Question

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

$\overrightarrow{\mathbf{x}}\mathbf{=}\left[\begin{array}{c}23\\ 29\end{array}\right]\mathbf{,}{\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{=}\left[\begin{array}{c}46\\ 58\end{array}\right]\mathbf{,}{\overrightarrow{\mathbf{v}}}_{\mathbf{2}}\mathbf{=}\left[\begin{array}{c}61\\ 67\end{array}\right]$

Short answer

The coordinates of the vector is, ${\overrightarrow{x}}_{\mathfrak{I}}=\left[\begin{array}{c}\frac{1}{2}\\ 0\end{array}\right]$.

Step-by-step solution

Consider the vectors

The vectors are,

$\overrightarrow{x}=\left[\begin{array}{c}23\\ 29\end{array}\right],{\overrightarrow{v}}_1=\left[\begin{array}{c}46\\ 58\end{array}\right],{\overrightarrow{v}}_2=\left[\begin{array}{c}61\\ 67\end{array}\right]$.

Check for the coordinates of the vector

The formula is, $\overrightarrow{x}=c_1{\overrightarrow{v}}_1+c_2{\overrightarrow{v}}_2$.

$$\begin{gathered} \left[\begin{array}{c}23\\ 29\end{array}\right]=c_1\left[\begin{array}{c}46\\ 58\end{array}\right]+c_2\left[\begin{array}{c}61\\ 67\end{array}\right] \\ 46c_1+61c_2=23,58c_1+67c_2=29 \\ {c}_1=\frac{1}{2},c_2=0 \end{gathered}$$

Substitute these values in the formula.

$$\begin{gathered} \overrightarrow{x}=c_1{\overrightarrow{v}}_1+c_2{\overrightarrow{v}}_2 \\ \left[\begin{array}{c}23\\ 29\end{array}\right]=\frac{1}{2}\left[\begin{array}{c}46\\ 58\end{array}\right]+0\left[\begin{array}{c}61\\ 67\end{array}\right] \\ {\overrightarrow{x}}_{\mathfrak{I}}=\left[\begin{array}{c}\frac{1}{2}\\ 0\end{array}\right] \end{gathered}$$

Final answer

The coordinates of the vector is, ${\overrightarrow{x}}_{\mathfrak{I}}=\left[\begin{array}{c}\frac{1}{2}\\ 0\end{array}\right]$.