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A=lu

skoker

New member
Feb 2, 2012
14
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
Feb 7, 2012
2,676
The obvious restriction in that factorisation is that you must have $a\ne 0$.
 

skoker

New member
Feb 2, 2012
14
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
Jan 29, 2012
661
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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