# Let us define a linear transformation T:R2 → R2 by T(x) = $$\begin{bmatrix}0 & -1\\ 1 & 0\end{bmatrix}\begin{bmatrix}x_1\\ x_2\end{bmatrix}$$ = $$\begin{bmatrix}-x_2\\ x_1\end{bmatrix}$$. Find the image under T of $$u = \begin{bmatrix}4\\ 1\end{bmatrix}$$, $$v = \begin{bmatrix}2\\ 3\end{bmatrix}$$ and u + v = $$\begin{bmatrix}6\\ 4\end{bmatrix}$$

Here, Given transformation T:R2→R2 defined by

T(X) = $$\begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}\begin{bmatrix}x1\\ x2\end{bmatrix}$$

A = $$\begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}$$ is standard matrix of transforation.

Now,

T(U) = AU

= $$\begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}\begin{bmatrix}4\\ 1\end{bmatrix}$$

=$$\begin{bmatrix}-1\\ 4\end{bmatrix}$$

Also, T(v) = Av

= $$\begin{bmatrix}0 & -1\\ 1 & 0\end{bmatrix}\begin{bmatrix}2\\ 3\end{bmatrix}$$

=$$\begin{bmatrix}-3\\ 2\end{bmatrix}$$

Also

T(u+v) = A(u+v)

= $$\begin{bmatrix}0 & -1\\ 1 & 0\end{bmatrix}\begin{bmatrix}6\\ 4\end{bmatrix}$$

=$$\begin{bmatrix}-4\\ 6\end{bmatrix}$$