Question

In Exercises 9 and 10, find the change-of-coordinates matrix from \(B\) to the standard basis in \({\mathbb{R}^n}\).

10. \(B = \left\{ {\left[ {\begin{array}{*{20}{c}}3\\{ - 1}\\4\end{array}} \right],\left[ {\begin{array}{*{20}{c}}2\\0\\{ - 5}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}8\\{ - 2}\\7\end{array}} \right]} \right\}\)

Short answer

The change-of-coordinates matrix from \(B\) to the standard basis in \({\mathbb{R}^3}\) is \({P_B} = \left[ {\begin{array}{*{20}{c}}3&2&8\\{ - 1}&0&{ - 2}\\4&{ - 5}&7\end{array}} \right]\).

Step-by-step solution

State the change of coordinates

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 \({\mathop{\rm x}\nolimits} = {P_B}{\left[ {\mathop{\rm x}\nolimits} \right]_B}\). \({P_B}\) denotes thechange-of-coordinates matrixfrom \(B\) to the standard basis in \({\mathbb{R}^n}\).

Determine the change-of-coordinates matrix from \(B\)

The change-of-coordinates matrix from\(B\)to the standard basis in\({\mathbb{R}^3}\)is shown below.

\[\begin{array}{c}{P_B} = \left[ {\begin{array}{*{20}{c}}{{{\mathop{\rm b}\nolimits} _1}}&{{{\mathop{\rm b}\nolimits} _2}}&{{{\mathop{\rm b}\nolimits} _3}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}3&2&8\\{ - 1}&0&{ - 2}\\4&{ - 5}&7\end{array}} \right]\end{array}\]

Thus, the change-of-coordinates matrix from \(B\) to the standard basis in \({\mathbb{R}^3}\) is \({P_B} = \left[ {\begin{array}{*{20}{c}}3&2&8\\{ - 1}&0&{ - 2}\\4&{ - 5}&7\end{array}} \right]\).