Factor the following matrix into PA = LU. Factor it also into A = L₁P₁U₁ (hold the exchange of row 3 until 3 times row 1 is subtracted from row 2):

$$A = \begin{vmatrix} 0 & 1 & 2 \\ 0 & 3 & 8 \\ 2 & 1 & 1 \\ \end{vmatrix}$$

To factor into PA = LU, observe that:

$$\begin{vmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{vmatrix} \space A = \begin{vmatrix} 2 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 3 & 8 \\ \end{vmatrix} = L \space \begin{vmatrix} 2 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \\ \end{vmatrix}$$

This means:

$$P = \begin{vmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{vmatrix}$$ $$L = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 3 & 1 \\ \end{vmatrix}$$ $$U = \begin{vmatrix} 2 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \\ \end{vmatrix}$$

To factor into A = L₁P₁U₁, observe that:

$$\begin{vmatrix} 0 & 1 & 2 \\ 0 & 3 & 8 \\ 2 & 1 & 1 \\ \end{vmatrix} \rightarrow L_{1} \begin{vmatrix} 0 & 1 & 2 \\ 0 & 0 & 2 \\ 2 & 1 & 1 \\ \end{vmatrix} = L_{1} P_{1} U_{1}$$

which means:

$$L_{1} = \begin{vmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & 1 \\ \end{vmatrix}$$ $$U_{1} = \begin{vmatrix} 2 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \\ \end{vmatrix}$$ $$P_{1} = \begin{vmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\ \end{vmatrix}$$

# posted by rot13(Unafba Pune) @ 7:40 PM