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

 

[blue_raver22]

Yes, this satisfies the A=LU factorization since we have shown that A can be written as the product of a lower triangular matrix L and an upper triangular matrix U. The only restriction on this factorization is that the matrix A must be invertible, meaning that its determinant must not equal zero. This is because in order to find the LU factorization, we need to perform row operations on A, and row operations can only be performed on invertible matrices.