Question

Find a basis of the subspace of ${\mathbf{ℝ}}^{\mathbf{3}}$defined by the equation$\mathbf{2}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}\mathbf{3}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}{\mathbf{x}}_{\mathbf{3}}\mathbf{=}\mathbf{0}\mathbf{.}$

Short answer

The basis of the subspace V of ${\mathrm{ℝ}}^3$defined by the equation $2{\mathrm{x}}_1+3{\mathrm{x}}_2+{\mathrm{x}}_3=0$is $\left\{\left[\begin{array}{c}-3\\ 2\\ 0\end{array}\right],\left[\begin{array}{c}-1\\ 0\\ 2\end{array}\right]\right\}$.

Step-by-step solution

Finding the basis of 2x1+3x2+x3=0

Let V be a subspace${\mathrm{ℝ}}^3$ such that

$\mathrm{V}=\{({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3)\in {\mathrm{ℝ}}^3:2{\mathrm{x}}_1+3{\mathrm{x}}_2+{\mathrm{x}}_3=0\}$

$\Rightarrow \mathrm{V}=\{(2{\mathrm{x}}_1,2{\mathrm{x}}_2,2{\mathrm{x}}_3)\in {\mathrm{ℝ}}^3:2{\mathrm{x}}_1+3{\mathrm{x}}_2+{\mathrm{x}}_3=0\}$

$\Rightarrow \mathrm{V}=\{(-3{\mathrm{x}}_2-{\mathrm{x}}_3,2{\mathrm{x}}_2,2{\mathrm{x}}_3)\in {\mathrm{ℝ}}^3:2{\mathrm{x}}_1=-3{\mathrm{x}}_2-{\mathrm{x}}_3\}$

$\Rightarrow V=\left\{x_2\left[\begin{array}{c}-3\\ 2\\ 0\end{array}\right]+x_3\left[\begin{array}{c}-1\\ 0\\ 2\end{array}\right]\right\}$

$\Rightarrow$Basis of V=$\left\{\left[\begin{array}{c}-3\\ 2\\ 0\end{array}\right],\left[\begin{array}{c}-1\\ 0\\ 2\end{array}\right]\right\}$

Final Answer

The basis of the subspace of ${\mathrm{ℝ}}^3$defined by the equation $2x_1+3x_2+x_3=0$is

$\left\{\left[\begin{array}{c}-3\\ 2\\ 0\end{array}\right],\left[\begin{array}{c}-1\\ 0\\ 2\end{array}\right]\right\}$