Page 258 - Advanced engineering mathematics

238 CHAPTER 7 Matrices and Linear Systems

Notice that the highlighting has played no role in producing U. These highlighted parts of
columns will be used to form L, which will be 3 × 3 lower triangular. The highlighted elements
are all in the first three columns, and all fall on or below the main diagonal. Now form the 3 × 3
lower triangular matrix
⎛ ⎞
2 0 0
D = 2 2 0 ⎠ .
⎝
6 −11/2
D includes the highlighted first column from A, the highlighted elements of the second column
in B, and the highlighted element of the third column in C, with zeros filled in above the
main diagonal. This is not yet L. For this, we want 1 along the main diagonal. Thus, mul-
tiply column one of D by 1/2, the second column by 1/2, and the third column by 2. This
yields L:
⎛ ⎞
1 0 0
L = 1 1 0 ⎠ .
⎝
3 −1/2 1
It is routine to check that LU = A.
This procedure can be carried out in general. First form U, exploiting leading elements
of columns of A and an elementary row operation to obtain zeros below these elements, then
retaining the elements of these columns on and above the main diagonal to form the ele-
ments of U above its main diagonal. Fill in the rest of U, below the main diagonal, with
zeros.
The highlighting strategy is a way of recording the elements to be used in forming columns
of L on and below its main diagonal. After placing these elements, and filling in zeros above
the main diagonal, multiply each column by a scalar to obtain 1’s along the main diagonal. The
resulting matrix is L.

The process of factoring A into a product of lower and upper triangular matrices is called
LU factorization.

What is the point to LU factorization? In real-world applications, matrices may be extremely
large and the numbers will not all be small integers. A great deal of arithmetic is involved in
manipulating such matrices. Upper and lower triangular matrices involve less arithmetic (hence
save computer time and money), and systems of equations having triangular coefficient matrices
are easier to solve.
As a specific instance of a simplification with LU factorization, suppose we want to solve a
system AX = B. If we write A = LU, then the system is
AX = (LU)X = L(UX) = B.
Let UX = Y and solve the system
LY = B

for Y. Once we know Y, then the solution of AX = B is the solution of
UX = Y.
Both of these systems involve triangular coefficient matrices, hence may be easier to solve
than the original system.

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October 14, 2010 14:23 THM/NEIL Page-238 27410_07_ch07_p187-246