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 = \(\begin{bmatrix}6\\ 4\end{bmatrix}\)

 

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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)

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

=\(\begin{bmatrix}-4\\ 6\end{bmatrix}\)

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