WeBWorK Standalone Renderer

The standard basis $\mathcal{S} = \lbrace \mathbf{e_1}, \mathbf{e_2} \rbrace$ and two custom bases $\mathcal{B} = \lbrace \mathbf{b_1}, \mathbf{b_2} \rbrace$ and $\mathcal{C} = \lbrace \mathbf{c_1}, \mathbf{c_2} \rbrace$ for $\mathbb{R}^2$ are shown in the figures below. Suppose that $f : \mathbb{R}^2 \to \mathbb{R}^2$ is the linear transformation defined by

$f(x,y) = x {\left[\begin{array}{c} -3\cr -3 \end{array}\right]} + y {\left[\begin{array}{c} 1\cr 3 \end{array}\right]}$

relative to the standard basis in the domain and the standard basis in the codomain.

Standard basis $\mathcal{S} = \lbrace \mathbf{e_1}, \mathbf{e_2} \rbrace$		Standard basis $\mathcal{S} = \lbrace \mathbf{e_1}, \mathbf{e_2} \rbrace$
	$\stackrel{\displaystyle \lbrack f \rbrack_\mathcal{S}^\mathcal{S}}{\to}$

$\lbrack id \rbrack_\mathcal{B}^\mathcal{S} \ \uparrow$		$\lbrack id \rbrack_\mathcal{C}^\mathcal{S} \ \uparrow$

	$\stackrel{\displaystyle \lbrack f \rbrack_\mathcal{B}^\mathcal{C}}{\to}$
Custom basis $\mathcal{B} = \lbrace \mathbf{b_1}, \mathbf{b_2} \rbrace$		Custom basis $\mathcal{C} = \lbrace \mathbf{c_1}, \mathbf{c_2} \rbrace$

Part 1: Matrices for the linear transformation

Find the matrix $A$ for the linear transformation $f$ relative to the standard basis in the domain and in the codomain. That is, find the matrix $A$ such that $\lbrack f \rbrack_\mathcal{S}^\mathcal{S} (\mathbf{x}) = A \mathbf{x}$ .
$A =$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$ $\left]\Rule{0pt}{2.4em}{0pt}\right.$

Find the change of basis matrix from $\mathcal{B}$ -coordinates to standard $\mathcal{S}$ -coordinates. That is, find the matrix $B$ such that $\lbrack id \rbrack_\mathcal{B}^\mathcal{S} (\mathbf{x}) = B \mathbf{x}$ .
$B =$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$ $\left]\Rule{0pt}{2.4em}{0pt}\right.$

Find the change of basis matrix from $\mathcal{C}$ -coordinates to standard $\mathcal{S}$ -coordinates. That is, find the matrix $C$ such that $\lbrack id \rbrack_\mathcal{C}^\mathcal{S} (\mathbf{x}) = C \mathbf{x}$ .
$C =$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$ $\left]\Rule{0pt}{2.4em}{0pt}\right.$

Find the matrix $D$ for the linear transformation $f$ relative to the basis $\mathcal{B}$ in the domain and $\mathcal{C}$ in the codomain. That is, find the matrix $D$ such that $\lbrack f \rbrack_\mathcal{B}^\mathcal{C} (\mathbf{x}) = D \mathbf{x}$ .
$D =$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$ $\left]\Rule{0pt}{2.4em}{0pt}\right.$

Part 2: Mapping coordinate vectors

Part 3: Verifying the diagram is commutative

You can earn partial credit on this problem.