# Thread: formula for matrix determinant

1. ## formula for matrix determinant

please compute the determinant of this matrix..

$\displaystyle \left[ { \begin{array}{ccccc} k & 1 & 0 & \ldots & 0 \\ 1 & k & 1 & \ldots & 0 \\ 0 & 1 & k & \ldots & 0 \\ \vdots & \, & \ddots & \ddots & \vdots \\ 0 & 0 & \ldots & 1 & k \\ \end{array} } \right]$

can someone give the general formula.. thanks!!

2. Originally Posted by kalagota
please compute the determinant of this matrix..

$\displaystyle \left[ { \begin{array}{ccccc} k & 1 & 0 & \ldots & 0 \\ 1 & k & 1 & \ldots & 0 \\ 0 & 1 & k & \ldots & 0 \\ \vdots & \, & \ddots & \ddots & \vdots \\ 0 & 0 & \ldots & 1 & k \\ \end{array} } \right]$

can someone give the general formula.. thanks!!
Develop a recursive form for this determinant, for the $\displaystyle n \times n$ case put::

$\displaystyle K_n=\left[ { \begin{array}{ccccc} k & 1 & 0 & \ldots & 0 \\ 1 & k & 1 & \ldots & 0 \\ 0 & 1 & k & \ldots & 0 \\ \vdots & \, & \ddots & \ddots & \vdots \\ 0 & 0 & \ldots & 1 & k \\ \end{array} } \right]$

The expand the determinant about the 1 st row:

$\displaystyle \det(K_n) = k \det(K_{n-1}) - 1 \det (L_{n-1})$

where:

$\displaystyle L_n=\left[ { \begin{array}{ccccc} 1 & 1 & 0 & \ldots & 0 \\ 0 & k & 1 & \ldots & 0 \\ 0 & 1 & k & \ldots & 0 \\ \vdots & \, & \ddots & \ddots & \vdots \\ 0 & 0 & \ldots & 1 & k \\ \end{array} } \right]$

Now expand the determinant of $\displaystyle L_n$ about the first column:

$\displaystyle \det(L_n) = \det(K_{n-1})$

so we have:

$\displaystyle \det(K_n) = k \det(K_{n-1}) - 1 \det (K_{n-2})$

and hopefully you can take it from there.

ZB

3. Originally Posted by Constatine11
Develop a recursive form for this determinant, for the $\displaystyle n \times n$ case put::

$\displaystyle K_n=\left[ { \begin{array}{ccccc} k & 1 & 0 & \ldots & 0 \\ 1 & k & 1 & \ldots & 0 \\ 0 & 1 & k & \ldots & 0 \\ \vdots & \, & \ddots & \ddots & \vdots \\ 0 & 0 & \ldots & 1 & k \\ \end{array} } \right]$

The expand the determinant about the 1 st row:

$\displaystyle \det(K_n) = k \det(K_{n-1}) - 1 \det (L_{n-1})$

where:

$\displaystyle L_n=\left[ { \begin{array}{ccccc} 1 & 1 & 0 & \ldots & 0 \\ 0 & k & 1 & \ldots & 0 \\ 0 & 1 & k & \ldots & 0 \\ \vdots & \, & \ddots & \ddots & \vdots \\ 0 & 0 & \ldots & 1 & k \\ \end{array} } \right]$

Now expand the determinant of $\displaystyle L_n$ about the first column:

$\displaystyle \det(L_n) = \det(K_n-1)$

so we have:

$\displaystyle \det(K_n) = k \det(K_{n-1}) - 1 \det (K_{n-2})$

and hopefully you can take it from there.

ZB

? i can't see how you got $\displaystyle L_n$.. can you show it? thanks

4. Originally Posted by kalagota
? i can't see how you got $\displaystyle L_n$.. can you show it? thanks
Delete the second column and first row from $\displaystyle K_{n+1}$.

( $\displaystyle -\det(L_n)$ is the cofactor of the $\displaystyle 1$ in the first row of $\displaystyle K_{n+1}$ )

ZB

5. Originally Posted by Constatine11
Delete the second column and first row from $\displaystyle K_{n+1}$.

( $\displaystyle -\det(L_n)$ is the cofactor of the $\displaystyle 1$ in the first row of $\displaystyle K_{n+1}$ )

ZB
ok.. i'm just used to do the expansion about rows.. thanks anyway.. (although its too late.. Ü)

6. Originally Posted by kalagota
ok.. i'm just used to do the expansion about rows.. thanks anyway.. (although its too late.. Ü)
By the way I have just coded up the solution to the recurrence and compared it to the determinant of some low dimensional examples and they agree.

ZB