MHB Find LU-Factorization of A with 1's Along Main Diagonal of L

[skoker]

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]

The obvious restriction in that factorisation is that you must have $a\ne 0$.

 

[skoker]

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.

 

[Fernando Revilla]

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