Fundamentals of Matrix Algebra, 2011a

Chapter 2
Matrix Arithmec
. Key Idea 8 Soluons of Consistent Systems
Let A⃗x = ⃗b be a consistent system of linear equaons.
1. If A⃗x = ⃗0 has exactly one . soluon (⃗x = ⃗0), then A⃗x =
⃗b has exactly one soluon.
2. If A⃗x = ⃗0 has infinite soluons, then A⃗x = ⃗b has infinite
soluons.
A key word in the above statement is consistent. If A⃗x = ⃗b is inconsistent (the
linear system has no soluon), then it doesn’t maer how many soluons A⃗x = ⃗0 has;
A⃗x = ⃗b has no soluon.
Enough fun, enough theory. We need to pracce.
. Example 47 .Let
A =
[ 1 −1 1
] 3
4 2 4 6
and ⃗b =
[ 1
10
]
.
Solve the linear systems A⃗x = ⃗0 and A⃗x = ⃗b for ⃗x, and write the soluons in vector
form.
S We’ll tackle A⃗x = ⃗0 first. We form the associated augmented matrix,
put it into reduced row echelon form, and interpret the result.
[ 1 −1 1 3
] 0
4 2 4 6 0
−→
rref
[ 1 0 1 2
] 0
0 1 0 −1 0
x 1 = −x 3 − 2x 4
x 2 = x 4
x 3 is free
x 4 is free
To write our soluon in vector form, we rewrite x 1 and x 2 in ⃗x in terms of x 3 and x 4 .
⎡ ⎤ ⎡
x 1
⃗x = ⎢ x 2
⎥
⎣ x 3
⎦ = ⎢
⎣
x 4
−x 3 − 2x 4
x 4
x 3
x 4
Finally, we “pull apart” this vector into two vectors, one with the “x 3 stuff” and one
⎤
⎥
⎦
90