Let $S$ be the parallelogram determined by the vectors : $b_1 = (-3,4), b_2 = (-3,8)$ and,
$A = \begin{bmatrix} 6 & -8 \\ -2 & 8 \\ \end{bmatrix} $
Compute the image $S$ under the mapping $x \mapsto Ax$
I don't really care for the answer I just want to know the mechanics and how to do the arithmetic. I have no clue.
Recall that the norm of the cross product gives the area spanned by two vectors (which in general will be a parallelogram). The vectors $b_1$ and $b_2$ get mapped to the vectors:
$$ Ab_1 \;\; =\;\; \begin{bmatrix} 6 & -8 \\ -2 & 8 \\ \end{bmatrix} \left [ \begin{array}{c} -3\\ 4\\ \end{array} \right ] \;\; =\;\; \left [ \begin{array}{c} -50 \\ 38 \\ \end{array} \right ] $$ and $$ Ab_2 \;\; =\;\; \begin{bmatrix} 6 & -8 \\ -2 & 8 \\ \end{bmatrix} \left [\begin{array}{c} -3 \\ 8 \\ \end{array} \right ] \;\; =\;\; \left [\begin{array}{c} -82 \\ 70 \\ \end{array} \right ]. $$
The area of these can easily be found by arranging them as the columns of a new matrix and taking their determinant:
$$ \text{Area} \;\; =\;\; \left |\det \left [ \begin{array}{cc} -50 & -82 \\ 38 & 70 \\ \end{array} \right ] \right | \;\; =\;\; |-3500 + 2624| \;\; =\;\; 876. $$
Another way of seeing this is by recalling that the formula for the area of a parallelogram is base x height, which is equivalent to the quantity $||b_1||\cdot ||b_2||\sin\theta$ where $\theta$ is the angle between the vectors. In this case, finding $\sin\theta$ is the most challenging part to this, but it can easily be found from the dot product as
$$ \sin\theta \;\; =\;\; \sqrt{1-\cos^2\theta} \;\; =\;\; \sqrt{1 - \left (\frac{b_1\cdot b_2}{||b_1||\cdot ||b_2||} \right )^2}. $$