- #1

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:

Solve the following system of equations:

[tex]2x_{1} - 2x_{2} - 3x_{3} = -2[/tex]

[tex]3x_{1} - 3x_{2} - 2x_{3} + 5x_{4} = 7[/tex]

[tex]x_{1} - x_{2} - 2x_{3} - x_{4} = -3[/tex]

So, clearly there won't be a unique solution. I do the work and get this matrix:

[tex]

\[ \left( \begin{array}{cccc|c}

1 & -1 & 0 & 0 & 5 \\

0 & 0 & 1 & 0 & 4 \\

0 & 0 & 0 & 1 & 0 \end{array} \right)\]

[/tex]

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 [tex]\in \mathbb{R}[/tex]}

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!

**Homework Statement**Solve the following system of equations:

[tex]2x_{1} - 2x_{2} - 3x_{3} = -2[/tex]

[tex]3x_{1} - 3x_{2} - 2x_{3} + 5x_{4} = 7[/tex]

[tex]x_{1} - x_{2} - 2x_{3} - x_{4} = -3[/tex]

**The attempt at a solution**So, clearly there won't be a unique solution. I do the work and get this matrix:

[tex]

\[ \left( \begin{array}{cccc|c}

1 & -1 & 0 & 0 & 5 \\

0 & 0 & 1 & 0 & 4 \\

0 & 0 & 0 & 1 & 0 \end{array} \right)\]

[/tex]

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 [tex]\in \mathbb{R}[/tex]}

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!

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