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Exercise 3.3.1

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We have

\begin{gathered} [\begin{matrix} 2 & 4 & 6 & 4 & b_{1} \\ 2 & 5 & 7 & 6 & b_{2} \\ 2 & 3 & 5 & 2 & b_{3} \end{matrix}] \to [\begin{matrix} 2 & 4 & 6 & 4 & b_{1} \\ 0 & 1 & 1 & 2 & b_{2} - b_{1} \\ 0 & - 1 & - 1 & - 2 & b_{3} - b_{1} \end{matrix}] \to [\begin{matrix} 2 & 4 & 6 & 4 & b_{1} \\ 0 & 1 & 1 & 2 & b_{2} - b_{1} \\ 0 & 0 & 0 & 0 & b_{3} + b_{2} - {2 b}_{1} \end{matrix}] . \end{gathered}

$A x = b$ has a solution when $b_{3} + b_{2} - 2 b_{1} = 0$ ; the column space contains all combinations of $(2, 2, 2)$ and $(4, 5, 3) .$ This is the plane $b_{3} + b_{2} - 2 b_{1} = 0$ (!). The nullspace contains all combinations of $s_{1} = (- 1, - 1, 1, 0)$ and $s_{2} = (2, - 2, 0, 1); x_{complete} =$ $x_{p} + c_{1} s_{1} + c_{2} s_{2}$

[R d] = [\begin{matrix} 1 & 0 & 1 & - 2 & 4 \\ 0 & 1 & 1 & 2 & - 1 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}]

gives the particular solution $x_{p} = (4, - 1, 0, 0)$ .

mathchakra

2022-01-23 14:18

Exercise 3.3.1

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