Solve the system Ax = b using the given LU factorization of A A=[4−223]=[10121][4−204],b=[08]

Step 1 Given that A=[4−223]=[10121]×[4−204] and b=[08] Compare the above equation with A=LU and obtain that, L=[10121] and U=[4−204] Now solve the equation Ax=b by taking Ux=y as follows. Step 2 Ax=b LUx=b Ly=b [10121][y1y2]=[08] y1=0 12y1+y2=8⇒y2=8 Now substitute y=[y1y2]=[08] in Ux=y and solve as follows. Step 3 Ux=y [4−204][x1x2]=[y1y2] [4−204][x1x2]=[08] 4x1−2x2=0 4x2=8 This gives, x2=84=2 4x1−2(2)=0 4x1=4 x1=1 Thus, the solution is x=[12].

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2021-06-04

Solve the system Ax = b using the given LU factorization of A

$A = [\begin{matrix} 4 & - 2 \\ 2 & 3 \end{matrix}] = [\begin{matrix} 1 & 0 \\ \frac{1}{2} & 1 \end{matrix}] [\begin{matrix} 4 & - 2 \\ 0 & 4 \end{matrix}], b = [\begin{matrix} 0 \\ 8 \end{matrix}]$

Answer & Explanation

Brittany Patton

Skilled2021-06-05Added 100 answers

Step 1
Given that $A = [\begin{matrix} 4 & - 2 \\ 2 & 3 \end{matrix}] = [\begin{matrix} 1 & 0 \\ \frac{1}{2} & 1 \end{matrix}] \times [\begin{matrix} 4 & - 2 \\ 0 & 4 \end{matrix}] a n d b = [\begin{matrix} 0 \\ 8 \end{matrix}]$
Compare the above equation with $A = L U$ and obtain that,
$L = [\begin{matrix} 1 & 0 \\ \frac{1}{2} & 1 \end{matrix}] a n d U = [\begin{matrix} 4 & - 2 \\ 0 & 4 \end{matrix}]$
Now solve the equation $A x = b$ by taking $U x = y$ as follows.
Step 2
$A x = b$
LU $x = b$
$L y = b$
$[\begin{matrix} 1 & 0 \\ \frac{1}{2} & 1 \end{matrix}] [\begin{matrix} y_{1} \\ y_{2} \end{matrix}] = [\begin{matrix} 0 \\ 8 \end{matrix}]$
$y_{1} = 0$
$\frac{1}{2} y_{1} + y_{2} = 8 \Rightarrow y_{2} = 8$
Now substitute $y = [\begin{matrix} y_{1} \\ y_{2} \end{matrix}] = [\begin{matrix} 0 \\ 8 \end{matrix}]$ in $U x = y$ and solve as follows.
Step 3
$U x = y$
$[\begin{matrix} 4 & - 2 \\ 0 & 4 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = [\begin{matrix} y_{1} \\ y_{2} \end{matrix}]$
$[\begin{matrix} 4 & - 2 \\ 0 & 4 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = [\begin{matrix} 0 \\ 8 \end{matrix}]$
$4 x_{1} - 2 x_{2} = 0$
$4 x_{2} = 8$
This gives,
$x_{2} = \frac{8}{4} = 2$
$4 x_{1} - 2 (2) = 0$
$4 x_{1} = 4$
$x_{1} = 1$
Thus, the solution is $x = [\begin{matrix} 1 \\ 2 \end{matrix}]$ .

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