Gauss & Gauss-Jordan method

[foukr]

Hello,

solve the following set of equations using Gauss and Gauss-Jordan elimination.

This is my set of equations:

x +3y+4z = 0
2x+2y+2z = 0
2y+3z = 0
2x -2y -4z = 0

I have this solution but it is correct? I dont have any possibilities to check my solution.

 

[Deleted member 4993]

Hello,

solve the following set of equations using Gauss and Gauss-Jordan elimination.

This is my set of equations:

x +3y+4z = 0
2x+2y+2z = 0
2y+3z = 0
2x -2y -4z = 0

I have this solution but it is correct? I dont have any possibilities to check my solution.
View attachment 14114

You have 4 equations and 3 unknowns.

However, out of 4 equations only two are independent. So there are no unique solutions.

 

[HallsofIvy]

"I don't have any possibility of checking my solution". Why not? You can certainly check if the x, y, z you got satisfy the equations by putting those values in the equations. But what solution did you get? I see that you have y= 0 and z= 0 but I see no value for x. What did you get for x? If y= z= 0 then your equations become 2x= 0, 2x= 0, 0= 0, and x= 0. It is easy to see that x= y= z= 0 is a solution but, as Subhotosh Kahn said, since you have reduced to only two independent equations in three unknowns, there are, in fact, infinitely many solutions.

From your reduced matrix, you have x+ 3y+ 4z= 0 and -4y- 6z= 0. From -4y- 6z= 0, z= -(2/3)y. So x+ 3y+ 4z= x+ 3y- (8/3)y= x+ (1/3)y= 0 so x= -(1/3)y. We can write the solution set as x= -(1/3)t, y= t, z= -(2/3)t for any t or, if you don't like fractions, replacing t by 3s, x= -s, y= 3s, z= -2s. You can think of the set of all solutions as a straight line in an xyz coordinate system, passing through the origin.

 

[Steven G]

Hello,

solve the following set of equations using Gauss and Gauss-Jordan elimination.

This is my set of equations:

x +3y+4z = 0
2x+2y+2z = 0
2y+3z = 0
2x -2y -4z = 0

I have this solution but it is correct? I dont have any possibilities to check my solution.
View attachment 14114

To make life easier when you are reducing you can (and in my opinion should!) try to make the numbers smaller by (in this particular case) dividing rows 2 and row 4 by 2.