Show that B = {[1 2 -1], [1 1 1]} is a basis for the plane -3x1 + 2x2 + x3 = 0.

Question

Show that [latex]\mathfrak{B} = \left\{\left [ \begin{matrix} 1 \ 2 \ -1 \end{matrix} ight ] , \left [ \begin{matrix} 1 \ 1 \ 1 \end{matrix} ight ] ight\}[/latex] is a basis for the plane [latex] -3x_{1} + 2x_{2} + x_{3} = 0[/latex].

Accepted Answer

To prove that B is a basis for the plane, we must show that B is linearly independent and spans the plane.
We first observe that B is clearly linearly independent since neither vector is a scalar multiple of the other.
For spanning, first observe that any vector [latex]\vec{x} [/latex] in the plane must satisfy the condition of the plane. Hence, every vector in the plane has the form

[latex] \vec{x} = \left [ \begin{matrix} x_{1} \ x_{2} \ 3x_{1} - 2x_{2} \end{matrix} ight ][/latex]

since [latex] x_{3} = 3x_{1} - 2x_{2}[/latex]. Therefore, we now just need to show that the equation

[latex] t_{1} \left [ \begin{matrix} 1 \ 2 \ -1 \end{matrix} ight ] + t_{2} \left [ \begin{matrix} 1 \ 1 \ 1 \end{matrix} ight ] = \left [ \begin{matrix} x_{1} \ x_{2} \ 3x_{1} - 2x_{2} \end{matrix} ight ][/latex]

is always consistent. Row reducing the corresponding augmented matrix gives

[latex] \left [ \begin{matrix} 1 & 1 \ 2 & 1 \ -1 & 1 \end{matrix} \left | \begin{matrix} x_{1} \ x_{2} \ 3x_{1} - 2x_{2} \end{matrix} ight. ight ] hicksim \left [ \begin{matrix} 1 & 1 \ 0 & 1 \ 0 & 0 \end{matrix} \left | \begin{matrix} x_{1} \ 2x_{1} - x_{2} \ 0 \end{matrix} ight. ight ][/latex]

The system is consistent for all [latex]x_{1}, x_{2} \in \mathbb{R}[/latex] and hence B also spans the plane.

Question 2.3.8: Show that B = {[1 2 -1], [1 1 1]} is a basis for the plane -...

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