In Exercises 7-10, let \(B = \left\{ {{{\mathop{\rm b}\nolimits} _1},{{\mathop{\rm b}\nolimits} _2}} \right\}\) and \(C = \left\{ {{{\mathop{\rm c}\nolimits} _1},{{\mathop{\rm c}\nolimits} _2}} \right\}\) be bases for \({\mathbb{R}^2}\). In each exercise, find the change-of-coordinates matrix from \(B\) to \(C\) and the change-of-coordinates matrix from \(C\) to \(B\).

8. \({{\mathop{\rm b}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}{ - 1}\\8\end{array}} \right),{{\mathop{\rm b}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}1\\{ - 5}\end{array}} \right),{{\mathop{\rm c}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right),{{\mathop{\rm c}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}1\\1\end{array}} \right)\).

Let \(B = \left\{ {{{\mathop{\rm b}\nolimits} _1},...,{{\mathop{\rm b}\nolimits} _n}} \right\}\)and \(C = \left\{ {{{\mathop{\rm c}\nolimits} _1},...,{{\mathop{\rm c}\nolimits} _n}} \right\}\) be bases of a vector space \(V\). Then according toTheorem 15,there is a unique \(n \times n\) matrix \(\mathop P\limits_{C \leftarrow B} \) such that

\({\left( {\mathop{\rm x}\nolimits} \right)_C} = \mathop P\limits_{C \leftarrow B} {\left( {\mathop{\rm x}\nolimits} \right)_B}\). …(1)

The columns of \(\mathop P\limits_{C \leftarrow B} \) are the \(C - \)coordinate vectors of the vectors in the basis \(B\). That is, \(\mathop P\limits_{C \leftarrow B} = \left( {\begin{array}{*{20}{c}}{{{\left( {{{\mathop{\rm b}\nolimits} _1}} \right)}_C}}&{{{\left( {{{\mathop{\rm b}\nolimits} _2}} \right)}_C}}& \cdots &{{{\left( {{{\mathop{\rm b}\nolimits} _n}} \right)}_C}}\end{array}} \right)\).

Write the augmented matrix as shown below:

Perform an elementary row operation to produce a row-reduced echelon form of the matrix.

At row 2, multiply row 1 by 4 and subtract it from row 2.

\( \sim \left( {\begin{array}{*{20}{c}}1&1&{ - 1}&1\\0&{ - 3}&{12}&{ - 9}\end{array}} \right)\)

At row 2, multiply row 2 by \( - \frac{1}{3}\).

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

At row 1, subtract row 2 from row 1.

\( \sim \left( {\begin{array}{*{20}{c}}1&0&3&{ - 2}\\0&1&{ - 4}&3\end{array}} \right)\)

Therefore, \(\mathop P\limits_{C \leftarrow B} = \left( {\begin{array}{*{20}{c}}3&{ - 2}\\{ - 4}&3\end{array}} \right)\).

Thus, the change-of-coordinates matrix from \(B\) to \(C\) is \(\mathop P\limits_{C \leftarrow B} = \left( {\begin{array}{*{20}{c}}3&{ - 2}\\{ - 4}&3\end{array}} \right)\).

It is known that \({\left( {\mathop P\limits_{C \leftarrow B} } \right)^{ - 1}}\) is the matrix that converts \(C - \)coordinates into \(B - \)coordinates. That is, \({\left( {\mathop P\limits_{C \leftarrow B} } \right)^{ - 1}} = \mathop P\limits_{B \leftarrow C} \).

\(\begin{aligned} \mathop P\limits_{B \leftarrow C} &= {\left( {\mathop P\limits_{C \leftarrow B} } \right)^{ - 1}}\\ &= {\left( {\begin{array}{*{20}{c}}3&{ - 2}\\{ - 4}&3\end{array}} \right)^{ - 1}}\\ &= \frac{1}{1}\left( {\begin{array}{*{20}{c}}3&2\\4&3\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}3&2\\4&3\end{array}} \right)\end{aligned}\)

Thus, the change-of-coordinates matrix from \(C\) to \(B\) is \(\mathop P\limits_{B \leftarrow C} = \left( {\begin{array}{*{20}{c}}3&2\\4&3\end{array}} \right)\).