# Help solving linear system

#### amcgl064

##### New member
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.

#### Sudharaka

##### Well-known member
MHB Math Helper
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.
Hi amcgl064,

The answer you have obtained for the row echelon form is incorrect. The correct answer is,

$\left(\begin{matrix}1&-1&-1&4\\0&1&\frac{1}{2}&-1\\0&0&0&0\end{matrix}\right)$

Please refer >>this<< for a basic introduction about basic variables and free variables. I hope you can do the rest.

Kind Regards,
Sudharaka.