Albarellak

2021-05-04

Solve the system $Ax=b$ using the given LU factorization of A
$A=\left[\begin{array}{cc}-2& 1\\ 2& 5\end{array}\right]=\left[\begin{array}{cc}1& 0\\ -1& 1\end{array}\right]\left[\begin{array}{cc}-2& 1\\ 0& 6\end{array}\right],b=\left[\begin{array}{c}5\\ 1\end{array}\right]$

sweererlirumeX

Step 1
$A=\left[\begin{array}{cc}-2& 1\\ 2& 5\end{array}\right]=\left[\begin{array}{cc}1& 0\\ -1& 1\end{array}\right]\left[\begin{array}{cc}-2& 1\\ 0& 6\end{array}\right],b=\left[\begin{array}{c}5\\ 1\end{array}\right]$
Step 2
$A=\left[\begin{array}{cc}1& 0\\ -1& 1\end{array}\right]\left[\begin{array}{cc}-2& 1\\ 0& 6\end{array}\right]=LU$
Let $Y=UX.\left(\begin{array}{c}{y}_{1}\\ {y}_{2}\end{array}\right)=\left[\begin{array}{cc}-2& 1\\ 0& 6\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]$
Let $LY=b.\left[\begin{array}{cc}1& 0\\ -1& 1\end{array}\right]\left(\begin{array}{c}{y}_{1}\\ {y}_{2}\end{array}\right)=\left[\begin{array}{c}5\\ 1\end{array}\right]$
Solve:
${y}_{1}=5$
$-{y}_{1}+{y}_{2}=1⇒-5+{y}_{2}=1⇒{y}_{2}=6$
Now solve $Y=UX.{y}_{1}=-2{x}_{1}+{x}_{2}.{y}_{2}=0{x}_{1}+6{x}_{2}$
Substituting the value of
$6=6{x}_{2}⇒{x}_{2}=1$
$5=-2{x}_{1}+1⇒-2{x}_{1}=4⇒{x}_{1}=-2$
$\left({x}_{1},{x}_{2}\right)=\left(-2,1\right)$