Question

Gauss-Jordan Reduction
Solve the following set of homogeneous equations by Gauss-Jordan reduction of the matrix of coefficients (without the column of zeros from the right-hand side.

5x+5y-5z=0
3x+4y-7z=0
2y-8z=0
-2y-3y+6z=0

I know to do Gauss-Jordan reduction for three equations, but not four. How do I do this - I'm pretty confused

Answer 1

Hi princessanna, how do you use Gauss-Jordan reduction to solve three equations, do you use matrix and elementary row operations?

Answer 2

Hi 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.

Answer 3

the system of equations can be written in a matrix form:

5 5 -5
3 4 -7
0 2 -8
-2 -3 6

Answer 4

Hi princessanna, it doesn't matter if the matrix is square or not, it is still applicable.

Answer 5

Well I don't know how to do the method :( because I only know the way when all the 1's are diagonal?

Answer 6

Ok, 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?

Answer 7

Yes that would be great, thank you :)

Answer 8

1 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)

Answer 9

It's pretty hard to read and understand :(

Answer 10

From 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)

Answer 11

R_{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..

Answer 12

Is it better now? Do they helpful to you?

Answer 13

By 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)