Fundamentals of Matrix Algebra, 2011a

Chapter 2
Matrix Arithmec
In previous secons we were fine stang that the result as
x 1 = 2, x 2 = 0, x 3 = 1,
but we were asked to find ⃗x; therefore, we state the soluon as
⎡ ⎤
.
⃗x =
⎣ 2 0 ⎦ .
1
This probably seems all well and good. While asking one to solve the equaon
A⃗x = ⃗b for ⃗x seems like a new problem, in reality it is just asking that we solve a
system of linear equaons. Our variables x 1 , etc., appear not individually but as the
entries of our vector ⃗x. We are simply wring an old problem in a new way.
In line with this new way of wring the problem, we have a new way of wring
the soluon. Instead of lisng, individually, the values of the unknowns, we simply list
them as the elements of our vector ⃗x.
These are important ideas, so we state the basic principle once more: solving the
equaon A⃗x = ⃗b for⃗x is the same thing as solving a linear system of equaons. Equivalently,
any system of linear equaons can be wrien in the form A⃗x = ⃗b for some
matrix A and vector ⃗b.
Since these ideas are equivalent, we’ll refer to A⃗x = ⃗b both as a matrix–vector
equaon and as a system of linear equaons: they are the same thing.
We’ve seen two examples illustrang this idea so far, and in both cases the linear
system had exactly one soluon. We know from Theorem 1 that any linear system has
either one soluon, infinite soluons, or no soluon. So how does our new method of
wring a soluon work with infinite soluons and no soluons?
Certainly, if A⃗x = ⃗b has no soluon, we simply say that the linear system has no
soluon. There isn’t anything special to write. So the only other opon to consider is
the case where we have infinite soluons. We’ll learn how to handle these situaons
through examples.
. Example 44 .Solve the linear system A⃗x = ⃗0 for ⃗x and write the soluon in
vector form, where [ ]
[ ]
1 2
0
A = and ⃗0 = .
2 4
0
S
(Note: we didn’t really need to specify that
[ ] 0 ⃗0 = ,
0
but we did just to eliminate any uncertainty.)
To solve this system, put the augmented matrix into reduced row echelon form,
which we do below.
[ 1 2
] 0
2 4 0
−→
rref
[ 1 2
] 0
0 0 0
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