Most, but not all. square matrices have an LU-Decomposition. Consider the matrix

Question

Most, but not all. square matrices have an LU-Decomposition. Consider the matrix g = [0 2 1 3 1 1 6 -8 3 Why does G fail to have an LU-Deeomposition? Explain in complete sentences. Write G as a product PLU where L is lower triangular. U is upper triangular, and P is a permutation matrix (i.e. a product of type-1 elementary matrices). In class we discussed a step-by-step produce for using LU-decompositions to solves systems. Create, and clearly explain, a step-by-step procedure for using PLU-Decompositions to solve systems. Use the procedure you created to solve the following system Gx = [5 18 17]

Explanation / Answer

a) Suppose G admits a LU decomposition , with L =

and U =

This would imply

                                ap =0, aq=2 , ar =1, bp =3.

So neither a nor p can be zero from the last two equations, but the first equation would imply otherwise.

So LU decomposition is not possible for the matrix G.

b) Let G =PLU, where the permutation matrix exchanges the first and third rows, so P =

Then G = P H, where H is the matrix obtained by swapping the first and third rows of G.

It can be easily checked that H admits LU decomposition , with L and U given by

L =

and U =

C) ,D). The steps are as follows;First solve L[x y z]' = Pb', and then U[p q r]= [x y z]'.Then [p q r] is the vector which solves the given equation G[p q r]' =b'.

In our case [x y z ] are the solutions of L[x y z]' = [17 18 5]' , which yield

                                                                     x = 17, y=9.5 and z = 1.2.

To obtain the solution [p q r]' , solve the system

                                                              U[p q r]' = [ 17 9.5 1.2]'

to obtain the desired solution : p =5, q =2, r =1

a 0 0 b c 0 d e f

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