How do you find the matrix before elementary row operations have been applied?

Question

The following row operations are applied (in order) to a $3 \times 3$ matrix $A$: $\space R2\rightarrow R2 - R1,\; R3 \rightarrow R3 - 4R1, \;R3 \rightarrow R3 - 2R2$.

The matrix which results is $$ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ \end{bmatrix}. $$ Solve the system of equations $Ax= \begin{bmatrix} 2\\ -1\\ 2 \\ \end{bmatrix}. $

For the question above, I was wondering how I could find the original matrix $A$ to be able to then solve the system of equations? How do I "work backwards" with the given elementary row operations? How do I even show my working in terms of proper notation?

Answer 1

Every step you describe can be reversed. Note that I will use the notation $R_i^{(n)}$ for the new $i$-th row (i.e., after the operation is applied) and $R_i^{(o)}$ for the old $i$-th row (before the operation).

Let us look at the example $R3\to R3-2R2$. In that case, you have

$$\begin{align} R_2^{(n)} &= R_2^{(o)} R_3^{(n)} &= R_3^{(o)} - 2\cdot R_2^{(o)}\\ \end{align} $$

It is relatively simple to see that this also means that you have

$$ \begin{align} R_2^{(o)} &= R_2^{(n)} R_3^{(o)} &= R_3^{(n)} + 2\cdot R_2^{(o)} = R_3^{(n)} + 2\cdot R_2^{(n)} \end{align} $$

which means that the reverse of the operation $$R_3\to R_3 - 2R_2$$ is the operation $$R_3 \to R_3 + 2R_2.$$

Answer 2

You apply the same operations by the same order to$$\begin{bmatrix}2\\-1\\2\end{bmatrix}.$$You will get\begin{align}\begin{bmatrix}2\\-1\\2\end{bmatrix}&\rightarrow\begin{bmatrix}2\\-3\\2\end{bmatrix}\\&\rightarrow\begin{bmatrix}2\\-3\\-6\end{bmatrix}\\&\rightarrow\begin{bmatrix}2\\-3\\0\end{bmatrix}.\end{align}And now you solve the system$$\begin{bmatrix}1&2&3\\0&1&1\\0&0&1\end{bmatrix}x=\begin{bmatrix}2\\-3\\0\end{bmatrix}.$$