Linear stuff: I never get the awnser

[cmab]

Linear stuff: I never get the awnser...!

[3 1 1 1 0]
[5 -1 1 -1 0]

Where the variables are X1 X2 X3 X4.

I always get somethignt that is far away from the answer.
The answer should be x1= -s x2=-t-s x3=4s x4 = t

Help please!

 

[amcavoy]

[3 1 1 1 0]
[5 -1 1 -1 0]

Where the variables are X1 X2 X3 X4.

I always get somethignt that is far away from the answer.
The answer should be x1= -s x2=-t-s x3=4s x4 = t

Help please!

What did you do to begin?

 

[cmab]

What did you do to begin?

At first, i putted the first number in the frist row as 1.

so i get [1 1/3 1/3 1/3 0]
than i eliminate the 5 to make it a 0.
But its after that i got ****ed!

mayday!

 

[HallsofIvy]

[3 1 1 1 0]
[5 -1 1 -1 0]

Where the variables are X1 X2 X3 X4.

I always get somethignt that is far away from the answer.
The answer should be x1= -s x2=-t-s x3=4s x4 = t

Help please!

What was the question?? I think you have two equations in 4 unknowns and both equations are equal to 0. (It is mildly confusing that you tell us the variables are X1, X2, X3, X4 but then write x1, x2, x3, and x4!)

Do you understand that there are many different ways to write the answer depending on exactly how you do this? It might be that your answers that are "far away from the answer" are, in fact, exactly on the answer!

You could do this by "row reducing" but with a simple problem like this I think just "substitution" is best.

The equations are 3x1+ x2+ x3+ x4= 0 and 5x1- x2+ x3- x4= 0. I might do something like add the two equations and get 8x1+ 2x3= 0 so that x3= -4x1.
Now, I choose (arbitrarily) to make x1= s. Then x3= -4s. Putting those into the two equations I get 3s+ x2- 4x+ x4= 0 or x2+ x4= s and
5s- x2- 4s- x4= 0 or x2+ x4= s. Since those two equations are exactly the same, I choose (again arbitrarily) to let x2= t and solve for x4= s-t.
My solution is x1= s, x2= t, x3= -4s, x4= s-t.

Is that "far away from the answer", which was x1= -s x2=-t-s x3=4s x4 = t? No, not at all. It might make a little more sense if I don't use the same letters as in your given answer: In my answer use "u" instead of "s" and "v" instead of "t". Then my answer is x1= u, x2=v, x3=-4u, x4= u- v. Looking at the "given" answer, I see that x1=u= -s. If I just replace u by -s, I will have both x1= -s and x3= -4u= 4s as in the "given" answer. MY x2 was v while the "given" x2 is -t-s. That would be the same if v= -t-s. In that case, my x4= u- t would become x4= (-s)-(-t-s)= -s+ t+ s= t, exactly as given.

That means that the set of points from my answer and the "given" answer are exactly the same! For example, If you take s= 1, t= 1 in the "given" answer you get x1= -1, x2= -2, x3= 4, x4= 1.
If you take s= -1, v= -2 (in my original form. In terms of u, v, I am taking u= -s= -1, v= -t-s= -2.) so that x= s= -1, x2= t= -2, x3=-4s= 4, x4= s-t= 1, just as before.

Because there are 2 (independent) equations in 4 unknowns, we have 4-2= 2 "degrees of freedom". We are free to choose any two parameters, arbitrarily, and write x1, x2, x3, x4 in terms of those two parameters. Of course, the equations you get will depend upon those arbitrary choices.

 

[cmab]

Thanks bud, i'll look at it later on. If i have any problem, i'll post.