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anonymous
 5 years ago
GaussJordan Reduction
Solve the following set of homogeneous equations by GaussJordan reduction of the matrix of coefficients (without the column of zeros from the righthand side.
5x+5y5z=0
3x+4y7z=0
2y8z=0
2y3y+6z=0
I know to do GaussJordan reduction for three equations, but not four. How do I do this  I'm pretty confused
anonymous
 5 years ago
GaussJordan Reduction Solve the following set of homogeneous equations by GaussJordan reduction of the matrix of coefficients (without the column of zeros from the righthand side. 5x+5y5z=0 3x+4y7z=0 2y8z=0 2y3y+6z=0 I know to do GaussJordan reduction for three equations, but not four. How do I do this  I'm pretty confused

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hi princessanna, how do you use GaussJordan reduction to solve three equations, do you use matrix and elementary row operations?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hi sklee, yes I use the matrix method. But I can't use it unless the matrix is a square, which it isn't for this question.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the system of equations can be written in a matrix form: 5 5 5 3 4 7 0 2 8 2 3 6

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hi princessanna, it doesn't matter if the matrix is square or not, it is still applicable.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well I don't know how to do the method :( because I only know the way when all the 1's are diagonal?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, can i assume you have no problem to do the elementary row operations? As i will show you only the matrix that has been reduced, is that fine for you?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes that would be great, thank you :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01 1 1 0 1 4 0 0 0 0 0 0 The operations that i have used are R_{1}(1/5), R_{3}(1/2), R_{1,2}(3), R_{1,4}(2), R_{2,3}(1) and R_{2,4}(1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's pretty hard to read and understand :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0From this reduced form, you can use backward substitution (that is called Gauss elimination method) or you can reduce more till it is in row reduced form. 1 0 3 0 1 4 0 0 0 0 0 0 The operation that has been used is R_{2,1}(1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0R_{1} (1/5) means you multiply row 1 with 1/5 R_{3} (1/2) means you multiply row 3 with 1/2 R_{1,2} (3) means you multiply row 1 with 3 and add it to row 2 to get the new row 2 etc..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is it better now? Do they helpful to you?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0By letting z = a, where a is any parameter, we get x = 3z = 3a and y = 4z = 4a. So the solution is (x,y,z) = (3a,4a,a) = a(3,4,1)
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