Question

In Exercise 4, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

4. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{2}}\\{\bf{0}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{3}}\\{ - {\bf{5}}}\\{\bf{2}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{4}}\\{ - {\bf{7}}}\\{\bf{3}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{4}}}\\{\bf{8}}\\{ - {\bf{7}}}\end{array}} \right)\)

Short answer

Vector \(x = \left( {\begin{array}{*{20}{c}}0\\1\\{ - 5}\end{array}} \right)\)

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

\(\begin{array}{c}x = - 4\left( {\begin{array}{*{20}{c}}{ - 1}\\2\\0\end{array}} \right) + 8\left( {\begin{array}{*{20}{c}}3\\{ - 5}\\2\end{array}} \right) + \left( { - 7} \right)\left( {\begin{array}{*{20}{c}}4\\{ - 7}\\3\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}4\\{ - 8}\\0\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{24}\\{ - 40}\\{16}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - 28}\\{49}\\{ - 21}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{4 + 24 - 28}\\{ - 8 - 40 + 49}\\{0 + 16 - 21}\end{array}} \right)\\x = \left( {\begin{array}{*{20}{c}}0\\1\\{ - 5}\end{array}} \right).\end{array}\)

Draw a conclusion

Hence, vector \(x = \left( {\begin{array}{*{20}{c}}0\\1\\{ - 5}\end{array}} \right)\).