[M] Let \(H = {\bf{span}}\left\{ {{{\bf{v}}_{\bf{1}}},{{\bf{v}}_{\bf{2}}},{{\bf{v}}_{\bf{3}}}} \right\}\) and \({\rm B} = \left\{ {{{\bf{v}}_{\bf{1}}},{{\bf{v}}_{\bf{2}}},{{\bf{v}}_{\bf{3}}}} \right\}\). Show that \(B\)is a basis for \(H\)and \(x\) is in H, and find the \({\rm B} - \)coordinate vector of x, when

\({{\bf{v}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}{ - {\bf{6}}}\\{\bf{4}}\\{ - {\bf{9}}}\\{\bf{4}}\end{array}} \right]\), \({{\bf{v}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{\bf{8}}\\{ - {\bf{3}}}\\{\bf{7}}\\{ - {\bf{3}}}\end{array}} \right]\), \({{\bf{v}}_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{ - {\bf{9}}}\\{\bf{5}}\\{ - {\bf{8}}}\\{\bf{3}}\end{array}} \right]\), \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{7}}\\{ - {\bf{8}}}\\{\bf{3}}\end{array}} \right]\)

The matrix formed using the column vectors is

\(A = \left[ {\begin{array}{*{20}{c}}{ - 6}&8&{ - 9}&4\\4&{ - 3}&5&7\\{ - 9}&7&{ - 8}&{ - 8}\\4&{ - 3}&3&3\end{array}} \right]\).

Consider matrix \(A = \left[ {\begin{array}{*{20}{c}}{ - 6}&8&{ - 9}&4\\4&{ - 3}&5&7\\{ - 9}&7&{ - 8}&{ - 8}\\4&{ - 3}&3&3\end{array}} \right]\).

Use the code in MATLAB to obtain the row-reducedechelon form as shown below:

\[\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left[ {{\rm{ }}\begin{array}{*{20}{c}}{ - 6}&8&{ - 9}&{4;\,\,\begin{array}{*{20}{c}}4&{ - 3}&5&{7;\,\,\begin{array}{*{20}{c}}{ - 9}&7&{ - 8}&{ - 8;\,\,\begin{array}{*{20}{c}}4&{ - 3}&3&3\end{array}}\end{array}}\end{array}}\end{array}} \right];\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\]

\(\left[ {\begin{array}{*{20}{c}}{ - 6}&8&{ - 9}&4\\4&{ - 3}&5&7\\{ - 9}&7&{ - 8}&{ - 8}\\4&{ - 3}&3&3\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&0&3\\0&1&0&5\\0&0&1&2\\0&0&0&0\end{array}} \right]\)

Columns \({{\bf{v}}_1}\), \({{\bf{v}}_2}\), and \({{\bf{v}}_3}\) form basis \(B\) for the subspace H in which they span.

The equation \({c_1}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2} + {c_2}{{\bf{v}}_3} = {\bf{x}}\) is consistent. The values of \({c_1}\), \({c_2}\), and \({c_3}\) from the echelon form are \({c_1} = 3\), \({c_2} = 5\), and \({c_3} = 2\).

So, the \(\beta - \)coordinates of \({\bf{x}}\) are \(\left( {3,5,2} \right)\).