Let $\bold{v} = \begin{pmatrix} 5 \\ -3 \end{pmatrix}$ and $\bold{w} = \begin{pmatrix} 11 \\ -2 \end{pmatrix}$.  Find the area of the parallelogram with vertices $\bold{0}$, $\bold{v}$, $\bold{w}$, and $\bold{v} + \bold{w}$.
I know that the area of a parallelogram is the magnitude of the cross product of its adjacent sides, so I need to find $\bold{v} \times \bold{w}$. I also know that the cross product of two 2-dimensional vectors is a scalar, not a vector, and it is given by the determinant of the matrix formed by the vectors as rows or columns. So, I can write $\bold{v} \times \bold{w} = \begin{vmatrix} 5 & -3 \\ 11 & -2 \end{vmatrix}$. To evaluate this determinant, I can use the rule that it is the difference of the products of the diagonal entries, so I get $\bold{v} \times \bold{w} = 5 \cdot (-2) - (-3) \cdot 11 = -10 + 33 = 23$. Therefore, the area of the parallelogram is $|\bold{v} \times \bold{w}| = |23| = 23$ square units. 