 0 & 0 & \cdots & 0 \\ 0 & 2 & 5 & 2 & 0 & \cdots & 0 \\ \vdots & \vdots& \vdots& \vdots & \vdots & \vdots & \vdots \\ 0 & \cdots & \cdots & 0 & 2 & 5 & 2 \\ 0 & \cdots & \cdots & \cdots & \cdots & 2 & 5 \end{pmatrix}$$

is equal to $\frac13(4^{n+1}-1)$?


More generally:

How does one compute the determinant of the following tridiagonal matrix (where the three diagonals are constant)?

$$M_n(a,b,c) = \begin{pmatrix} a & b & 0 & 0 & 0 & \cdots & 0 \\ c & a & b & 0 & 0 & \cdots & 0 \\ 0 & c & a & b & 0 & \cdots & 0 \\ \vdots & \vdots& \vdots& \vdots & \vdots& \vdots & \vdots \\ 0 & \cdots & \cdots & 0 & c & a & b \\ 0 & \cdots & \cdots & \cdots & \cdots & c & a \end{pmatrix}$$

Here $a,b,c$ can be taken to be real numbers, or complex numbers.

In other words, the matrix $M_n(a,b,c) = (m_{ij})_{1 \le i,j \le n}$ is such that $$m_{ij} = \begin{cases} a & i = j, \\ b & i = j - 1, \\ c & i = j + 1, \\ 0 & \text{otherwise.} \end{cases}$$

There does not seem to be an easy pattern to use induction: the matrix is not a diagonal block matrix of the type $M = \bigl(\begin{smallmatrix} A & C \\ 0 & B \end{smallmatrix}\bigr)$ (where we could use $\det(M) = \det(A) \det(B)$ for the induction step), and there are no lines or columns with only one nonzero entry, so Laplace expansion gets complicated quickly.

Is there a general pattern that one could use? Or is the answer only known on a case-by-case basis? It's possible to compute the determinant by hand for small $n$:

$$\begin{align} \det(M_1(a,b,c)) & = \begin{vmatrix} a \end{vmatrix} = a \\ \det(M_2(a,b,c)) & = \begin{vmatrix} a & b \\ c & a \end{vmatrix} = a^2 - bc \\ \det(M_3(a,b,c)) & = \begin{vmatrix} a & b & 0 \\ c & a & b \\ 0 & c & a \end{vmatrix} = a^3 - 2abc \end{align}$$

But there is no readily apparent pattern and the computation becomes very difficult when $n$ gets large.

  • 0
    I would suggest induction. It's going to be a little fiddly though.2012-12-29
  • 0
    Induction would work using [block determinants](http://en.wikipedia.org/wiki/Determinant#Block_matrices).2012-12-30
  • 0
    A more general case, where the diagonals aren't constant, is discussed [here](http://math.stackexchange.com/questions/575748/determinant-of-symmetric-tridiagonal-matrices).2015-02-15
  • 1
    The determinant of such a matrix is sometimes called a [(generalized) continuant](http://en.wikipedia.org/wiki/Continuant_(mathematics)).2015-02-17

5 Answers 5