Find a basis and the dimension of the solution space of the homogeneous system

$x_{1} + 2x_{2} + 2x_{3} + x_{4} + 4x_{5} = 0 \\$ $3x_{1} + 7x_{2} + 7x_{3} + 3x_{4} + 13x_{5} = 0 \\$ $2x_{1} + 5x_{2} + 5x_{3} + 2x_{4} + 9x_{5} = 0$

Question

Find a basis and the dimension of the solution space of the homogeneous system

[latex]x_{1} + 2x_{2} + 2x_{3} + x_{4} + 4x_{5} = 0 \[/latex] [latex]3x_{1} + 7x_{2} + 7x_{3} + 3x_{4} + 13x_{5} = 0 \[/latex] [latex]2x_{1} + 5x_{2} + 5x_{3} + 2x_{4} + 9x_{5} = 0[/latex]

Accepted Answer

We found in Exercise 2.2.4 that the general solution of this system is

[latex]\vec{x} = t_{1} \left [ \begin{matrix} 0 \ -1 \ 1 \ 0 \ 0 \end{matrix} ight ] + t_{2} \left [ \begin{matrix} -1 \ 0 \ 0 \ 1 \ 0 \end{matrix} ight ] + t_{3} \left [ \begin{matrix} -2 \ -1 \ 0 \ 0 \ 1 \end{matrix} ight ] , \ \ \ \ \ [/latex] [latex] t_{1}, t_{2}, t_{3} \in \mathbb{R}[/latex]

This shows that a spanning set for the solution space is

[latex]\mathfrak{B} = \left\{\left [ \begin{matrix} 0 \ -1 \ 1 \ 0 \ 0 \end{matrix} ight ] , \left [ \begin{matrix} -1 \ 0 \ 0 \ 1 \ 0 \end{matrix} ight ] , \left [ \begin{matrix} -2 \ -1 \ 0 \ 0 \ 1 \end{matrix} ight ] ight\}[/latex]

It is not difficult to verify that B is also linearly independent. Consequently, B is a basis for the solution space, and hence the dimension of the solution space is 3.

Question 2.3.10: Find a basis and the dimension of the solution space of the ...

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