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

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

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

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

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

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 \({{\bf{P}}_3}\).