Thread: Help solving linear equation system

1. Help solving linear equation system

Problem:
Using only elementary row operations, solve the following system of linear equations:

{3x - y + 5z + 4w = -11}
{-2x + y -3z -3w = 8}

My steps: I keep going in circles. I can't seem to get rid of enough variables to make a dent in this.

R1 +R2 -> R1 {x + 2z + w = -3}
{-2x + y -3 z - 3w = 8}

R2 + 2R1 -> R2 {x +2z +w = -3}
{y + z - w = 2}

R1 - 2R2 -> R1 {x - 2y + 3w = -1}
{y + z - w = 2}

2R2 -> R2 {x - 2y + 3w = -1}
{2y + 2z + 2w = 4}

R1 + R2 -> R1 {x + 2z + w = 3}
{2y + 2z - 2w = 4}

R2 - R1 -> R2 {x + 2z + w = 3}
{-x + 2y = 1}

R1 + R2 -> R2 {x + 2z + w = 3}
{2y + 2z + w = 2}

Do you see how I'm going in circles? Am I even taking the correct first step?

2. Hello,

You should know what you mean by "solving" the system of linear equations.
Since there are only 2 equations for 4 variables,
the space of solutions is 2-dimensional.
(If there were 2 equations for 2 variables, the solution space would often be 0-dimensional, yielding only one solution.)
In this case, the solution space is parametrized by 2 variables.
Assuming we take z and w for parameters,
you are asked to express (x, y, z, w) in terms of z and w.

R1 +R2 -> R1
{x + 2z + w = -3}
{-2x + y -3 z - 3w = 8}

R2 + 2R1 -> R2
{x +2z +w = -3}
{y + z - w = 2}

Thus,
(x, y, z, w)=(-2z-w-3, -z+w+2, z, w)
is the solution.
The solution space is
{(-3, 2, 0, 0) + z(-2, -1, 1, 0) + w(-1, 1, 0, 1)| z, w: arbitrary}.

Bye.

3. Thank you so much. I didn't even put 2 and 2 together to realize that this required a parametric solution. Thanks again!