Fundamentals of Matrix Algebra, 2011a

1.4 Existence and Uniqueness of Soluons
The constants and coefficients of a matrix work together to determine whether a
given system of linear equaons has one, infinite, or no soluon. The concept will be
fleshed out more in later chapters, but in short, the coefficients determine whether a
matrix will have exactly one soluon or not. In the “or not” case, the constants determine
whether or not infinite soluons or no soluon exists. (So if a given linear system
has exactly one soluon, it will always have exactly one soluon even if the constants
are changed.) Let’s look at an example to get an idea of how the values of constants
and coefficients work together to determine the soluon type.
. Example 15 For what values of k will the given system have exactly one solu-
on, infinite soluons, or no soluon?
x 1 + 2x 2 = 3
3x 1 + kx 2 = 9
S We answer this queson by forming the augmented matrix and
starng the process of pung it into reduced row echelon form. Below we see the
augmented matrix and one elementary row operaon that starts the Gaussian eliminaon
process.
[ 1 2
] 3
3 k 9
[ ]
−−−−−−−−−−−→ 1 2 3
−3R 1 + R 2 → R 2
0 k − 9 0
This is as far as we need to go. In looking at the second row, we see that if k = 9,
then that row contains only zeros and x 2 is a free variable; we have infinite soluons.
If k ≠ 9, then our next step would be to make that second row, second column entry
a leading one. We don’t parcularly care about the soluon, only that we would
have exactly one as both x 1 and x 2 would correspond to a leading one and hence be
dependent variables.
Our final analysis is then this. If k ≠ 9, there is exactly one soluon; if k = 9, there
are infinite soluons. In this example, it is not possible to have no soluons. .
As an extension of the previous example, consider the similar augmented matrix
where the constant 9 is replaced with a 10. Performing the same elementary row
operaon gives
[ 1 2
] 3
3 k 10
[ ]
−−−−−−−−−−−→ 1 2 3
−3R 1 + R 2 → R 2 .
0 k − 9 1
As in the previous example, if k ≠ 9, we can make the second row, second column
entry a leading one and hence we have one soluon. However, if k = 9, then our last
row is [0 0 1], meaning we have no soluon.
We have been studying the soluons to linear systems mostly in an “academic”
seng; we have been solving systems for the sake of solving systems. In the next sec-
on, we’ll look at situaons which create linear systems that need solving (i.e., “word
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