Finding general solution of a linear system

[dumbledore]

Find the general solution to the following system of equations and indicate which variables are free and which are basic.

Putting it in augmented matrix form to start we have:



1 -1 -1 4 | -3
1 0 -1/2 3 | -1
1 1 0 2 | 1

Now performing the following fundamental row operations:

R1<-->R2
R2+R3-->R2
-2R3+R2-->R2
-R3+R1-->R3
R2/-2
R2+R3-->R2
-3R3+R1-->R1

And finally I end with the augmented matrix:

1 0 -2 0 | 5
0 1 0 0 | 0
0 0 -1/2 1 |-2

Can someone please tell me if I got the correct matrix at the end and if so how do I determine which variables are free and which are basic?

Thank you.

 

[mmm4444bot]

Find the general solution to the following system of equations and indicate which variables are free and which are basic.

Putting it in augmented matrix form to start we have:

1 -1 -1 4 | -3
1 0 -1/2 3 | -1
1 1 0 2 | 1

Now performing the following fundamental row operations:

R1<-->R2
R2+R3-->R2 Why this one?
-2R3+R2-->R2
-R3+R1-->R3 This is not a valid row operation.

A row may be replaced by adding a multiple of another row to it.

We may not replace a row with a multiple of itself when adding another row to it.

You determine which variables are free or basic by examining the solution with respect to the definitions of free and basic variables. Did you look up their definitions, yet? :cool:

 

[HallsofIvy]

Are you required to do this with matrices?

The given equations are equivalent to \(\displaystyle x_1- x_2- x_3+ 4x_4= -3\), \(\displaystyle x_1+ \frac{5}{2}x_4= -1\), and \(\displaystyle x_1+ x_2+ 2x_4= 1\). If you subtract the first equation from the second, you get \(\displaystyle x_2+ x_3-\frac{3}{2}x_4= 2\). If you subtract the first equation from the third, you get \(\displaystyle 2x_2+ x_3- 2x_4= 4\). Now, subtract the first of those two equations from the second to finally get \(\displaystyle x_2- \frac{1}{2}x_4= 0\) so that \(\displaystyle x_4= 2x_2\). You can then replace \(\displaystyle x_4\) back into either of the previous equations to find \(\displaystyle x_3\) in terms of \(\displaystyle x_2\) also and then put those into any of the original equations to find \(\displaystyle x_1\) in terms of \(\displaystyle x_2\).