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.