Method If an nxn matrix A has an LU-factorization, then the solution of AX = b can be determined by a Forward substitution followed by a Back substitution [A] = [L][U] where [L] = lower triangular matrix all of whose diagonal entries are different from zero. [U] = upper triangular matrix all of whose diagonal entries are different from zero. LU Decomposition

How can this be used? Given [A][X] = [b] 1.Decompose [A] into [L] and [U] (LU)X = b L(UX) = b ; Let UX = z then LZ = b 1.Use Forward substitution to Solve [L][Z] = [b] for [Z] 2.Use Back substitution to Solve [U][X] = [Z] for [X]

Exercise 2.5 ; Page # 136; Qn # 1 3

Ax = b (LU)X = b L(UX) = b LZ = b; Let UX = Z 4

[L][Z] = [b] Solve for [Z] using Forward Substitution :

Complete the forward substitution to solve for [Z]

[U][X] = [Z] Solve for [X] using Back Substitution :

The Solution is :

Method: [A] Decompose to [L] and [U] Storage of Multipliers Scheme :

Exercise 2.5; Page #136; Qn # 5 Finding the [U] matrix Step 1:

Finding the [U] Matrix Step 2: Matrix after Step 1:

Finding the [L] matrix

Finding the [L] Matrix

[L][Z] = [b] Solve for [Z] using Forward Substitution :

Complete the forward substitution to solve for [Z]

[U][X] = [Z] Solve for [X] using Back Substitution :

The Solution is :

Limitations to use this Procedure : Interchanging of any two rows not allowed. Only elementary row operation permitted is the one that subtract a multiple of one row to another. In matrix A, if then this procedure fails. In matrix if, or in if then this procedure fails.

Test for you? Exercise 2.5 ; Page # 136; Qn # 7 19