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.

6. \[{b_1} = \left[ {\begin{array}{*{20}{c}}{ - 3}\\1\\{ - 4}\end{array}} \right]\], \[{b_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}7\\5\\{ - 6}\end{array}} \right]\], \[x = \left[ {\begin{array}{*{20}{c}}{{\bf{11}}}\\0\\7\end{array}} \right]\]

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

The\(B\)-coordinate vectorof 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 \[{b_1} = \left[ {\begin{array}{*{20}{c}}{ - 3}\\1\\{ - 4}\end{array}} \right]\],\[{b_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}7\\5\\{ - 6}\end{array}} \right]\], and\[x = \left[ {\begin{array}{*{20}{c}}{11}\\0\\7\end{array}} \right]\].

Then, it can be represented as shown below:

\({c_1}\left[ {\begin{array}{*{20}{c}}{ - 3}\\1\\{ - 4}\end{array}} \right] + {c_2}\left[ {\begin{array}{*{20}{c}}7\\5\\{ - 6}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{11}\\0\\7\end{array}} \right]\)

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

\(\left[ {\begin{array}{*{20}{c}}{{{\bf{b}}_1}}&{{{\bf{b}}_2}}&{\bf{x}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 3}&7&{11}\\1&5&0\\{ - 4}&{ - 6}&7\end{array}} \right]\)

Exchange row 1 with row 3.

\(\left[ {\begin{array}{*{20}{c}}{ - 3}&7&{11}\\1&5&0\\{ - 4}&{ - 6}&7\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&5&0\\{ - 3}&7&{11}\\{ - 4}&{ - 6}&7\end{array}} \right]\)

Use the\({x_1}\)term in the first equation to eliminate the\( - 3{x_1}\)term from the second equation. Add 3 times row 1 to row 3.

\(\left[ {\begin{array}{*{20}{c}}1&5&0\\{ - 3}&7&{11}\\{ - 4}&{ - 6}&7\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&5&0\\0&{22}&{11}\\{ - 4}&{ - 6}&7\end{array}} \right]\)

Multiply row 2 by \(\frac{1}{{22}}\).

\(\left[ {\begin{array}{*{20}{c}}1&5&0\\0&{22}&{11}\\{ - 4}&{ - 6}&7\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&5&0\\0&1&{1/2}\\{ - 4}&{ - 6}&7\end{array}} \right]\)

Use the\({x_2}\)term in the second equation to eliminate the\(5{x_2}\)term from the first equation. Add\( - 5\)times row 2 to row 1.

\(\left[ {\begin{array}{*{20}{c}}1&5&0\\0&1&{1/2}\\{ - 4}&{ - 6}&7\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&{ - 5/2}\\0&1&{1/2}\\0&0&0\end{array}} \right]\)

Thus, the weights are \({c_1} = - \frac{5}{2}\) and \({c_2} = \frac{1}{2}\).

Obtain the B-coordinate vector of x.

\({\left[ {\bf{x}} \right]_B} = \left[ {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 5/2}\\{1/2}\end{array}} \right]\)

Thus, the B-coordinate vector of x is \({\left[ {\bf{x}} \right]_B} = \left[ {\begin{array}{*{20}{c}}{ - 5/2}\\{1/2}\end{array}} \right]\).