WeBWorK Standalone Renderer

Suppose $f : \mathbb{R}^{2} \to \mathbb{R}^{3}$ is a linear transformation such that $f \left( \begin{array}{c} 1 \\ 0 \end{array} \right) = \left( \begin{array}{r} 1 \\ 2 \\ 1 \end{array} \right), \ \ \ f \left( \begin{array}{c} 0 \\ 1 \end{array} \right) = \left( \begin{array}{r} -1 \\ 3 \\ -3 \end{array} \right).$ Then $\displaystyle f \left( \begin{array}{r} -3 \\ 5 \end{array} \right) =$ $\left[\Rule{0pt}{3.6em}{0pt}\right.$ $\left]\Rule{0pt}{3.6em}{0pt}\right.$ .

Suppose $f : \mathbb{R}^{12} \to \mathbb{R}^{2}$ is a linear transformation such that $f \left( \begin{array}{c} {\vec{e}}_4 \end{array} \right) = \left( \begin{array}{r} -3 \\ 4 \end{array} \right), \ \ \ f \left( \begin{array}{c} {\vec{e}}_7 \end{array} \right) = \left( \begin{array}{r} 4 \\ -1 \end{array} \right), \ \ \ f \left( \begin{array}{c} {\vec{e}}_8 \end{array} \right) = \left( \begin{array}{r} 4 \\ 2 \end{array} \right).$ Then $\displaystyle f ( 2 {\vec{e}}_4 + 7 {\vec{e}}_7 ) - f ( 6 {\vec{e}}_8 + 4 {\vec{e}}_7 )=$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$ $\left]\Rule{0pt}{2.4em}{0pt}\right.$ .

Let $V$ be a vector space and let ${\vec{v}}_1, {\vec{v}}_2, {\vec{v}}_3 \in V$ . Suppose $T : V \to \mathbb{R}^{2}$ is a linear transformation such that $T({\vec{v}}_1) = \left( \begin{array}{r} 3 \\ 4 \end{array} \right), \ \ \ T({\vec{v}}_2) = \left( \begin{array}{r} 1 \\ -1 \end{array} \right), \ \ \ T({\vec{v}}_3) = \left( \begin{array}{r} 1 \\ -4 \end{array} \right).$ Then $-5 T({\vec{v}}_1) + T(4 {\vec{v}}_2 + 6 {\vec{v}}_3) =$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$ $\left]\Rule{0pt}{2.4em}{0pt}\right.$ .