Linear Algebra, Theory And Applications, 2012a

5.3. SOLVING LINEAR SYSTEMS USING AN LU FACTORIZATION 125
the row operation which involves replacing a row by itself added to a multiple of another
row. The matrix L is what you get by updating the identity matrix as illustrated above.
You should note that for a square matrix, the number of row operations necessary to
reduce to LU form is about half the number needed to place the matrix in row reduced
echelon form. This is why an LU factorization is of interest in solving systems of equations.
5.3 Solving Linear Systems Using An LU Factorization
The reason people care about the LU factorization is it allows the quick solution of systems
of equations. Here is an example.
⎛ ⎞
⎛
Example 5.3.1 Suppose you want to find the solutions to ⎝ 1 2 3 2
⎞ x
4 3 1 1 ⎠ ⎜ y
⎟
⎝ z ⎠ =
1 2 3 0
⎛
w
⎝ 1 ⎞
2 ⎠ .
3
Of course one way is to write the augmented matrix and grind away. However, this
involves more row operations than the computation of an LU factorization and it turns out
that an LU factorization can give the solution quickly. Here is how. The following is an LU
factorization for the matrix.
⎛
⎝ 1 2 3 2
4 3 1 1
1 2 3 0
⎞ ⎛
⎠ = ⎝ 1 0 0 ⎞ ⎛
4 1 0 ⎠ ⎝ 1 2 3 2 ⎞
0 −5 −11 −7 ⎠ .
1 0 1 0 0 0 −2
Let Ux = y and consider Ly = b where in this case, b =(1, 2, 3) T .Thus
⎛
1 0 0
⎞ ⎛
y 1
⎞ ⎛ ⎞
1
⎝ 4 1 0 ⎠ ⎝ y 2
⎠ = ⎝ 2 ⎠
1 0 1 y 3 3
⎛ ⎞
which yields very quickly that y = −2 ⎠ . Now you can find x by solving Ux = y. Thus
⎝ 1
2
in this case,
⎛ ⎞
⎛
⎞ x ⎛ ⎞
1 2 3 2
⎝ 0 −5 −11 −7 ⎠ ⎜ y
1
⎟
⎝ z ⎠ = ⎝ −2 ⎠
0 0 0 −2
2
w
which yields
⎛
− 3 5 + 7 5 t
⎞
9
x = ⎜ 5
⎝
− 11
5 t
⎟
t ⎠ ,t∈ R.
−1
Work this out by hand and you will see the advantage of working only with triangular
matrices.
It may seem like a trivial thing but it is used because it cuts down on the number of
operations involved in finding a solution to a system of equations enough that it makes a
difference for large systems.