Introduction : In this article, we are discussing matrix representation of a linear transformation and solving a question related to this with explaining every steps.
Question (28): Let \(C =\{\left( \begin{array}{l}1\\2\end{array}\right), \left( \begin{array}{l}2\\1\end{array} \right)\}\) be a basis of \(R^2\) and \(T:R^2\rightarrow R^2\) be defined by \(T\left( \begin{array}{l}x\\y\end{array} \right)=\left( \begin{array}{l}x + y\\x – 2y\end{array} \right)\) . If \(T[C]\) represents the basis of \(T\) with respect to \(C\) then which of the following is true?\[\]
(A) \(T[C]=\left( {\begin{array}{*{20}{c}}{ – 3}&{ – 2}\\3&1\end{array}} \right)\) (B) \(T[C]=\left( {\begin{array}{*{20}{c}}{ 3}&{ – 2}\\-3&1\end{array}} \right)\)(C)\(T[C]=\left( {\begin{array}{*{20}{c}}{ – 3}&{ – 1}\\3&2\end{array}} \right)\) (D)\(T[C]=\left( {\begin{array}{*{20}{c}}{ 3}&{ – 1}\\-3&2\end{array}} \right)\)
Solution:
\[T\left( \begin{array}{l}1\\2\end{array}\right) =\left( \begin{array}{l}3\\-3\end{array} \right) = -3\left( \begin{array}{l}1\\2\end{array}\right)+3\left( \begin{array}{l}2\\1\end{array} \right)\]
\[T\left( \begin{array}{l}2\\1\end{array} \right) =\left( \begin{array}{l}3\\0\end{array} \right) =-1\left( \begin{array}{l}1\\2\end{array}\right)+2\left( \begin{array}{l}2\\1\end{array} \right) \]
\[\Rightarrow T[C]=\left( {\begin{array}{*{20}{c}}{ – 3}&{ 3}\\-1&2\end{array}} \right)^{T}= \left( {\begin{array}{*{20}{c}}{ – 3}&{ – 1}\\3&2\end{array}} \right) \]
Therefore option \((C)T[C]=\left( {\begin{array}{*{20}{c}}{ – 3}&{ – 1}\\3&2\end{array}} \right)\) is correct.