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Chapter 4: Q12Q (page 191)

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

Expert verified

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

01

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:

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

02

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

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Most popular questions from this chapter

Exercises 23-26 concern a vector space V, a basis \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\) and the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\).

Show the coordinate mapping is one to one. (Hint: Suppose \({\left( {\bf{u}} \right)_B} = {\left( {\bf{w}} \right)_B}\) for some u and w in V, and show that \({\bf{u}} = {\bf{w}}\)).

In Exercise 1, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

1. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{3}}\\{ - {\bf{5}}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - {\bf{4}}}\\{\bf{6}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{\bf{3}}\end{array}} \right)\)

A scientist solves a nonhomogeneous system of ten linear equations in twelve unknowns and finds that three of the unknowns are free variables. Can the scientist be certain that, if the right sides of the equations are changed, the new nonhomogeneous system will have a solution? Discuss.

Let \({\mathop{\rm u}\nolimits} = \left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]\). Find \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^3}\) such that \(\left[ {\begin{array}{*{20}{c}}1&{ - 3}&4\\2&{ - 6}&8\end{array}} \right] = {{\mathop{\rm uv}\nolimits} ^T}\) .

Let V be a vector space that contains a linearly independent set \(\left\{ {{u_{\bf{1}}},{u_{\bf{2}}},{u_{\bf{3}}},{u_{\bf{4}}}} \right\}\). Describe how to construct a set of vectors \(\left\{ {{v_{\bf{1}}},{v_{\bf{2}}},{v_{\bf{3}}},{v_{\bf{4}}}} \right\}\) in V such that \(\left\{ {{v_{\bf{1}}},{v_{\bf{3}}}} \right\}\) is a basis for Span\(\left\{ {{v_{\bf{1}}},{v_{\bf{2}}},{v_{\bf{3}}},{v_{\bf{4}}}} \right\}\).
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