Question

In Exercise 3, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

3. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{4}}}\\{\bf{3}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{5}}\\{\bf{2}}\\{ - {\bf{2}}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{4}}\\{ - {\bf{7}}}\\{\bf{0}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{0}}\\{ - {\bf{1}}}\end{array}} \right)\)

Short answer

Vector \(x = \left( {\begin{array}{*{20}{c}}{ - 1}\\{ - 5}\\9\end{array}} \right)\)

Step-by-step solution

Use the definition

The coordinates of x relative to basis\({\rm B} = \left\{ {{b_{\bf{1}}},{b_{\bf{2}}},...,{b_n}} \right\}\)are the weights\({c_{\bf{1}}},{c_{\bf{2}}},...,{c_n}\),such that \(x = {c_{\bf{1}}}{b_{\bf{1}}} + {c_{\bf{2}}}{b_{\bf{2}}} + ... + {c_n}{b_n}\). Then,\({\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\\ \vdots \\{{c_n}}\end{array}} \right)\).

Find x

By the above definition, you get

\[\begin{array}{c}x = 3\left[ {\begin{array}{*{20}{c}}1\\{ - 4}\\3\end{array}} \right] + 0\left[ {\begin{array}{*{20}{c}}5\\2\\{ - 2}\end{array}} \right] + \left( { - 1} \right)\left[ {\begin{array}{*{20}{c}}4\\{ - 7}\\0\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}3\\{ - 12}\\9\end{array}} \right] + 0 + \left[ {\begin{array}{*{20}{c}}{ - 4}\\7\\0\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{3 - 4}\\{ - 12 + 7}\\9\end{array}} \right]\\x = \left[ {\begin{array}{*{20}{c}}{ - 1}\\{ - 5}\\9\end{array}} \right].\end{array}\]

Draw a conclusion

Hence, vector \(x = \left( {\begin{array}{*{20}{c}}{ - 1}\\{ - 5}\\9\end{array}} \right)\).