In Exercises 3–6, the vector x is in a subspace Hwith a basis \(B = \left\{ {{b_{\bf{1}}},{{\bf{b}}_{\bf{2}}}} \right\}\). Find the B-coordinate vector of x.

3. \[{b_1} = \left[ {\begin{array}{*{20}{c}}1\\{ - 4}\end{array}} \right]\], \[{b_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\7\end{array}} \right]\], \[{\bf{x}} = \left[ {\begin{array}{*{20}{c}}{ - {\bf{3}}}\\7\end{array}} \right]\]

For the basis of asubspace H, let the set be\(B = \left\{ {{{\bf{b}}_1},...,{{\bf{b}}_p}} \right\}\). For the weights\({c_1},...,{c_p}\), the coordinate of x is represented as\({\bf{x}} = {c_1}{{\bf{b}}_1} + \cdots + {c_p}{{\bf{b}}_p}\).

The\(B\)-coordinatevector of x is represented as shown below:

\({\left[ {\bf{x}} \right]_B} = \left[ {\begin{array}{*{20}{c}}{{c_1}}\\ \vdots \\{{c_p}}\end{array}} \right]\)

For theweights \({c_1}\)and\({c_2}\), and the vector x in H, it must satisfy the equation\({\bf{x}} = {c_1}{{\bf{b}}_1} + {c_2}{{\bf{b}}_2}\).

Consider thevectors \[{{\bf{b}}_1} = \left[ {\begin{array}{*{20}{c}}1\\{ - 4}\end{array}} \right]\],\[{{\bf{b}}_2} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\7\end{array}} \right]\], and\[{\bf{x}} = \left[ {\begin{array}{*{20}{c}}{ - 3}\\7\end{array}} \right]\].

Then, it can be represented as shown below:

\({c_1}\left[ {\begin{array}{*{20}{c}}1\\{ - 4}\end{array}} \right] + {c_2}\left[ {\begin{array}{*{20}{c}}{ - 2}\\7\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 3}\\7\end{array}} \right]\)

Write theaugmented matrix \(\left[ {\begin{array}{*{20}{c}}{{{\bf{b}}_1}}&{{{\bf{b}}_2}}&{\bf{x}}\end{array}} \right]\)as shown below:

\(\left[ {\begin{array}{*{20}{c}}{{{\bf{b}}_1}}&{{{\bf{b}}_2}}&{\bf{x}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ - 2}&{ - 3}\\{ - 4}&7&7\end{array}} \right]\)

Use the\({x_1}\)term in the first equation to eliminate the\( - 4{x_1}\)term from the second equation. Add 4 times row 1 to row 2.

\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&{ - 3}\\{ - 4}&7&7\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&{ - 2}&{ - 3}\\0&{ - 1}&{ - 5}\end{array}} \right]\)

Use the\( - {x_2}\)term in the second equation to eliminate the\( - 2{x_2}\)term from the first equation. Add 2 times row 2 to row 1.

\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&{ - 3}\\0&{ - 1}&{ - 5}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&7\\0&1&5\end{array}} \right]\)

Thus, the weights are \({c_1} = 7\)and \({c_2} = 5\).

Obtain the B-coordinate vectorof x.

\({\left[ {\bf{x}} \right]_B} = \left[ {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}7\\5\end{array}} \right]\)

Thus, the B-coordinate vector of x is \({\left[ {\bf{x}} \right]_B} = \left[ {\begin{array}{*{20}{c}}7\\5\end{array}} \right]\).