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

9. \(B = \left\{ {\left[ {\begin{array}{*{20}{c}}2\\{ - 9}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}1\\8\end{array}} \right]} \right\}\)

Short Answer

Expert verified

The change-of-coordinates matrix from \(B\) to the standard basis in \({\mathbb{R}^2}\) is \({P_B} = \left[ {\begin{array}{*{20}{c}}2&1\\{ - 9}&8\end{array}} \right]\).

Step by step solution

01

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 \({\mathbb{R}^n}\).

02

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

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

\(\begin{array}{c}{P_B} = \left[ {\begin{array}{*{20}{c}}{{{\mathop{\rm b}\nolimits} _1}}&{{{\mathop{\rm b}\nolimits} _2}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}2&1\\{ - 9}&8\end{array}} \right]\end{array}\)

Thus, the change-of-coordinates matrix from \(B\) to the standard basis in \({\mathbb{R}^2}\) is \({P_B} = \left[ {\begin{array}{*{20}{c}}2&1\\{ - 9}&8\end{array}} \right]\).

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