A =and use FORWARD SUBSTITUTION AND BACKWARD SUBSTITUTION to solve the system-10-40-4L =U =Find the LU factorization of-2 -3 2 3-4 -10 8 10#-21 -4 12 11 -6 -18-2 -3-4 -10x1 =x2 =-2 1 -4A = LU wherex3 =X4 =2 80 +831012 11 -6 -18X1x2x3X4=90

Given: A=-2-323-4-10810-21-41211-6-18 -2-323-4-10810-21-41211-6-18x1x2x3x4=-10-40-490 We have to…

A = L = Find the LU factorization of -2 -3 -4 -10 8 U = 4 x₁ = x₂ = x3 = X4 = 2 2 A∞ N A = LU where -2 1 -4 1 and use FORWARD SUBSTITUTION AND BACKWARD SUBSTITUTION to solve the system -2 -3 2 3 -10 -4 -10 8 10 -40 -2 1 -4 1 -4 3 10 11 -6 -18 2 11 -6 -18 X1 x2 x3 X4 = 90

A = L = Find the LU factorization of -2 -3 -4 -10 8 U = 4 x₁ = x₂ = x3 = X4 = 2 2 A∞ N A = LU where -2 1 -4 1 and use FORWARD SUBSTITUTION AND BACKWARD SUBSTITUTION to solve the system -2 -3 2 3 -10 -4 -10 8 10 -40 -2 1 -4 1 -4 3 10 11 -6 -18 2 11 -6 -18 X1 x2 x3 X4 = 90

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

A =
and use FORWARD SUBSTITUTION AND BACKWARD SUBSTITUTION to solve the system
-10
-40
-4
L =
U =
Find the LU factorization of
-2 -3 2 3
-4 -10 8 10
#
-2
1 -4 1
2 11 -6 -18
-2 -3
-4 -10
x1 =
x2 =
-2 1 -4
A = LU where
x3 =
X4 =
2 80 +
8
3
10
1
2 11 -6 -18
X1
x2
x3
X4
=
90

Expert Solution

Step 1

Given:

$A = [\begin{matrix} - 2 & - 3 & 2 & 3 \\ - 4 & - 10 & 8 & 10 \\ - 2 & 1 & - 4 & 1 \\ 2 & 11 & - 6 & - 18 \end{matrix}]$

$[\begin{matrix} - 2 & - 3 & 2 & 3 \\ - 4 & - 10 & 8 & 10 \\ - 2 & 1 & - 4 & 1 \\ 2 & 11 & - 6 & - 18 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix}] = [\begin{matrix} - 10 \\ - 40 \\ - 4 \\ 90 \end{matrix}]$

We have to find the LU factorization and use backward and forward substitution to solve the system.

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