# Determinant

• Oct 13th 2012, 11:43 AM
Tala
Determinant
How do I find the determinant for this matrix:

first row: (x+y 3 2 1)
second row: ( 0 1 0 0)
third row: ( 0 2 x+y 3)
fourth row: ( 0 3 0 1)

(x,y) ∈ R2

Any ideas ?
• Oct 13th 2012, 12:48 PM
Soroban
Re: Determinant
Hello, Tala!

Are you waiting for Divine Intervention?
You haven't been taught to evaluate determinants in general?

Quote:

$\text{Evluate: }\:\begin{vmatrix}\:x+y & 3 & 2 & 1\: \\ 0 & 1 & 0 & 0\: \\ 0 & 2 & x+y & 3\: \\ 0 & 3 & 0 & 1\: \end{vmatrix}$

Reading down the first column and using cofactors:

$D \;=\;(x+y)\begin{vmatrix}1&0&0\\2&x\!+\!y&3\\3&0&1 \end{vmatrix} \,-\, 0\begin{vmatrix}3&2&1\\2&x\!+\!y&3\\3&0&1 \end{vmatrix} \,+\, 0\begin{vmatrix}3&2&1\\1&0&0\\3&0&1\end{vmatrix} \,-\, 0\begin{vmatrix}3&2&1 \\ 1&0&0 \\ 2&x\!+\!y&3 \end{vmatrix}$

$D \;=\;(x+y)\big[1(x+y-0) - 0 + 0\big] \:-\: 0 \:+\: 0 \:-\: 0$

$D \;=\;(x+y)(x+y)$

$D \;=\;(x+y)^2$
• Oct 13th 2012, 01:42 PM
Tala
Re: Determinant
No I actually haven't been taught to evaluate determinants in general only by Gauss-Jordan elimination. I do know how to find the determinant and I got the same result as you. I'm just wondering if there's anything smart I can do with the function ?
• Oct 13th 2012, 02:12 PM
HallsofIvy
Re: Determinant
Okay, by Gauss-Jordan elimination.

Since the first column is $\begin{bmatrix}x+y \\ 0 \\ 0 \\ 0\end{bmatrix}$, we don't have to do anything to the first column. To get "0"s below the diagonal in the second column, subtract twice the second row from the third row and subtract three times the second row from the fourth row:
$\begin{bmatrix}x+ y & 3 & 2 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & x+ y & 3\\ 0 & 0 & 0 & 1\end{bmatrix}$

Since that is now a triangular matrix, we can practically "read" the determinant of the diagonal. That was easy, wasn't it?
• Oct 13th 2012, 02:25 PM
Tala
Re: Determinant
So the function (x+y)2 is the determinant ?