How to express solution to system of equations

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SUMMARY

The discussion focuses on solving a system of equations using linear algebra techniques. The user presents a system with three equations and four variables, leading to a matrix representation that indicates a non-unique solution. The correct solution format provided in the textbook includes parameters r and s, indicating the presence of free variables. The user seeks clarification on how to derive the solution set from the reduced row echelon form of the matrix.

PREREQUISITES
  • Understanding of linear algebra concepts, specifically systems of equations.
  • Familiarity with matrix operations and row reduction techniques.
  • Knowledge of vector spaces and parameterization of solutions.
  • Ability to interpret solutions in the context of free variables.
NEXT STEPS
  • Study the process of converting a system of equations into reduced row echelon form (RREF).
  • Learn about parameterization of solutions in linear algebra, focusing on free variables.
  • Explore examples of non-unique solutions in linear systems using MATLAB or Python.
  • Review the concept of vector spaces and linear combinations in the context of solution sets.
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Students in linear algebra courses, educators teaching systems of equations, and anyone looking to deepen their understanding of matrix theory and solution parameterization.

dewert
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Sorry if this is too basic for this forum, but it IS from a 2nd-year linear algebra course. I'm probably just being stupid, and missing something obvious, but here goes:

Homework Statement

Solve the following system of equations:

2x_{1} - 2x_{2} - 3x_{3} = -2
3x_{1} - 3x_{2} - 2x_{3} + 5x_{4} = 7
x_{1} - x_{2} - 2x_{3} - x_{4} = -3

The attempt at a solution

So, clearly there won't be a unique solution. I do the work and get this matrix:
<br /> \[ \left( \begin{array}{cccc|c}<br /> 1 &amp; -1 &amp; 0 &amp; 0 &amp; 5 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 &amp; 4 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1 &amp; 0 \end{array} \right)\]<br />

However, the answer in the back of the book is
{r(1,1,0,0) + s(-3,0,-2,1) + (5,0,4,0) | r,s \in \mathbb{R}}

My question is how to get this from the solution matrix. Regardless of whether the solution is right or wrong, I don't know where this is coming from. I can conceivably see that (5,0,4,0) can be obtained by choosing x_2 = 0, but that's about it.

Thanks!
 
Last edited:
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Are you sure you don't get something else, like, say,
<br /> \[ \left( \begin{array}{cccc|c}<br /> 2 &amp; -2 &amp; 0 &amp; 3 &amp; 5 \\<br /> 0 &amp; 0 &amp; 1 &amp; 2 &amp; 4 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \end{array} \right)\]<br />
 

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