91Ó°ÊÓ

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

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

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

and

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

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

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

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

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.