Fundamentals of Matrix Algebra, 2011a

2.4 Vector Soluons to Linear Systems
Thus our soluon to the linear system A⃗x = ⃗b is
⃗x = ⃗x p + x 2 ⃗v.
Let us see how exactly this soluon works; let’s see why A⃗x equals ⃗b. Mulply A⃗x:
A⃗x = A(⃗x p + x 2 ⃗v)
= A⃗x p + A(x 2 ⃗v)
= A⃗x p + x 2 (A⃗v)
= A⃗x p + x 2
⃗0
= A⃗x p + ⃗0
= A⃗x p
= ⃗b
We know that the last line is true, that A⃗x p = ⃗b, since we know that⃗x was a soluon
to A⃗x = ⃗b. The whole point is that ⃗x p itself is a soluon to A⃗x = ⃗b, and we could find
more soluons by adding vectors “that go to zero” when mulplied by A. (The subscript
p of “⃗x p ” is used to denote that this vector is a “parcular” soluon.)
Stated in a different way, let’s say that we know two things: that A⃗x p
A⃗v = ⃗0. What is A(⃗x p +⃗v)? We can mulply it out:
= ⃗b and
A(⃗x p +⃗v) = A⃗x p + A⃗v
= ⃗b + ⃗0
= ⃗b
and see that A(⃗x p +⃗v) also equals ⃗b.
So we wonder: does this mean that A⃗x = ⃗b will have infinite soluons? Aer all,
if ⃗x p and ⃗x p +⃗v are both soluons, don’t we have infinite soluons?
No. If A⃗x = ⃗0 has exactly one soluon, then ⃗v = ⃗0, and ⃗x p = ⃗x p +⃗v; we only have
one soluon.
So here is the culminaon of all of our fun that started a few pages back. If ⃗v is
a soluon to A⃗x = ⃗0 and ⃗x p is a soluon to A⃗x = ⃗b, then ⃗x p + ⃗v is also a soluon to
A⃗x = ⃗b. If A⃗x = ⃗0 has infinite soluons, so does A⃗x = ⃗b; if A⃗x = ⃗0 has only one
soluon, so does A⃗x = ⃗b. This culminang idea is of course important enough to be
stated again.
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