Fundamentals of Matrix Algebra, 2011a

1.4 Existence and Uniqueness of Soluons
By seng x 2 = 0 = x 4 , we have the soluon x 1 = 4, x 2 = 0, x 3 = 7, x 4 = 0.
By seng x 2 = 1 and x 4 = −5, we have the soluon x 1 = 15, x 2 = 1, x 3 = −8,
x 4 = −5. It is easier to read this when are variables are listed vercally, so we repeat
these soluons:
One parcular soluon is: Another parcular soluon is:
.
x 1 = 4
x 2 = 0
x 3 = 7
x 4 = 0.
x 1 = 15
x 2 = 1
x 3 = −8
x 4 = −5.
. Example 13 .Find the soluon to a linear system whose augmented matrix in
reduced row echelon form is
[ ]
1 0 0 2 3
0 1 0 4 5
and give two parcular soluons.
S
Converng the two rows into equaons we have
x 1 + 2x 4 = 3
x 2 + 4x 4 = 5.
We see that x 1 and x 2 are our dependent variables, for they correspond to the
leading 1s. Therefore, x 3 and x 4 are independent variables. This situaon feels a lile
unusual, 5 for x 3 doesn’t appear in any of the equaons above, but cannot overlook it;
it is sll a free variable since there is not a leading 1 that corresponds to it. We write
our soluon as:
x 1 = 3 − 2x 4
x 2 = 5 − 4x 4
x 3 is free
x 4 is free.
To find two parcular soluons, we pick values for our free variables. Again, there
is no “right” way of doing this (in fact, there are . . . infinite ways of doing this) so we
give only an example here.
5 What kind of situaon would lead to a column of all zeros? To have such a column, the original matrix
needed to have a column of all zeros, meaning that while we acknowledged the existence of a certain
variable, we never actually used it in any equaon. In praccal terms, we could respond by removing the
corresponding column from the matrix and just keep in mind that that variable is free. In very large systems,
it might be hard to determine whether or not a variable is actually used and one would not worry about it.
When we learn about eigenvectors and eigenvalues, we will see that under certain circumstances this
situaon arises. In those cases we leave the variable in the system just to remind ourselves that it is there.
31