In Exercise 6, find the coordinate vector of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).

6. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{2}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{ - {\bf{6}}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{0}}\end{array}} \right)\)

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

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}{{b_1}}&{{b_2}}\end{array}} \right){\left( x \right)_{\rm B}} = x\\\left( {\begin{array}{*{20}{c}}1&5\\{ - 2}&{ - 6}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}4\\0\end{array}} \right)\end{array}\)

Its augmented matrix is \(\left( {\begin{array}{*{20}{c}}{{b_1}}&{{b_2}}&x\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&5&4\\{ - 2}&{ - 6}&0\end{array}} \right)\).

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

\( \sim \left( {\begin{array}{*{20}{c}}1&5&4\\0&4&8\end{array}} \right)\)

At row 2, divide row 2 by 4.

\( \sim \left( {\begin{array}{*{20}{c}}1&5&4\\0&1&2\end{array}} \right)\)

At row 1, multiply row 2 by 5 and subtract it from row 1, i.e., \({R_1} \to {R_1} - 5{R_2}\).

\( \sim \left( {\begin{array}{*{20}{c}}1&0&{ - 6}\\0&1&2\end{array}} \right)\)

This implies \({c_1} = - 6,\) and \({c_2} = 2\).

\(\begin{array}{c}{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 6}\\2\end{array}} \right)\end{array}\)

\({\left( x \right)_{\rm B}}\)