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Chapter 4: Q34E (page 191)

In Exercises 33 and 34, determine whether the sets of polynomials form a basis for \({{\bf{P}}_3}\). Justify your conclusions.

(M) \({\bf{5}} - {\bf{3}}t + {\bf{4}}{t^{\bf{2}}} + {\bf{2}}{t^{\bf{3}}}\), \({\bf{9}} + t + {\bf{8}}{t^{\bf{2}}} - {\bf{6}}{t^{\bf{3}}}\), \({\bf{6}} - {\bf{2}}t + {\bf{5}}{t^{\bf{2}}}\), \({t^{\bf{3}}}\)

Short Answer

Expert verified

Set S is not the basis for \({{\bf{P}}_3}\).

Step by step solution

01

Write the polynomial in the standard vector form

\(\left\{ {5 - 3t + 4{t^2} + 2{t^3},\;9 + t + 8{t^2} - 6{t^3},6 - 2\;t + 5{t^2},\;{t^3}} \right\} = \left\{ {\left( {\begin{array}{*{20}{c}}5\\{ - 3}\\4\\2\end{array}} \right),\;\;\left( {\begin{array}{*{20}{c}}9\\1\\8\\{ - 6}\end{array}} \right),\;\;\left( {\begin{array}{*{20}{c}}6\\{ - 2}\\5\\0\end{array}} \right),\;\;\left( {\begin{array}{*{20}{c}}0\\0\\0\\1\end{array}} \right)} \right\}\)

02

Form the matrix using the vectors

The matrix formed by using the vectors of the polynomials is:

\(A = \left( {\begin{array}{*{20}{c}}5&9&6&0\\{ - 3}&1&{ - 2}&0\\4&8&5&0\\2&{ - 6}&0&1\end{array}} \right)\)

03

Write the matrix in the echelon form

Consider matrix\(A = \left( {\begin{array}{*{20}{c}}5&9&6&0\\{ - 3}&1&{ - 2}&0\\4&8&5&0\\2&{ - 6}&0&1\end{array}} \right)\).

Use code in the MATLAB to obtain the row-reducedechelon formas shown below:

\(\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left( {{\rm{ }}\begin{array}{*{20}{c}}5&9&6&{0;\;\;\begin{array}{*{20}{c}}{ - 3}&1&{ - 2}&{0;\;\;\begin{array}{*{20}{c}}4&8&5&{0;\;\;\begin{array}{*{20}{c}}2&{ - 6}&0&1\end{array}}\end{array}}\end{array}}\end{array}} \right);\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\)

\(\left( {\begin{array}{*{20}{c}}5&9&6&0\\{ - 3}&1&{ - 2}&0\\4&8&5&0\\2&{ - 6}&0&1\end{array}} \right) \sim \left( {\begin{array}{*{20}{c}}1&0&{\frac{3}{4}}&0\\0&1&{\frac{1}{4}}&0\\0&0&0&1\\0&0&0&0\end{array}} \right)\)

It can be observed from the echelon form that the last row is zero, which shows the set is linearly dependent.

Therefore, set S is not the basis for \({{\bf{P}}_3}\).

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Most popular questions from this chapter

Use Exercise 28 to explain why the equation\(Ax = b\)has a solution for all\({\rm{b}}\)in\({\mathbb{R}^m}\)if and only if the equation\({A^T}x = 0\)has only the trivial solution.

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

22. (M) \(A = \left( {\begin{array}{*{20}{c}}0&1&0&0\\0&0&1&0\\0&0&0&1\\{ - 1}&{ - 13}&{ - 12.2}&{ - 1.5}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}1\\0\\0\\{ - 1}\end{array}} \right)\).

(M) Determine whether w is in the column space of \(A\), the null space of \(A\), or both, where

\({\mathop{\rm w}\nolimits} = \left( {\begin{array}{*{20}{c}}1\\1\\{ - 1}\\{ - 3}\end{array}} \right),A = \left( {\begin{array}{*{20}{c}}7&6&{ - 4}&1\\{ - 5}&{ - 1}&0&{ - 2}\\9&{ - 11}&7&{ - 3}\\{19}&{ - 9}&7&1\end{array}} \right)\)

Consider the following two systems of equations:

\(\begin{array}{c}5{x_1} + {x_2} - 3{x_3} = 0\\ - 9{x_1} + 2{x_2} + 5{x_3} = 1\\4{x_1} + {x_2} - 6{x_3} = 9\end{array}\) \(\begin{array}{c}5{x_1} + {x_2} - 3{x_3} = 0\\ - 9{x_1} + 2{x_2} + 5{x_3} = 5\\4{x_1} + {x_2} - 6{x_3} = 45\end{array}\)

It can be shown that the first system of a solution. Use this fact and the theory from this section to explain why the second system must also have a solution. (Make no row operations.)

Consider the polynomials \({{\bf{p}}_{\bf{1}}}\left( t \right) = {\bf{1}} + t\), \({{\bf{p}}_{\bf{2}}}\left( t \right) = {\bf{1}} - t\), \({{\bf{p}}_{\bf{3}}}\left( t \right) = {\bf{4}}\), \({{\bf{p}}_{\bf{4}}}\left( t \right) = {\bf{1}} + {t^{\bf{2}}}\), and \({{\bf{p}}_{\bf{5}}}\left( t \right) = {\bf{1}} + {\bf{2}}t + {t^{\bf{2}}}\), and let H be the subspace of \({P_{\bf{5}}}\) spanned by the set \(S = \left\{ {{{\bf{p}}_{\bf{1}}},\,{{\bf{p}}_{\bf{2}}},\;{{\bf{p}}_{\bf{3}}},\,{{\bf{p}}_{\bf{4}}},\,{{\bf{p}}_{\bf{5}}}} \right\}\). Use the method described in the proof of the Spanning Set Theorem (Section 4.3) to produce a basis for H. (Explain how to select appropriate members of S.)
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