I am currently working on LU factorization of a matrix, and I'm struggling with the following problem:
$$A=\begin{bmatrix}1&2&3&1&2\\ 1&4&2&3&1\\ 2&2&-1&1&1\end{bmatrix}$$
I have tried to factorize this matrix into a lower triangular matrix and an upper triangular matrix, but I have not been successful. Here are the steps I have taken so far:
I start by subtracting the first row from the second and third rows, to get:
$$ \left[\begin{array}{ccccc} 1 & 2 & 3 & 1 & 2 \\ 0 & 2 & -1 & 2 & -1 \\ 0 & -2 & -7 & -1 & -3 \end{array}\right] $$ Then, I swapped the second and third rows to get: $$ \left[\begin{array}{ccccc} 1 & 2 & 3 & 1 & 2 \\ 0 & -2 & -7 & -1 & -3 \\ 0 & 2 & -1 & 2 & -1 \end{array}\right] $$
I then subtracted the second row from the third row, to get: $$ \left[\begin{array}{ccccc} 1 & 2 & 3 & 1 & 2 \\ 0 & -2 & -7 & -1 & -3 \\ 0 & 0 & 6 & 3 & 2 \end{array}\right] $$ At this point, I got stuck and could not proceed further. I would appreciate any help or guidance on how to proceed with this problem. Thank you in advance!
First, row-reduce $A$ (put it in upper-triangular form $U$):
$\begin{bmatrix}1&2&3&1&2\\1&4&2&3&1\\2&2&-1&1&1\end{bmatrix} \overrightarrow{\substack{r_2^*=r_2-r_1\\r_3^*=r_3-2r_1}}\begin{bmatrix}1&2&3&1&2\\0&2&-1&2&-1\\0&-2&-7&-1&-3\end{bmatrix}\overrightarrow{r_3^*=r_3+r_2}\begin{bmatrix}1&2&3&1&2\\0&2&-1&2&-1\\0&0&-8&1&-4\end{bmatrix}$
Next, gather all the steps of row-reduction into elementary matrices. Elementary matrices are square with the same number of rows as $A$. They have 1's on the diagonal, with an entry in the $i^{\text{th}}$ row (the row where the "new" row occurs) and $j^{\text{th}}$ column (the row of the former matrix whose multiple is being added) with value $k$ (corresponding the multiple of which is being added). Here we have:
$E_1=\begin{bmatrix} 1&0&0\\-1&1&0\\0&0&1\end{bmatrix}, \;\; E_2=\begin{bmatrix} 1&0&0\\0&1&0\\-2&0&1\end{bmatrix}, \;\; E_3=\begin{bmatrix} 1&0&0\\0&1&0\\0&1&1\end{bmatrix}$
So $E_3E_2E_1A=U$ (make sure you understand the ordering!) and thus $A=(E_3^{-1}E_2^{-1}E_1^{-1})U=LU$. The inverse of the elementary matrices just flips the sign of the off-diagonal entries, and the product just collects the off-diagonal entries into one matrix (you can convince yourself of this fact by playing around with examples or by carefully looking at the rules of matrix multiplication).
Finally, we arrive at our answer:
$L=\begin{bmatrix} 1&0&0\\1&1&0\\2&-1&1\end{bmatrix}$
Indeed, we can verify that $A=LU$ by quickly checking that $\begin{bmatrix}1&2&3&1&2\\1&4&2&3&1\\2&2&-1&1&1\end{bmatrix}=\begin{bmatrix} 1&0&0\\1&1&0\\2&-1&1\end{bmatrix}\begin{bmatrix}1&2&3&1&2\\0&2&-1&2&-1\\0&0&-8&1&-4\end{bmatrix}$