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3 1 2 Let A = 6 3 4 3 1 5 We know that A can be decomposed as A = LU where Lis lower triangular matrix and U is upper triangular matrix. The matrix A is decomposed as followings Г3 1 2 6 3 4 Г3 1 0 1 b23 Lo o b33. 1 2 a21 a22 0 3 1 A32 Fill in the blank in the below.

3 1 2 Let A = 6 3 4 3 1 5 We know that A can be decomposed as A = LU where Lis lower triangular matrix and U is upper triangular matrix. The matrix A is decomposed as followings Г3 1 2 6 3 4 Г3 1 0 1 b23 Lo o b33. 1 2 a21 a22 0 3 1 A32 Fill in the blank in the below.

Advanced Engineering Mathematics
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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3 1 2 Let A =|6 3 4 3 1 5 We know that A can be decomposed as A = LU where L is lower triangular matrix and U is upper triangular matrix. The matrix A is decomposed as followings 3 1 2 1 3 1 2 0 1 b23 [o 0 b33. 6 3 4 a21 a22 3 1 1 a32 Fill in the blank in the below. A21 = a22 = a32 b23 b33 || ||
Transcribed Image Text:3 1 2 Let A =|6 3 4 3 1 5 We know that A can be decomposed as A = LU where L is lower triangular matrix and U is upper triangular matrix. The matrix A is decomposed as followings 3 1 2 1 3 1 2 0 1 b23 [o 0 b33. 6 3 4 a21 a22 3 1 1 a32 Fill in the blank in the below. A21 = a22 = a32 b23 b33 || ||
Expert Solution
Step 1

Given that

312 634 315=100 a21a220 1a321312 0 1b23 0 0e

To find the values of a21,a22, a32, b23, b33.

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