The area of the parallelogram generated by the vectors $\mathbf{a}$ and $\mathbf{b}$ is 8.  Find the area of the parallelogram generated by the vectors $2 \mathbf{a} + 3 \mathbf{b}$ and $\mathbf{a} - 5 \mathbf{b}.$
Since the area of the parallelogram generated by the vectors $\mathbf{a}$ and $\mathbf{b}$ is 8,
\[\|\mathbf{a} \times \mathbf{b}\| = 8.\]Then the area of the parallelogram generated by the vectors $2 \mathbf{a} + 3 \mathbf{b}$ and $\mathbf{a} - 5 \mathbf{b}$ is
\[\|(2 \mathbf{a} + 3 \mathbf{b}) \times (\mathbf{a} - 5 \mathbf{b})\|.\]Expanding the cross product, we get
\begin{align*}
(2 \mathbf{a} + 3 \mathbf{b}) \times (\mathbf{a} - 5 \mathbf{b}) &= 2 \mathbf{a} \times \mathbf{a} - 10 \mathbf{a} \times \mathbf{b} + 3 \mathbf{b} \times \mathbf{a} - 15 \mathbf{b} \times \mathbf{b} \\
&= \mathbf{0} - 10 \mathbf{a} \times \mathbf{b} - 3 \mathbf{a} \times \mathbf{b} - \mathbf{0} \\
&= -13 \mathbf{a} \times \mathbf{b}.
\end{align*}Thus, $\|(2 \mathbf{a} + 3 \mathbf{b}) \times (\mathbf{a} - 5 \mathbf{b})\| = 13 \|\mathbf{a} \times \mathbf{b}\| = \boxed{104}.$
104
Level 4
Precalculus
Problem: The area of the parallelogram generated by the vectors $\mathbf{a}$ and $\mathbf{b}$ is 8.  Find the area of the parallelogram generated by the vectors $2 \mathbf{a} + 3 \mathbf{b}$ and $\mathbf{a} - 5 \mathbf{b}.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the area of the parallelogram generated by the vectors $2 \mathbf{a} + 3 \mathbf{b}$ and $\mathbf{a} - 5 \mathbf{b}$, we first need to understand how the area of a parallelogram formed by two vectors can be calculated. The area is given by the magnitude of the cross product of the two vectors.

Let $\mathbf{u} = 2 \mathbf{a} + 3 \mathbf{b}$ and $\mathbf{v} = \mathbf{a} - 5 \mathbf{b}$. The area of the parallelogram formed by $\mathbf{u}$ and $\mathbf{v}$ is given by:

Area = ||$\mathbf{u} \times \mathbf{v}$||.

We can express the cross product in terms of $\mathbf{a}$ and $\mathbf{b}$. First, we calculate $\mathbf{u} \times \mathbf{v}$:

$\mathbf{u} \times \mathbf{v} = (2 \mathbf{a} + 3 \mathbf{b}) \times (\mathbf{a} - 5 \mathbf{b})$.

Using the distributive property of the cross product, we expand this:

= $2 \mathbf{a} \times \mathbf{a} - 10 \mathbf{a} \times \mathbf{b} + 3 \mathbf{b} \times \mathbf{a} - 15 \mathbf{b} \times \mathbf{b}$.

Since the cross product of any vector with itself is zero, we have:

= $0 - 10 \mathbf{a} \times \mathbf{b} + 3 \mathbf{b} \times \mathbf{a} + 0$.

We know that $\mathbf{b} \times \mathbf{a} = -(\mathbf{a} \times \mathbf{b})$, thus:

= $-10 \mathbf{a} \times \mathbf{b} - 3 \mathbf{a} \times \mathbf{b}$.

Combining the terms gives:

= $-13 \mathbf{a} \times \mathbf{b}$.

Now, we find the magnitude of this vector:

||$\mathbf{u} \times \mathbf{v}$|| = 13 ||$\mathbf{a} \times \mathbf{b}$||.

The area of the parallelogram formed by $\mathbf{a}$ and $\mathbf{b}$ is given as 8, which means:

||$\mathbf{a} \times \mathbf{b}$|| = 8.

Substituting this value into our expression for the area gives:

Area = 13 * 8 = 104.

Thus, the final answer is $104$.<|END_OF_LATENT|> The final answer is $104$.