A little LU Decomposition

Compute the LU decomposition of
$$\left( \begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix} \right).$$

Let $A=\begin{pmatrix} 1&2\\3&4\end{pmatrix}$.
Then a lower triangular matrix is $$L=\begin{pmatrix} 1&0\\3&1\end{pmatrix}$$
and an upper triangular matrix is $$U=\begin{pmatrix} 1&2\\0&-2\end{pmatrix}$$ .
We can check our answer by solving $A=LU$ $$A=\begin{pmatrix} 1&0\\3&1\end{pmatrix}\begin{pmatrix} 1&2\\0&-2\end{pmatrix}$$ $$=\begin{pmatrix} 1&2\\3&4\end{pmatrix}$$
which is in fact $A$.

@Cheryl What about the pivot matrix $P$?

Doesn't P have to be either $$\begin{pmatrix} 1&0\\0&1\end{pmatrix}$$ or $$\begin{pmatrix} 0&1\\ 1&0\end{pmatrix}$$

Can we count the identity matrix as P or does it HAVE to permutate?

@ctyra Yes, $P$ must be one of those two matrix. More generally, an $n$-dimensional permutation matrix is just an identity matrix with it's rows permuted. The permutation can be any permutation - including the identity permutation, which produces the identity matrix.

Shouldn't we pivot to swap row 1 and row 2 in $A$? That way the row with the largest absolute value in the first column is up top? Or did I completely misunderstand that in class?

Swapping the rows, I got
$$U= \begin{bmatrix} 3 & 4 \\ 0 & \frac{2}{3} \end{bmatrix}$$

$$L= \begin{bmatrix} 1 & 0 \\ \frac{1}{3} & 1 \end{bmatrix}$$

$$P= \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$

Multiplying all these together gave me correct $A$...