In Exercises 27-30, use coordinate vectors to test the linear independence of the sets of polynomials. Explain your work.

\({\left( {{\bf{2}} - t} \right)^{\bf{3}}}\), \({\left( {{\bf{3}} - t} \right)^2}\), \({\bf{1}} + {\bf{6}}t - {\bf{5}}{t^{\bf{2}}} + {t^{\bf{3}}}\)

The vectors of the given polynomials can be written as follows:

\(\begin{array}{c}{\left( {2 - t} \right)^3} = 8 - 12t + 6{t^2} - {t^3}\\ \equiv \left( {\begin{array}{*{20}{c}}8\\{ - 12}\\6\\{ - 1}\end{array}} \right)\end{array}\),

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

and

\(1 + 6t - 5{t^2} + {t^3} \equiv \left( {\begin{array}{*{20}{c}}1\\6\\{ - 5}\\1\end{array}} \right)\)

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

\(A = \left( {\begin{array}{*{20}{c}}8&9&1\\{ - 12}&{ - 6}&6\\6&1&{ - 5}\\{ - 1}&0&1\end{array}} \right)\)

\(\left( {\begin{array}{*{20}{c}}8&9&1\\{ - 12}&{ - 6}&6\\6&1&{ - 5}\\{ - 1}&0&1\end{array}} \right) \sim \left( {\begin{array}{*{20}{c}}1&0&{ - 1}\\0&1&1\\0&0&0\\0&0&0\end{array}} \right)\)

From the echelon form, it can be observed that for three variables, there are two equations. Hence, one free variable is present.

So, the polynomials are linearly dependent.