formula for matrix determinant

• Dec 2nd 2007, 09:55 PM
kalagota
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!!
• Dec 5th 2007, 04:38 AM
Constatine11
Quote:

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
• Dec 5th 2007, 05:37 AM
kalagota
Quote:

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
• Dec 5th 2007, 06:04 AM
Constatine11
Quote:

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
• Dec 5th 2007, 06:14 AM
kalagota
Quote:

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)
• Dec 5th 2007, 07:55 AM
Constatine11
Quote:

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