Unformatted text preview: · · ann xn bn If b = 0, then the system is homogeneous; otherwise, it is nonhomogeneous.
2 / 13 Nonsingular Case
If the coefﬁcient matrix A is nonsingular (i.e., det A = 0), then A is invertible, and we can solve Ax = b as
follows:
Ax = b =⇒ A−1 (Ax) = A−1 b
=⇒ (A−1 A)x = A−1 b =⇒ Ix = A−1 b
=⇒ x = A−1 b.
The solution is therefore unique. Also, if b = 0, the unique solution to Ax = 0 is: x = A−1 0 = 0.
Hence, if det A = 0, the only solution to Ax = 0 is the trivial solution x = 0.
3 / 13 2 How to Solve
Even if A−1 exists, we need a method for solving Ax = b.
Construct the augmented matrix: A  b . Then apply row operations, trying to convert A → I. When A−1 exists, this is same as multiplying A  b by A−1 :
A  b −→ I  c = A−1 A  b = A−1 A  A−1 b
= I  A− 1 b =⇒ c = A−1 b =⇒ x = c. If det A = 0, row operations on A  b will lead to one (or both) of the following rows:
(a) 0 · · · 0  1 is equivalent to 0 = 1. (Two or more equations contradict each other.) T...
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This note was uploaded on 03/19/2014 for the course APMA 2130 taught by Professor Bernardfulgham during the Fall '09 term at UVA.
 Fall '09
 BERNARDFULGHAM

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