Fundamentals of Matrix Algebra, 2011a

2.4 Vector Soluons to Linear Systems
with the “x 4 stuff.”
⃗x =
=
⎡
⎢
⎣
⎡
⎢
⎣
−x 3 − 2x 4
x 4
x 3
−x 3
0
x 3
0
⎡
= x 3
⎢
⎣
x 4
⎤ ⎡
⎥
⎦ + ⎢
⎣
−1
0
1
0
⎤
= x 3 ⃗u + x 4 ⃗v
⎤
⎥
⎦
⎥
⎦ + x 4
−2x 4
x 4
0
x 4
⎡
⎢
⎣
⎤
⎥
⎦
−2
1
0
1
⎤
⎥
⎦
We use ⃗u and⃗v simply to give these vectors names (and save some space).
It is easy to confirm that both ⃗u and ⃗v are soluons to the linear system A⃗x = ⃗0.
(Just mulply A⃗u and A⃗v and see that both are ⃗0.) Since both are soluons to a homogeneous
system of linear equaons, any linear combinaon of ⃗u and ⃗v will be a
soluon, too.
.
Now let’s tackle A⃗x = ⃗b. Once again we put the associated augmented matrix into
reduced row echelon form and interpret the results.
[ 1 −1 1 3
] 1
4 2 4 6 10
−→
rref
[ 1 0 1 2
] 2
0 1 0 −1 1
x 1 = 2 − x 3 − 2x 4
x 2 = 1 + x 4
x 3 is free
x 4 is free
Wring this soluon in vector form gives
⎡ ⎤ ⎡
x 1
⃗x = ⎢ x 2
⎥
⎣ x 3
⎦ = ⎢
⎣
x 4
2 − x 3 − 2x 4
1 + x 4
x 3
x 4
Again, we pull apart this vector, but this me we break it into three vectors: one with
⎤
⎥
⎦ .
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