I read this chapter in my book and thought I understood it, but I don't. I tried working a problem to test my understanding and I just don't know how to get started.
Given the following matrices:
$A=\begin{bmatrix} 1 & 2 & -3 \\ 0 & 1 & 2 \\ -1 & 2 & 0 \\ \end{bmatrix}$ $B=\begin{bmatrix} -1 & 2 & 0 \\ 0 & 1 & 2 \\ 1 & 2 & -3 \\ \end{bmatrix}$
Find an elementary matrix $E$ such that $EA = B$
What I think I understand... a matrix is elementary when a single row operation forms an $I_n$ matrix. I don't understand how this applies though. Please help!
$\endgroup$ 51 Answer
$\begingroup$The unique matrix that satisfies $EA = B$ is the matrix that "swaps" the first and third rows. It is given as
$$E=\begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ \end{bmatrix}.$$
Edit: Due to a question in the comments, here comes a bit longer explanation.
(1) The rows of matrix $B$ and $A$ are the same, except for the fact that we have to swap the first and the third row.
(2) $E$, defined above, is the special matrix that swaps the first and third rows of any $3 \times 3$ matrix $O$ when multiplied by it from the left. This can, for example, be seen by simple matrix multiplication
$$\begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ \end{bmatrix}\begin{bmatrix} p_1 & p_2 & p_3 \\ q_1 & q_2 & q_3 \\ r_1 & r_2 & r_3 \\ \end{bmatrix} = \begin{bmatrix} r_1 & r_2 & r_3 \\ q_1 & q_2 & q_3 \\ p_1 & p_2 & p_3 \\ \end{bmatrix}. $$
Hence, as a particular case, we also have $EA=B$. (Moreover, since $A$ and $B$ are non-singular matrices, the solution to the matrix equation $XA=B$ is unique: $X=BA^{-1}$, calculating this we would again get $X=E$.)
(3) Elementary matrices (see definition here) differs from the identity matrix by one single elementary row operation. After swapping the first and third row of $E$ (which is an elementary row operation) we arrive to matrix
$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix},$$
which is exactly the identity matrix. Hence $E$ is an elementary matrix.
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