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.

5. \[{b_1} = \left[ {\begin{array}{*{20}{c}}1\\5\\{ - 3}\end{array}} \right]\], \[{b_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 7}\\5\end{array}} \right]\], \[x = \left[ {\begin{array}{*{20}{c}}4\\{10}\\{ - 7}\end{array}} \right]\]

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 vector 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 the vectors\[{b_1} = \left[ {\begin{array}{*{20}{c}}1\\5\\{ - 3}\end{array}} \right]\],\[{b_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 7}\\5\end{array}} \right]\], and\[x = \left[ {\begin{array}{*{20}{c}}4\\{10}\\{ - 7}\end{array}} \right]\].

Then, it can be represented as shown below:

\({c_1}\left[ {\begin{array}{*{20}{c}}1\\5\\{ - 3}\end{array}} \right] + {c_2}\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 7}\\5\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}4\\{10}\\{ - 7}\end{array}} \right]\)

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

\(\left[ {\begin{array}{*{20}{c}}{{{\bf{b}}_1}}&{{{\bf{b}}_2}}&{\bf{x}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ - 3}&4\\5&{ - 7}&{10}\\{ - 3}&5&{ - 7}\end{array}} \right]\)

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

\(\left[ {\begin{array}{*{20}{c}}1&{ - 3}&4\\5&{ - 7}&{10}\\{ - 3}&5&{ - 7}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&{ - 3}&4\\5&{ - 7}&{10}\\0&{ - 4}&5\end{array}} \right]\)

Use the\({x_1}\)term in the first equation to eliminate the\(5{x_1}\)term from the second equation. Add\( - 5\)times row 1 to row 2.

\[\left[ {\begin{array}{*{20}{c}}1&{ - 3}&4\\5&{ - 7}&{10}\\0&{ - 4}&5\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&{ - 3}&4\\0&8&{ - 10}\\0&{ - 4}&5\end{array}} \right]\]

Multiply row 2 by 2, and then add row 2 and row 3 to eliminate \( - 4{x_2}\) from the third equation.

\[\left[ {\begin{array}{*{20}{c}}1&{ - 3}&4\\0&8&{ - 10}\\0&{ - 4}&5\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&{ - 3}&4\\0&4&{ - 5}\\0&0&0\end{array}} \right]\]

Multiply row 2 by \(\frac{1}{4}\).

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

Use the\({x_2}\)term in the second equation to eliminate the\( - 3{x_2}\)term from the second equation. Add\(3\)times row 2 to row 1.

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

Thus, the weights are \({c_1} = \frac{1}{4}\)and \({c_2} = - \frac{5}{4}\).

Obtain the B-coordinate vector of x.

\({\left[ {\bf{x}} \right]_B} = \left[ {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{1/4}\\{ - 5/4}\end{array}} \right]\)

Thus, the B-coordinate vector of x is \({\left[ {\bf{x}} \right]_B} = \left[ {\begin{array}{*{20}{c}}{1/4}\\{ - 5/4}\end{array}} \right]\).