# Finding the inverse of a matrix using elementary matricies

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:

1. divide $R_{2}$ by $3$,
2. add $2R_{1}$ to $R_{4}$,
3. swap $R_{2}$ and $R_{4}$, then
4. 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 $Y^{-1}$

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That's how I understand it too. – Daniel Fischer Jul 8 '13 at 22:02
Sounds right. If you think of the matrix-vector multiplication function $x \mapsto Yx$ as something that takes $x$ and performs steps 1 through 4 on it, then the inverse $x \mapsto Y^{-1} x$ needs to undo those steps, in reverse order. – Christopher A. Wong Jul 8 '13 at 22:40

1. Add $R_3$ to $R_4$
2. Swap $R_2$ and $R_4$
3. Subtract $2R_1$ from $R_4$
4. Multiply $R_2$ by $3$
Apply these transformations to the identity matrix to find the inverse of $Y$.