Answer
we get a linear combination 1;
${\bf{v}} = 5{{\bf{v}}_1} - 2{{\bf{v}}_{\bf{2}}}$
we get a Linear combination 2;
${\bf{v}} = 10{{\bf{v}}_1} - 3{{\bf{v}}_{\bf{2}}} + {{\bf{v}}_3}$
Work Step by Step
Expressing the vectors in form of an equation
${x_1}{{\bf{v}}_{\bf{1}}} + {x_2}{{\bf{v}}_2} + {x_3}{{\bf{v}}_3} = {\bf{v}}$
and $\mathbf{v}=\begin{bmatrix}1\\1\end{bmatrix}$
$x_{1}\begin{bmatrix}1\\-3\end{bmatrix}+x_{2}\begin{bmatrix}2\\-8\end{bmatrix}+x_{3}\begin{bmatrix}-3\\7\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix}$
$\begin{bmatrix}{{x_1} + 2{x_2} - 3{x_3}}\\{ - 3{x_1} - 8{x_2} + 7{x_3}}\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix} $
By forming augmented matrix and row reducing we get
$\begin{bmatrix}1&2&{ - 3}&1\\{ - 3}&{ - 8}&7&1\end{bmatrix}\sim \begin{bmatrix}1&0&{ - 5}&5\\0&{ 1}&{ 1}&-2\end{bmatrix}$
$\begin{bmatrix}{{x_1} + 0{x_2} - 5{x_3}}\\{ 0{x_1} +1{x_2} + 1{x_3}}\end{bmatrix}=\begin{bmatrix}5\\-2\end{bmatrix} $
by considering $x_{3}$
we have $\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{5 + 5{x_3}}\\{ - 2 - {x_3}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}5\\{ - 2}\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}5\\{ - 1}\\1\end{array}} \right]{x_3}$
Let $x_{3}=0$
$\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}5\\{ - 2}\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}5\\{ - 1}\\1\end{array}} \right]\left( 0 \right)\\\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}5\\{ - 2}\\0\end{array}} \right]\end{array}$
we get a linear combination 1;
${\bf{v}} = 5{{\bf{v}}_1} - 2{{\bf{v}}_{\bf{2}}}$
Let $x_{3}=1$
$\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}5\\{ - 2}\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}5\\{ - 1}\\1\end{array}} \right]\left( 1 \right)\\\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{10}\\{ - 3}\\1\end{array}} \right]\end{array}$
we get a Linear combination 2;
${\bf{v}} = 10{{\bf{v}}_1} - 3{{\bf{v}}_{\bf{2}}} + {{\bf{v}}_3}$
