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Chapter 4: Q13E (page 191)

In \({{\rm{P}}^2}\), find the change-of-coordinates matrix from the basis \(B = \left\{ {1 - 2t + {t^2},\,3 - 5t + 4{t^2},\,2t + 3{t^2}} \right\}\) to the standard basis \(C = \left\{ {1,t,{t^2}} \right\}\)Then find the \(B\)-coordinate for\(\left\{ {1 - 2t} \right\}\).

Short Answer

Expert verified

\(\left( {\begin{array}{*{20}{c}}1&0&0&{\,\,\,\,\,\,5}\\0&1&0&{\,\,\, - 2}\\0&0&0&{\,\,\,\,\,\,1}\end{array}} \right)\), \({\left( x \right)_B} = \left( \begin{array}{l}\,\,\,5\\ - 2\\\,\,\,1\end{array} \right)\)

Step by step solution

01

Write the \(C\)- coordinate vector of \({b_1}\), \({b_2}\), \({b_3}\) and then write \(\mathop P\limits_{C \to B} \) matrix

The \(C\)-coordinates vectors of \({b_1}\),\({b_2}\),\({b_3}\) are\({\left( {{b_1}} \right)_c} = \left( \begin{array}{l}\,\,\,1\,\\ - 2\\\,\,\,1\end{array} \right)\), \({\left( {{b_2}} \right)_c} = \left( \begin{array}{l}\,\,\,3\\ - 5\\\,\,\,4\end{array} \right)\), \({\left( {{b_3}} \right)_c} = \left( \begin{array}{l}\,\,\,0\,\\\,\,\,2\\\,\,\,3\end{array} \right)\). Thus, the \(\mathop P\limits_{C \to B} \) matrix can be written as \(\mathop P\limits_{C \to B} = \left( {\begin{array}{*{20}{c}}1&3&0\\{ - 2}&{ - 5}&2\\1&4&3\end{array}} \right)\).

02

Use the result \({\left( x \right)_C} = P{\left( x \right)_B}\)

If \(x\)is the vector \(\left\{ { - 1 + 2t} \right\}\), then according to the result \({\left( x \right)_c} = \mathop P\limits_{C \to B} {\left( x \right)_B}\), you can write

\(\mathop P\limits_{C \to B} {\left( x \right)_B} = \left( \begin{array}{l} - 1\\\,\,\,2\\\,\,\,0\end{array} \right)\).

03

Substitute the matrix for\(\mathop P\limits_{C \to B} \) and apply row reduction on the augmented matrix

\(\begin{array}{l}\left( {\begin{array}{*{20}{c}}1&3&0&{ - 1}\\{ - 2}&{ - 5}&2&{\,\,\,2}\\1&4&3&{\,\,0}\end{array}} \right) \sim \left( {\begin{array}{*{20}{c}}1&0&0&{\,\,\,\,\,\,5}\\0&1&0&{\,\,\, - 2}\\0&0&0&{\,\,\,\,\,\,1}\end{array}} \right)\\ \sim \end{array}\)

04

Compare the resulting matrix with \(P{\left( x \right)_B}\) and write the column matrix\({\left( x \right)_B}\)

\({\left( x \right)_B} = \left( \begin{array}{l}\,\,\,5\\ - 2\\\,\,\,1\end{array} \right)\)

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Most popular questions from this chapter

In Exercise 5, find the coordinate vector \({\left( x \right)_{\rm B}}\) of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).

5. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{3}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{5}}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{\bf{1}}\end{array}} \right)\)

In Exercise 7, find the coordinate vector \({\left( x \right)_{\rm B}}\) of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).

7. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{1}}}\\{ - {\bf{3}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{3}}}\\{\bf{4}}\\{\bf{9}}\end{array}} \right),{b_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{2}}}\\{\bf{4}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{\bf{8}}\\{ - {\bf{9}}}\\{\bf{6}}\end{array}} \right)\)

Is it possible that all solutions of a homogeneous system of ten linear equations in twelve variables are multiples of one fixed nonzero solution? Discuss.

Question: Determine if the matrix pairs in Exercises 19-22 are controllable.

20. \(A = \left( {\begin{array}{*{20}{c}}{.8}&{ - .3}&0\\{.2}&{.5}&1\\0&0&{ - .5}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}1\\1\\0\end{array}} \right)\).

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 \({\bf{u}} = {\bf{w}}\)).
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