212 CHAPTER 10. A FEW FACTORIZATIONS

The matrix on the right is upper triangular and so the LU factorization of the matrix M′ hasbeen obtained above.

Thus M′ = PM = LU where L and U are given above. Notice that P2 = I and therefore,M = P2M = PLU and so 1 2 3 2

1 2 3 04 3 1 1

=

 1 0 00 0 10 1 0

 1 0 0

4 1 01 0 1

 1 2 3 2

0 −5 −11 −70 0 0 −2

This process can always be followed and so there always exists a PLU factorization of

a given matrix even though there isn’t always an LU factorization.

Example 10.6.2 Use the PLU factorization of M≡

 1 2 3 21 2 3 04 3 1 1

 to solve the system

Mx = b where b =(1,2,3)T .

Let Ux = y and consider PLy = b. In other words, solve, 1 0 00 0 10 1 0

 1 0 0

4 1 01 0 1

 y1

y2

y3

=

 123

 .

Multiplying both sides by P gives 1 0 04 1 01 0 1

 y1

y2

y3

=

 132

and so

y =

 y1

y2

y3

=

 1−11

 .

Now Ux = y and so it only remains to solve

 1 2 3 20 −5 −11 −70 0 0 −2



x1

x2

x3

x4

=

 1−11

which yields (

x1 x2 x3 x4

)T=

=(

75 t + 1

59

10 −115 t t − 1

2

)T: t ∈ R.