(M) Let \(H = {\bf{Span}}\left\{ {{{\bf{v}}_{\bf{1}}},\;{{\bf{v}}_{\bf{2}}}} \right\}\) and \(B = \left\{ {{{\bf{v}}_{\bf{1}}},\,{{\bf{v}}_{\bf{2}}}} \right\}\). Show that x is in the H and find the B coordinate vector of x, for

\({{\bf{v}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{{\bf{11}}}\\{ - {\bf{5}}}\\{{\bf{10}}}\\{\bf{7}}\end{array}} \right)\), \({{\bf{v}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{{\bf{14}}}\\{ - {\bf{8}}}\\{{\bf{13}}}\\{{\bf{10}}}\end{array}} \right)\), \({\bf{x}} = \left( {\begin{array}{*{20}{c}}{{\bf{19}}}\\{ - {\bf{13}}}\\{{\bf{18}}}\\{{\bf{15}}}\end{array}} \right)\)

\(\left( {\begin{array}{*{20}{c}}{{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{\bf{x}}\end{array}} \right)\)

The augmented matrixcan be written as:

\(\left( {\begin{array}{*{20}{c}}{{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{\bf{x}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{11}&{14}&{19}\\{ - 5}&{ - 8}&{ - 13}\\{10}&{13}&{18}\\7&{10}&{15}\end{array}} \right)\)

Consider matrix \(A = \left( {\begin{array}{*{20}{c}}{11}&{14}&{19}\\{ - 5}&{ - 8}&{ - 13}\\{10}&{13}&{18}\\7&{10}&{15}\end{array}} \right)\).

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

\(\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left( {{\rm{ }}\begin{array}{*{20}{c}}{11}&{14}&{19;\;\;\begin{array}{*{20}{c}}{ - 5}&{ - 8}&{ - 13;\;\;\begin{array}{*{20}{c}}{10}&{13}&{18;\;\;\begin{array}{*{20}{c}}7&{10}&{15}\end{array}}\end{array}}\end{array}}\end{array}} \right);\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\)

\(\left( {\begin{array}{*{20}{c}}{11}&{14}&{19}\\{ - 5}&{ - 8}&{ - 13}\\{10}&{13}&{18}\\7&{10}&{15}\end{array}} \right) \sim \left( {\begin{array}{*{20}{c}}1&0&{ - \frac{5}{3}}\\0&1&{\frac{8}{3}}\\0&0&0\\0&0&0\end{array}} \right)\)

For\({\bf{x}}\), using the augmented matrix, you get:

\({\bf{x}} = {c_1}{{\bf{x}}_1} + {c_2}{{\bf{x}}_2}\)

So, the values of\({c_1}\)and\({c_2}\)are:

\({c_1} = - \frac{5}{3}\)and\({c_2} = \frac{8}{3}\)

Therefore, the B-coordinate for vector \({\bf{x}}\) is \({\left( {\bf{x}} \right)_B} = \left( {\begin{array}{*{20}{c}}{ - \frac{5}{3}}\\{\frac{8}{3}}\end{array}} \right)\).