In Exercises 1 and 2, find the vector x determined by the given

coordinate vector \({\left[ {\bf{x}} \right]_{\rm B}}\)and the given basis B. Illustrate your answer with a figure, as in the solution of Practice Problem 2.

1. \(B = \left\{ {\left[ {\begin{array}{*{20}{c}}1\\1\end{array}} \right],\left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]} \right\}\), \({\left[ x \right]_B} = \left[ {\begin{array}{*{20}{c}}3\\2\end{array}} \right]\)

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

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

Compare\({\left[ x \right]_B} = \left[ {\begin{array}{*{20}{c}}3\\2\end{array}} \right]\)with\({\left[ x \right]_B} = \left[ {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\end{array}} \right]\). So,\({c_1} = 3\)and.

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

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

Then, it can be represented as shown below:

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

Thus, the vector is\({\bf{x}} = \left[ {\begin{array}{*{20}{c}}7\\1\end{array}} \right]\).

The illustrated figure is shown below: