Question

In Exercises 11 and 12, use an inverse matrix to find for the given \({\mathop{\rm x}\nolimits} \) and \(B\).

11. \(B = \left\{ {\left[ {\begin{array}{*{20}{c}}3\\{ - 5}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 4}\\6\end{array}} \right]} \right\},{\mathop{\rm x}\nolimits} = \left[ {\begin{array}{*{20}{c}}2\\{ - 6}\end{array}} \right]\)

Short answer

The \(B\)-coordinate vector is \({\left[ {\mathop{\rm x}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}6\\4\end{array}} \right]\).

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 matrixfrom \(B\) to the standard basis in \({\mathbb{R}^n}\).

Multiplication by \(P_B^{ - 1}\) 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}  

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

\[\begin{aligned}{c}{P_B} = \left[ {\begin{aligned}{*{20}{c}}{{{\mathop{\rm b}\nolimits} _1}}&{{{\mathop{\rm b}\nolimits} _2}}\end{aligned}} \right]\\ = \left[ {\begin{aligned}{*{20}{c}}3&{ - 4}\\{ - 5}&6\end{aligned}} \right]\end{aligned}\]

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.

\(\begin{array}{c}{\left[ {\mathop{\rm x}\nolimits} \right]_B} = P_B^{ - 1}{\mathop{\rm x}\nolimits} \\ = {\left[ {\begin{array}{*{20}{c}}3&{ - 4}\\{ - 5}&6\end{array}} \right]^{ - 1}}\left[ {\begin{array}{*{20}{c}}2\\{ - 6}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 3}&{ - 2}\\{ - \frac{5}{2}}&{ - \frac{3}{2}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2\\{ - 6}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 6 + 12}\\{ - 5 + 9}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}6\\4\end{array}} \right]\end{array}\)\[{\left[ {\mathop{\rm x}\nolimits} \right]_B}\]

Thus, the \(B\)-coordinate vector is \({\left[ {\mathop{\rm x}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}6\\4\end{array}} \right]\).