Question

In Exercises1 through 18, determine whether the vector $\overrightarrow{\mathbf{x}}$ is in the span V of the vectors $\overrightarrow{{\mathbf{v}}_{\mathbf{1}}}\mathbf{,}\mathbf{,}\mathbf{,}\mathbf{,}\mathbf{,}\overrightarrow{{\mathbf{v}}_{\mathbf{m}}}$(proceed “by inspection” if possible, and use the reduced row-echelon form if necessary). If localid="1664192105355" $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{x}}$is in V, find the coordinates of $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{x}}$with respect to the basis $\mathfrak{I}\mathbf{=}\stackrel{\mathbf{\rightharpoonup }}{{\mathbf{v}}_{\mathbf{1}}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\stackrel{\mathbf{\rightharpoonup }}{{\mathbf{v}}_{\mathbf{m}}}$of V, and write the coordinate vector $\left[\overrightarrow{x}\right]\mathfrak{I}$.

localid="1664193284963" $\overrightarrow{\mathbf{x}}\mathbf{=}\mathbf{\left[}\begin{array}{c}\mathbf{1}\\ \mathbf{1}\\ \mathbf{1}\\ \mathbf{-}\mathbf{1}\end{array}\mathbf{\right]}\mathbf{;}\overrightarrow{{\mathbf{v}}_{\mathbf{1}}}\mathbf{=}\mathbf{\left[}\begin{array}{c}\mathbf{1}\\ \mathbf{0}\\ \mathbf{2}\\ \mathbf{0}\end{array}\mathbf{\right]}\mathbf{;}\overrightarrow{{\mathbf{v}}_{\mathbf{2}}}\mathbf{=}\mathbf{\left[}\begin{array}{c}\mathbf{0}\\ \mathbf{1}\\ \mathbf{3}\\ \mathbf{0}\end{array}\mathbf{\right]}\mathbf{;}\overrightarrow{{\mathbf{v}}_{\mathbf{3}}}\mathbf{=}\mathbf{\left[}\begin{array}{c}\mathbf{0}\\ \mathbf{0}\\ \mathbf{4}\\ \mathbf{1}\end{array}\mathbf{\right]}$

Short answer

The coordinates of the vector is, $\stackrel{\rightharpoonup }{x}\mathfrak{I}=\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]$.

Step-by-step solution

Consider the vectors.

The vectors are,

$\overrightarrow{x}=\left[\begin{array}{c}1\\ 1\\ 1\\ -1\end{array}\right];\overrightarrow{v_1}=\left[\begin{array}{c}1\\ 0\\ 2\\ 0\end{array}\right];\overrightarrow{v_2}=\left[\begin{array}{c}0\\ 1\\ 3\\ 0\end{array}\right];\overrightarrow{v_3}=\left[\begin{array}{c}0\\ 0\\ 4\\ 1\end{array}\right]$

Check for the coordinates of the vector.

The formula is, $\stackrel{\rightharpoonup }{x}=c_1\stackrel{\rightharpoonup }{v_1}+c_2\stackrel{\rightharpoonup }{v_2}$.

By inspection method,

$\left[\begin{array}{c}1\\ 0\\ 2\\ 0\end{array}\right]+\left[\begin{array}{c}0\\ 1\\ 3\\ 0\end{array}\right]-\left[\begin{array}{c}0\\ 0\\ 4\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ 1\\ -1\end{array}\right]$

Substitute these values in the formula.

role="math" localid="1664193247857" $$\begin{gathered} \stackrel{\rightharpoonup }{x}=c_1\stackrel{\rightharpoonup }{v_1}+c_2\stackrel{\rightharpoonup }{v_2} \\ \therefore \stackrel{\rightharpoonup }{x}\mathfrak{I}=\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right] \end{gathered}$$

Final answer

The coordinates of the vector is, $\stackrel{\rightharpoonup }{x}\mathfrak{I}=\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]$.