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Exercise 1.6.13

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We can substract the second equation by the two times of the first equation. And then we have

\begin{array}{l} x_{1} - 2 x_{2} + x_{3} & = 0 \\ x_{2} - x_{3} & = 0 \end{array}

Let $x_{3} = s$ and hence $x_{2} = s$ and $x_{1} = s$ . We have the solution would be ${(s, s, s) = s (1, 1, 1) : s \in ℝ}$ . And the basis would be ${(1, 1, 1)}$ .

jephianlin

2011-06-27 00:00

Exercise 1.6.13

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