Can somebody help me understand what exactly is being asked here? I understand how to construct elementary matrices from these row operations, but I'm unsure what the end goal is. Am I to assume that $Y$ is built from these row operations?
Let $Y$ be the $4\times 4$ matrix which applies the following operations, in the order given, when used as a left-multiplier:
- divide $R_{2}$ by $3$,
- add $2R_{1}$ to $R_{4}$,
- swap $R_{2}$ and $R_{4}$, then
- subtract $R_{4}$ from $R_{3}$.
Find $Y^{-1}$ without calculating $Y$.
If I were to venture a guess, I would say that it's implying that $E_{4}E_{3}E_{2}E_{1}=Y$, and therefore I need to find $Y^{-1}=E^{-1}_{1}E^{-1}_{2}E^{-1}_{3}E^{-1}_{4}$. But the wording of the question makes me not 100% sure.
I've found \begin{align*}
E_{4}E_{3}E_{2}E_{1}=\left[\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & -1 \\
0 & 0 & 0 & 1
\end{array}\right]\left[\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0
\end{array}\right]\left[\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
2 & 0 & 0 & 1
\end{array}\right]\left[\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & \frac{1}{3} & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{array}\right]
\end{align*}
so if my assumption is right, I'd just need to build the inverse of the matrices and multiply them in reverse order to get $Y^{-1}$
Best Answer
Yes, that is correct. The inverse of an elementary operation matrix is the operation that reverses that elementary operation:
Apply these transformations to the identity matrix to find the inverse of $Y$.