Question

In Exercises 11 and 12, use an inverse matrix to find for the given and \(B\).

12. \(B = \left\{ {\left( {\begin{array}{*{20}{c}}4\\5\end{array}} \right),\left( {\begin{array}{*{20}{c}}6\\7\end{array}} \right)} \right\},{\mathop{\rm x}\nolimits} = \left( {\begin{array}{*{20}{c}}2\\0\end{array}} \right)\)

Short answer

The \(B\)-coordinate vector is \({\left( {\mathop{\rm x}\nolimits} \right)_B} = \left( {\begin{array}{*{20}{c}}{ - 7}\\5\end{array}} \right)\).

Step-by-step solution

State the change of coordinate from B

Let \({P_B} = \left( {\begin{array}{*{20}{c}}{{{\mathop{\rm b}\nolimits} _1}}&{{{\mathop{\rm b}\nolimits} _2}}& \cdots &{{{\mathop{\rm b}\nolimits} _n}}\end{array}} \right)\), then thevector equation\({\mathop{\rm x}\nolimits} = {c_1}{{\mathop{\rm b}\nolimits} _1} + {c_2}{{\mathop{\rm b}\nolimits} _2} + ... + {c_n}{{\mathop{\rm b}\nolimits} _n}\)becomes equivalent to \({\mathop{\rm x}\nolimits} = {P_B}{\left( {\mathop{\rm x}\nolimits} \right)_B}\). \({P_B}\) represents the change-of-coordinates matrix from \(B\) to the standard basis in \({\mathbb{R}^n}\).

Multiplication by \(P_B^{ - 1}\) on the L.H.S converts \({\mathop{\rm x}\nolimits} \) into its \(B\)-coordinate vector:

\(P_B^{ - 1}{\mathop{\rm x}\nolimits} = {\left( {\mathop{\rm x}\nolimits} \right)_B}\)

Use an inverse matrix to determine \({\left( {\mathop{\rm x}\nolimits}  \right)_B}\)

The change-of-coordinates matrix from \(B\) to the standard basis in \({\mathbb{R}^2}\) is shown below.

\(\begin{array}{c}{P_B} = \left( {\begin{array}{*{20}{c}}{{{\mathop{\rm b}\nolimits} _1}}&{{{\mathop{\rm b}\nolimits} _2}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}4&6\\5&7\end{array}} \right)\end{array}\)

Multiply the L.H.S of the equation \({\mathop{\rm x}\nolimits} = {P_B}{\left( {\mathop{\rm x}\nolimits} \right)_B}\) by \(P_B^{ - 1}\) to convert \({\mathop{\rm x}\nolimits} \) into its \(B\)-coordinate vector.

\(\begin{array}{c}{\left( {\mathop{\rm x}\nolimits} \right)_B} = P_B^{ - 1}{\mathop{\rm x}\nolimits} \\ = {\left( {\begin{array}{*{20}{c}}4&6\\5&7\end{array}} \right)^{ - 1}}\left( {\begin{array}{*{20}{c}}2\\0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - \frac{7}{2}}&3\\{\frac{5}{2}}&{ - 2}\end{array}} \right)\left( {\begin{array}{*{20}{c}}2\\0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 7 + 0}\\{5 + 0}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 7}\\5\end{array}} \right)\end{array}\)\({\left( {\mathop{\rm x}\nolimits} \right)_B}\)

Thus, the \(B\)-coordinate vector is \({\left( {\mathop{\rm x}\nolimits} \right)_B} = \left( {\begin{array}{*{20}{c}}{ - 7}\\5\end{array}} \right)\).