# Help solving a linear system

## Main Question or Discussion Point

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

x_1 + 4x_4 + 3 = x_2 + x_3
x_1 + 3x_4 + 1 = \frac{1}{2}x_3
x_1 + x_2 + 2x_4 = 1

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.

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Simon Bridge
Homework Helper
You were so close to getting the LaTeX - all you needed was a double-hash on each side of each line:

$x_1 + 4x_4 + 3 = x_2 + x_3$
$x_1 + 3x_4 + 1 = \frac{1}{2}x_3$
$x_1 + x_2 + 2x_4 = 1$

The secret is to think about what you know at each stage...
You have four unknowns and three equations - so what does that mean?

Putting it in augmented matrix form to start we have:

$$\left ( \begin{array}{cccc} 1 & -1 & -1 & 4 \\ 2 & 0 & -1 & 6\\ 1 & 1 & 0 & 2 \end{array}\right | \left . \begin{array}{c} -3\\-2\\1 \end{array}\right )$$
...I took the liberty of doubling row 2 to get rid of the annoying fraction.
The row-reductions to echelon form follows more easily:
Can someone please tell me if I got the correct matrix at the end...
R2 <--- R2-2R1; R3 <--- R3-R1