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

sumtingwong59
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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?
 
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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$$
\left (\begin{array}{c}
a\\b\\c\\d\\e
\end{array}\right)
=
\left (\begin{array}{c}
-2b+e\\b\\-\frac 2 3 e\\-e\\e
\end{array}\right)
=
b\left (\begin{array}{c}
?\\?\\?\\?\\?
\end{array}\right)
+ e
\left (\begin{array}{c}
?\\?\\?\\?\\?
\end{array}\right)
$$Fill in the ?'s and you will have it.
 
Last edited:
LCKurtz said:
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 into 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?
 
sumtingwong59 said:
Do I just pick any number to plug into 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.
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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