- Thread starter
- #1

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

\( 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.