Compute the LU decomposition of

$$

\left(

\begin{matrix}

1 & 2 \\

3 & 4

\end{matrix}

\right).

$$

# A little LU Decomposition

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 $A$...