Here's the question you clicked on:
liliy
matrix: 2x+3y+ 7z= 13 3x+2y - 5z = -22 5x+7y - 3z = -28
multiply -3 to the first equation multiply 2 to the second equation
idea is to get rid of one variable at a time matrix - even better can you put it in the calculator?
hahah. no calculators.. this is linear algebra
yes, well the idea is still the same except with not using the variables
sorry it has been a long time since I have done linear algebra. I just remember it was my favorite course during my undergrad days....... :)
i dont understand the matrix . can you brief me, i feel like my numbers are always wrong and then the whole thing is wrong
if i get a matrix of numbers and when it simplifies: 1 0 3 8 0 0 9 3 0 0 8 9
these are arbitrarty numbers but if i get two rows that have 0 0 does that mean its bad?
|dw:1338861963247:dw| this is your goal in the end
right, but if you have any rows that are all zeros that means what?
k, like:: 2 4 6 -8 3 6 7 10
this one has a solution x=-1, y=-2, z=3
|dw:1338862146454:dw| you goal is to create that solution form (sorry I can't even remember its official name, may I be forgiven by the math gods) I would multipy the first row by (1/2) that would force that first term to turn into a 1 but don't forget to change all the other numbers in that row
I think that is called row multiplication
http://www.wyzant.com/Help/Math/Precalculus/Systems_Of_Equations/Matrix_Method.aspx
got it, but for the little example i just posted: 2 4 6 -8 3 6 7 10, this matrix, i get the bottom row is 0 0 .. and none of the multiple choices look like that, what am i doing wrong?
not sure, I put it in my calculator and it does have a solution
sorry reposted your question but label it, must use the matrix (linear algebra) good luck :) sorry