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) ∈ R^{2 Any ideas ? }
Hello, Tala!
Are you waiting for Divine Intervention?
You haven't been taught to evaluate determinants in general?
$\displaystyle \text{Evluate: }\:\begin{vmatrix}\:x+y & 3 & 2 & 1\: \\ 0 & 1 & 0 & 0\: \\ 0 & 2 & x+y & 3\: \\ 0 & 3 & 0 & 1\: \end{vmatrix}$
Reading down the first column and using cofactors:
$\displaystyle D \;=\;(x+y)\begin{vmatrix}1&0&0\\2&x\!+\!y&3\\3&0&1 \end{vmatrix} \,-\, 0\begin{vmatrix}3&2&1\\2&x\!+\!y&3\\3&0&1 \end{vmatrix} \,+\, 0\begin{vmatrix}3&2&1\\1&0&0\\3&0&1\end{vmatrix} \,-\, 0\begin{vmatrix}3&2&1 \\ 1&0&0 \\ 2&x\!+\!y&3 \end{vmatrix}$
$\displaystyle D \;=\;(x+y)\big[1(x+y-0) - 0 + 0\big] \:-\: 0 \:+\: 0 \:-\: 0$
$\displaystyle D \;=\;(x+y)(x+y)$
$\displaystyle D \;=\;(x+y)^2$
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 ?
Okay, by Gauss-Jordan elimination.
Since the first column is $\displaystyle \begin{bmatrix}x+y \\ 0 \\ 0 \\ 0\end{bmatrix}$, 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:
$\displaystyle \begin{bmatrix}x+ y & 3 & 2 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & x+ y & 3\\ 0 & 0 & 0 & 1\end{bmatrix}$
Since that is now a triangular matrix, we can practically "read" the determinant of the diagonal. That was easy, wasn't it?