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

## Homework Statement

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

## Homework Equations

3. The Attempt at a Solution [/B]
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?

## Answers and Replies

LCKurtz
Science Advisor
Homework Helper
Gold Member
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.

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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$$\left (\begin{array}{c} a\\b\\c\\d\\e \end{array}\right) = \left (\begin{array}{c} -2b+e\\b\\-\frac 2 3 e\\-e\\e \end{array}\right) = b\left (\begin{array}{c} ?\\?\\?\\?\\? \end{array}\right) + c \left (\begin{array}{c} ?\\?\\?\\?\\? \end{array}\right)$$Fill in the ?'s and you will have it.

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?

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

LCKurtz
Science Advisor
Homework Helper
Gold Member
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?

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.

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