Linear Algebra, 2020a

Section I. Solving Linear Systems 9
any solutions at all despite that in echelon form it has a 0 = 0 row.
2x − 2z = 6
y + z = 1
2x + y − z = 7
3y + 3z = 0
−→
y + z = 1
y + z = 1
2x − 2z = 6
3y + 3z = 0
−ρ 1 +ρ 3
2x − 2z = 6
−ρ 2 +ρ 3
−→
y + z = 1
0 = 0
0 =−3
In summary, Gauss’s Method uses the row operations to set a system up for
back substitution. If any step shows a contradictory equation then we can stop
with the conclusion that the system has no solutions. If we reach echelon form
without a contradictory equation, and each variable is a leading variable in its
row, then the system has a unique solution and we find it by back substitution.
Finally, if we reach echelon form without a contradictory equation, and there is
not a unique solution — that is, at least one variable is not a leading variable —
then the system has many solutions.
The next subsection explores the third case. We will see that such a system
must have infinitely many solutions and we will describe the solution set.
Note. In the exercises here, and in the rest of the book, you must justify all
of your answers. For instance, if a question asks whether a system has a
solution then you must justify a yes response by producing the solution and
must justify a no response by showing that no solution exists.
Exercises
̌ 1.17 Use Gauss’s Method to find the unique solution for each system.
(a) 2x + 3y = 13 (b) x − z = 0
x − y =−1 3x + y = 1
−x + y + z = 4
1.18 Each system is in echelon form. For each, say whether the system has a unique
solution, no solution, or infinitely many solutions.
(a) −3x + 2y = 0
−2y = 0
(e) 3x + 6y + z =−0.5
−z = 2.5
(i) x − y =−1
0 = 0
0 = 4
(b) x + y = 4
y − z = 0
(f) x − 3y = 2
0 = 0
(j) x + y − 3z =−1
y − z = 2
z = 0
0 = 0
(c) x + y = 4
y −z = 0
0 = 0
(g) 2x + 2y = 4
y = 1
0 = 4
(d) x + y = 4
0 = 4
(h) 2x + y = 0