Lesson 16 Linear Systems - Bruce E. Shapiro

104 LESSON 16. LINEAR SYSTEMS
1. Convert the system Ax = b into an equivalent form T x = b ′ where T is uppertriangular.
2. Solve for x using back-substitution.
It is possible to take this idea one step further. If we can reduce equation 16.3 to the
form
Dx = b ′′ (16.5)
where D is a diagonal matrix, then it is even easier to read off the solutions, namel,
x i = b ′ i/D ii . In this revised form the matrix D is said to be in Reduced Row
Echelon Form, and the revised algorithm is called Gauss-Jordan Elimination.
The revised algorithm is summarized:
1. Convert the system Ax = b into an equivalent form T x = b ′ , where T is
upper-triangular.
2. Convert the system T x = b ′ into an equivalent form Dx = b ′′ , where D is
diagonal.
3. Solve for the x i .
We will outline the first algorithm (row reduction followed by back-substitution). We
start by writing the linear system
a 11 x 1 + a 12 x 2 + · · · + a 1n x n = b 1 (16.6)
a 21 x 1 + a 22 x 2 + · · · + a 2n x n = b 2 (16.7)
.
a n11 x 1 + a n2 x 2 + · · · + a nn x n = b n (16.8)
From equation 16.6 we can solve for x 1 in terms of x 2 , . . . , x n ,
x 1 = (b 1 − a 12 x 2 − a 13 x 3 − · · · − a 1n x n )/a 11 (16.9)
so if we already know x 2 , . . . , x n we can solve for x 1 immediately. But if we eliminate
x 1 from each of the remaining equations, we have system of n − 1 equations in the
n − 1 variables x 2 , . . . , x n , which is easier to solve than the original system because
it is smaller. We get this system by subtracting an appropriate multiple of the first
equation from each of the remaining equations, namely we subtract
(a i1 /a 11 ) × (a 11 x 1 + a 12 x 2 + · · · + a 1n x n = b 1 ) (16.10)
Math 481A
California State University Northridge
2008, B.E.Shapiro
Last revised: November 16, 2011