# Thread: formula for matrix determinant

1. ## formula for matrix determinant

please compute the determinant of this matrix..

$
\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..

$
\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 $n \times n$ case put::

$
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:

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

where:

$
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 $L_n$ about the first column:

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

so we have:

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

and hopefully you can take it from there.

ZB

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

$
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:

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

where:

$
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 $L_n$ about the first column:

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

so we have:

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

and hopefully you can take it from there.

ZB

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

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

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

ZB

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

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

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