flightwingsd2

2022-06-26

I had a question about a problem that I was working on for my pre-calculus class.

Here's the problem statement:

The area of the parallelogram with vertices 0, $м$, $w$, and $v+w$ is 34. Find the area of the parallelogram with vertices 0, $Av$, $Aw$, and $Av+Aw$, where
$A=\left(\begin{array}{cc}3& -5\\ -1& -3\end{array}\right).$
I got the answer by doing something very tedious. I set $v=\left(\genfrac{}{}{0}{}{17}{0}\right)$ and $w=\left(\genfrac{}{}{0}{}{0}{2}\right)$, and did some really crazy matrix multiplication and a lot of plotting points of GeoGebra to get the answer of: $\overline{)476}$.

Now, I'm 100% sure that was not the fastest way, can someone tell me the non-bash way to do the problem? Angelo Murray

Expert

Hint: If $v=\left(\begin{array}{c}a\\ b\end{array}\right)$ and $w=\left(\begin{array}{c}c\\ d\end{array}\right)$ then the area of the original parallelogram is
$|det\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)|=|ad-bc|.$
Provide a similar expression for the area of the second parallelogram, and relate it to the above expression using properties of determinants. sedeln5w

Expert

You're probably supposed to know that the absolute value of the determinant of $A$ is the area scale factor of the linear transformation represented by the matrix $A$. With this knowledge, you can easily solve the problem in your head.
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