# A=lu

#### skoker

##### New member
find the LU-factorization of $A=\begin{bmatrix} a\ b\ \; \\ c\ d\ \; \end{bmatrix}$ that has 1's along the main diagonal of L.are there restrictions on the matrix A?

$$A=\begin{bmatrix} \;a\ b\ \; \\ \;c\ d\ \; \end{bmatrix}$$

$$U=-\frac{c}{a}r_1+r_2\rightarrow r_2\begin{bmatrix} \;a\ b\ \; \\ \;0\ d-cb\ \; \end{bmatrix} =\underset{E_1}{\begin{bmatrix} \;1\ 0\ \; \\ \;-\frac{c}{a}\ 1\ \; \end{bmatrix}}.\underset{A}{\begin{bmatrix} \;a\ b\ \; \\ \;c\ d\ \; \end{bmatrix}}$$

$$L=\underset{E^{-1}_1}{\begin{bmatrix} \;1\ 0\ \; \\ \;\frac{c}{a}\ 1\ \; \end{bmatrix}}$$

$$\therefore A=LU$$

first does this satisfy the the a=lu? also i am not sure the restrictions they are talking about? it seems to have no restrictions.

#### Opalg

##### MHB Oldtimer
Staff member
The obvious restriction in that factorisation is that you must have $a\ne 0$.

#### skoker

##### New member
that is true. i can not think of any of the matrix properties that would be a restriction with $$abcd \quad n \times n$$. A is consistent and invertible. so i would not have any problems i think.

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#### Fernando Revilla

##### Well-known member
MHB Math Helper
In general for $A\in\mathbb{R}^{n\times n}$ invertible we can get the factorization $PA=LU$ were $P$ is a permutation matrix. In our case, if $a=0$ then, $c\neq 0$ and you can choose $P=\begin{bmatrix}{0}&{1}\\{1}&{0}\end{bmatrix}$ .

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