# Find an LU factorization of the matrix A (with L unit lower triangular). A=[(5,4),(-4,-3)] L-? U-?

Find an LU factorization of the matrix A (with L unit lower triangular).
$A=\left[\begin{array}{cc}5& 4\\ -4& -3\end{array}\right]$
L-?
U-?
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StrycharzT
Step 1
Given that,
$A=\left[\begin{array}{cc}5& 4\\ -4& -3\end{array}\right]$
A=LU
$⇒A=LU=\left[\begin{array}{cc}1& 0\\ {L}_{21}& 1\end{array}\right]\left[\begin{array}{cc}{U}_{11}& {U}_{12}\\ 0& {U}_{22}\end{array}\right]=\left[\begin{array}{cc}{U}_{11}& {U}_{12}\\ {L}_{21}{U}_{11}& {L}_{21}{U}_{12}+{U}_{22}\end{array}\right]$
$⇒\left[\begin{array}{cc}5& 4\\ -4& -3\end{array}\right]=\left[\begin{array}{cc}{U}_{11}& {U}_{12}\\ {L}_{21}{U}_{11}& {L}_{21}{U}_{12}+{U}_{22}\end{array}\right]$
Step 2
By comparing the matrices,
$⇒{U}_{11}=5,{U}_{12}=4$
${L}_{21}{U}_{11}=-4⇒{L}_{21}\left(5\right)=-4⇒{L}_{21}=-\frac{4}{5}$
${L}_{21}{U}_{12}+{U}_{22}=-3⇒-\frac{4}{5}\left(4\right)+{U}_{22}=-3⇒{U}_{22}=-3+\frac{16}{5}⇒{U}_{22}=\frac{1}{5}$
We have,
$L=\left[\begin{array}{cc}1& 0\\ {L}_{21}& 1\end{array}\right],U=\left[\begin{array}{cc}{U}_{11}& {U}_{12}\\ 0& {U}_{22}\end{array}\right]$
Plug the know values in matrices, get
$L=\left[\begin{array}{cc}1& 0\\ -\frac{4}{5}& 1\end{array}\right],U=\left[\begin{array}{cc}5& 4\\ 0& \frac{1}{5}\end{array}\right]$