Question

Let A and b be the matrices A = 1 2 4 17

3 6 −12 3

2 3 −3 2

0 2 −2 6

and b = (17, 3, 3, 4) . (a) Explain why A does not have an LU factorization. (b) Use partial pivoting and find the permutation matrix P as well as the LU factors such that PA = LU. (c) Use the information in P, L, and U to solve Ax = b

Answer 1

Given matrix is : A = .

Now, =

This shows that A cannot be expressed as a product of a lower triangular matrix and an upper triangular matrix.

Therefore, A does not have an LU factorization.

b) Now we apply elementary row operations on A.

Step I : Interchange R₁ and R₂

Then A becomes =

Step II : R₂-(1/3)R₁=R₂, R₃-(2/3)R₁=R₃

Then A becomes =

Step III : Interchange R₂ and R₄

Then A becomes =

Step IV : R₃+(1/2)R₂=R₃

Then A becomes =

Step V : Interchabge R₃ and R₄

Then A becomes =

Step VI : R₄-(1/2)R₃=R₄

Then A becomes = = U

Therefore, A = E₁₂E₂₁(1/3)E₃₁(2/3)E₂₄E₃₂(-1/2)E₃₄E₄₃(1/2)U

Then, P^-1L = E₁₂E₂₁(1/3)E₃₁(2/3)E₂₄E₃₂(-1/2)E₃₄E₄₃(1/2)

i.e., P^-1L =

i.e., P^-1L =

Therefore, P = and L = .

c) Now we have AX = B where B =

i.e., PAX = PB

i.e., LUX = PB [Since PA = LU]

First we solve LY = PB where Y = .

Then, =

i.e., =

i.e., a = 3

b = 4

(1/3)a+c = 17

(2/3)a-(1/2)b+(1/2)c+d = 3

i.e., a = 3, b = 4, c = 16, d = -5

Therefore, Y =

Now we solve UX = Y where X = .

i.e., =

i.e., 3x+6y-12z+3u = 3

2y-2z+6u = 4

8z+16u = 16

-5u = -5

i.e., u = 1, z = 0, y = -1, x = 2

Therefore, X = .