Question

In Exercise 7, find the coordinate vector \({\left( x \right)_{\rm B}}\) of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).

7. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{1}}}\\{ - {\bf{3}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{3}}}\\{\bf{4}}\\{\bf{9}}\end{array}} \right),{b_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{2}}}\\{\bf{4}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{\bf{8}}\\{ - {\bf{9}}}\\{\bf{6}}\end{array}} \right)\)

Short answer

Coordinate vector \({\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{ - 1}\\{ - 1}\\3\end{array}} \right)\)

Step-by-step solution

Write the system

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

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}{{b_1}}&{{b_2}}\end{array}} \right){\left( x \right)_{\rm B}} = x\\\left( {\begin{array}{*{20}{c}}1&{ - 3}&2\\{ - 1}&4&{ - 2}\\{ - 3}&9&4\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\\{{c_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}8\\{ - 9}\\6\end{array}} \right)\end{array}\)

Find the reduced row echelon form

Its augmented matrix is \(\left( {\begin{array}{*{20}{c}}{{b_1}}&{{b_2}}&x\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&{ - 3}&2&8\\{ - 1}&4&{ - 2}&{ - 9}\\{ - 3}&9&4&6\end{array}} \right)\).

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

Also, at row 3, multiply row 1 by 3 and add it to row 3, i.e., \({R_3} \to {R_3} + 3{R_1}\).

\( \sim \left( {\begin{array}{*{20}{c}}1&{ - 3}&2&8\\0&1&0&{ - 1}\\0&0&{10}&{30}\end{array}} \right)\)

At row 3, divide row 3 by 10.

\( \sim \left( {\begin{array}{*{20}{c}}1&{ - 3}&2&8\\0&1&0&{ - 1}\\0&0&1&3\end{array}} \right)\)

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

\( \sim \left( {\begin{array}{*{20}{c}}1&0&2&5\\0&1&0&{ - 1}\\0&0&1&3\end{array}} \right)\)

At row 1, multiply row 3 by 2 and subtract it from row 1, i.e., \({R_1} \to {R_1} - 2{R_3}\).

\( \sim \left( {\begin{array}{*{20}{c}}1&0&0&{ - 1}\\0&1&0&{ - 1}\\0&0&1&3\end{array}} \right)\)

This implies \({c_1} = - 1,{c_2} = - 1,\) and \({c_3} = 3\).

Draw a conclusion

\(\begin{array}{c}{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\\{{c_3}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 1}\\{ - 1}\\3\end{array}} \right)\end{array}\)