Let us define a linear transformation T: mathbbR2 to mathbbR2 by T(x) | Mathematics II 2076

Mathematics II20765 marksNumerical: linear transformation imagethe matrix of a linear Transformation

Let us define a linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ 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 $u + v = \begin{bmatrix} 6 \\ 4 \end{bmatrix}$ . [5]

Let us define a linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ 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 $u + v = \begin{bmatrix} 6 \ 4 \end{bmatrix}$. [5]

Linear transformation $T:\mathbb{R}^2 \to \mathbb{R}^2$ : $T(x) = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} x1 \\ x2 \end{bmatrix} = \begin{bmatrix} -x2 \\ x1 \end{bmatrix}$ Vectors: - $u = \begin{bmatrix} 4 \\ 1 \end{bmatrix}$ - $v = \begin{bmatrix} 2 \\ 3 \end{bmatrix}$ -

Linear transformation $T:\mathbb{R}^2 \to \mathbb{R}^2$: $$T(x) = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}\begin{bmatrix} x1 \ x2 \end{bmatrix} = \begin{bmatrix} -x2 \ x1 \end{bmatrix}$$ Vectors: - $u = \begin{bmatrix} 4 \ 1 \end{bmatrix}$ - $v = \begin{bmatrix} 2 \ 3 \end{bmatrix}$ -

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