SOLUTION: Evaluate the determinant, using row or column operations whenever possible to simplify your work. [0 0 4 6] [2 1 1 3] [2 1 2 3] [3 0 1 7]

Algebra ->  Matrices-and-determiminant -> SOLUTION: Evaluate the determinant, using row or column operations whenever possible to simplify your work. [0 0 4 6] [2 1 1 3] [2 1 2 3] [3 0 1 7]       Log On


   

Question 1129413: Evaluate the determinant, using row or column operations whenever possible to simplify your work.
[0 0 4 6]
[2 1 1 3]
[2 1 2 3]
[3 0 1 7]
Found 2 solutions by t0hierry, Alan3354:
Answer by t0hierry(194) About Me  (Show Source):
You can put this solution on YOUR website!

[0 0 4 6]
[2 1 1 3]
[2 1 2 3]
[3 0 1 7]
We start from the first row and we get:
4* [2 1 3]
[2 1 3]
[3 0 7]
6* [2 1 1]
[2 1 2]
[3 0 1]
again from last row for the first term
3*0 + 7*0 = 0
same for the second term
3*1 + 0 = 3
Now for the sign: it's minus
so I find -3.

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Evaluate the determinant, using row or column operations whenever possible to simplify your work.
[0 0 4 6]
[2 1 1 3]
[2 1 2 3]
[3 0 1 7]
---------------
Row 1 has 2 zeroes, makes it easy.
----
4 times
|2   1   3|
|2   1   3|
|3   0   7|
2 equal rows --> 0
4*0 = 0
========================
-6 times
|2   1   1|
|2   1   2|
|3   0   1|
= -6*(2*(1-0) - 1*(2-6) + 1*(0-3))
= -6*(2 + 4 - 3)
= -18