# Linear: Find a set of basic solutions and show as linear combination

1. Jan 28, 2015

### sumtingwong59

1. The problem statement, all variables and given/known data
Find a set of basic solutions and express the general solution as a linear combination of these basic solutions

a + 2b - c + 2d + e = 0
a + 2b + 2c + e = 0
2a + 4b - 2c + 3d + e = 0

2. Relevant equations
3. The attempt at a solution

i reduced it to:
1 2 0 0 -1 0
0 0 1 0 2/3 0
0 0 0 1 1 0

a = -2s + t
c = -2/3t
d = -t

I'm just not sure how i find solutions now. It could be literally anything could it not?

2. Jan 28, 2015

### LCKurtz

Assuming your arithmetic is correct (I didn't check), you have let the free variables $b = s$ and $e = t$. I'm going to leave them as $b$ and $e$. Write your solution as$$\left (\begin{array}{c} a\\b\\c\\d\\e \end{array}\right) = \left (\begin{array}{c} -2b+e\\b\\-\frac 2 3 e\\-e\\e \end{array}\right) = b\left (\begin{array}{c} ?\\?\\?\\?\\? \end{array}\right) + e \left (\begin{array}{c} ?\\?\\?\\?\\? \end{array}\right)$$Fill in the ?'s and you will have it.

Last edited: Jan 29, 2015
3. Jan 28, 2015

### sumtingwong59

Do I just pick any number to plug in to b and e, and then pick different numbers and plug them into the c bracket?

4. Jan 28, 2015

### sumtingwong59

Do I just want to make it so each equation equals 0?

5. Jan 29, 2015

### LCKurtz

The $c$ in front of the last bracket was a typo; I have corrected it to $e$. Fill in the ?'s and there is nothing left to do. Every solution can be gotten for some value of $b$ and $e$. That is the general solution.

Last edited: Jan 29, 2015