In Exercise 5, find the coordinate vector \({\left( x \right)_{\rm B}}\) of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).

5. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{3}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{5}}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{\bf{1}}\end{array}} \right)\)

Note that \({\left( x \right)_{\rm B}}\)is the solution of the system.

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}{{b_1}}&{{b_2}}\end{array}} \right){\left( x \right)_{\rm B}} = x\\\left( {\begin{array}{*{20}{c}}1&2\\{ - 3}&{ - 5}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 2}\\1\end{array}} \right)\end{array}\)

Its augmented matrix is \(\left( {\begin{array}{*{20}{c}}{{b_1}}&{{b_2}}&x\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&2&{ - 2}\\{ - 3}&{ - 5}&1\end{array}} \right)\).

At row 2, multiply row 1 by 3 and add it to row 2, i.e., \({R_2} \to {R_2} + 3{R_1}\).

\( \sim \left( {\begin{array}{*{20}{c}}1&2&{ - 2}\\0&1&{ - 5}\end{array}} \right)\)

At row 1, multiply row 2 by 2 and subtract it from row 1, i.e., \({R_1} \to {R_1} - 2{R_2}\).

\( \sim \left( {\begin{array}{*{20}{c}}1&0&8\\0&1&{ - 5}\end{array}} \right)\)

This implies \({c_1} = 8,\) and \({c_2} = - 5\).

Hence,

\(\begin{array}{c}{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}8\\{ - 5}\end{array}} \right)\end{array}\)