EXAMPLE: Section 1.1, Problems 12, 13 and 28
12. Solve the system:
x1 −3x2 +4x3 = −4
3x1 −7x2 +7x3 = −8
−4x1 +6x2 −x3 =
7
SOLUTION:
1 −3 4 −4 −3r1 +r2 →r2 1 −3 4 −4
4r1 +r3 →r3
→
0 2 −5 4
3 −7 7 −8
−4 6 −1 7
0 −6 15 −9
At this point, we could divide row 2 by 2, but that would introduce fractions. Alternatively, keep that for now.
1 −3 4 −4
1 −3 4 −4
3r2 +r3 →r3
→
0 2 −5 4
0 2 −5 4
3
0 0
0
0 −6 15 −9
Because the last row is translated to read: “0 = 3”, this system is inconsistent (has no
solution).
13. Solve the system:
x1
−3x3 =
8
2x1 +2x2 +9x3 =
7
x2 +5x3 = −2
SOLUTION:
1 0 −3 8
1 0 −3 8
1 0 −3 8
−2r1 +r2 →r2
r3 ↔r2
−2r2 +r3 →r2
2
2
9
7
0
2
15
−9
→
→
→
0 1 5 −2
0 1 5 −2
0 1 5 −2
0 2 15 −9
1 0 −3 8
0
1 5 −2
0 0 5 −5
1
r →r2
5 2
→
−5r3 +r2 →r2
1 0 −3 8
1 0 0 5
3r3 +r1 →r1
0
1
5
−2
0
1 0 3
→
0 0 1 −1
0 0 1 −1
Therefore, the solution is unique, and
x1 = 5, x2 = 3, x3 = −1
(EXTRA PRACTICE: Verify that this is the solution by substitution of these numbers
into the original equations!) 28. Find a, b, c, d so that the system below is consistent for all values of f and g. Assume
that a 6= 0.
ax + by = f
cx + dy = g
SOLUTION: Form the augmented matrix:
"
a b f
c d g
#
c
Multiply the first row by − (which is always possible since a 6= 0), and add to the
a
second row:
#
"
a
b
f
0 d − bca g − fac
For this system to be consistent for any f, g, we must have that
d−
bc
ad − bc
6= 0 ⇒
6= 0 ⇒ ad − bc 6= 0
a
a
Side Remark 1: If d − bca = 0, it is possible to have a consistent system- if g − fac is also
0. But in this case, the system is not consistent for all f, g.
Side Remark 2: You might recall that ad − bc is a special number for the matrix
"
. It is called the determinant.
a b
c d
#