Question

In Exercises1 through 18, determine whether the vector $\overrightarrow{\mathbf{x}}$ is in the span V of the vectors $\stackrel{\mathbf{\rightharpoonup }}{{\mathbf{v}}_{\mathbf{1}}}\mathbf{,}\mathbf{,}\mathbf{,}\mathbf{,}\stackrel{\mathbf{\rightharpoonup }}{{\mathbf{v}}_{\mathbf{m}}}$(proceed “by inspection” if possible, and use the reduced row-echelon form if necessary). If $\overrightarrow{\mathbf{x}}$is in V, find the coordinates of $\overrightarrow{\mathbf{x}}$with respect to the basis role="math" localid="1664193712120" $\mathfrak{I}\mathbf{=}\stackrel{\mathbf{\rightharpoonup }}{{\mathbf{v}}_{\mathbf{1}}}\mathbf{,}\mathbf{,}\mathbf{,}\mathbf{,}\stackrel{\mathbf{\rightharpoonup }}{{\mathbf{v}}_{\mathbf{m}}}$of V, and write the coordinate vector $\left[\stackrel{\rightharpoonup }{x}\right]\mathfrak{I}$.

role="math" localid="1664193998673" $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{x}}\mathbf{=}\mathbf{\left[}\begin{array}{c}\mathbf{5}\\ \mathbf{4}\\ \mathbf{3}\\ \mathbf{2}\end{array}\mathbf{\right]}\mathbf{;}\stackrel{\mathbf{\rightharpoonup }}{{\mathbf{v}}_{\mathbf{1}}}\mathbf{=}\mathbf{\left[}\begin{array}{c}\mathbf{1}\\ \mathbf{1}\\ \mathbf{0}\\ \mathbf{1}\end{array}\mathbf{\right]}\mathbf{;}\stackrel{\mathbf{\rightharpoonup }}{{\mathbf{v}}_{\mathbf{2}}}\mathbf{=}\mathbf{\left[}\begin{array}{c}\mathbf{0}\\ \mathbf{1}\\ \mathbf{1}\\ \mathbf{0}\end{array}\mathbf{\right]}\mathbf{;}\stackrel{\mathbf{\rightharpoonup }}{{\mathbf{v}}_{\mathbf{3}}}\mathbf{=}\mathbf{\left[}\begin{array}{c}\mathbf{0}\\ \mathbf{-}\mathbf{1}\\ \mathbf{0}\\ \mathbf{1}\end{array}\mathbf{\right]}$

Short answer

As the coordinates of the vector are not available, so, $\stackrel{\rightharpoonup }{x}\mathfrak{I}$ isn’t in V.

Step-by-step solution

Consider the vectors.

The vectors are,

$\stackrel{\rightharpoonup }{\mathrm{x}}=\left[\begin{array}{c}5\\ 4\\ 3\\ 2\end{array}\right];\stackrel{\rightharpoonup }{{\mathrm{v}}_1}=\left[\begin{array}{c}1\\ 1\\ 0\\ 1\end{array}\right];\stackrel{\rightharpoonup }{{\mathrm{v}}_2}=\left[\begin{array}{c}0\\ 1\\ 1\\ 0\end{array}\right];\stackrel{\rightharpoonup }{{\mathrm{v}}_3}=\left[\begin{array}{c}0\\ -1\\ 0\\ 1\end{array}\right]$

Check for the coordinates of the vector.

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

Consider the matrix form and perform rrref,

$rref\left[\begin{array}{cccc}1& 0& 0& 5\\ 1& 1& -1& 4\\ 0& 1& 0& 3\\ 0& 0& 1& 2\end{array}\right]=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$

Final answer.

As the coordinates of the vector are not available, so, $\stackrel{\rightharpoonup }{x}\mathfrak{I}$ isn’t in V .