34.
(a)
If the three planes are paralk-1 and di.c;tinct, then they have no common point of
intersection, so the system has no solution.
(b)
If the three planes coincide, then each of the infi nitely n1any different point,;
(x ,y ,=) of thi.c; co,mnon plane provides a solution of the system.
(c)
If two of tl1e plane.- coincide and are parallel to the third plane, tl1en tl,e tlu ee
planes have no co,mnon point of intersection, so the system has no solution.
(d)
If two of the platles intersect in a line that is parallel to the third platle, then tile
tl1ree planes have no common point of intersection, so tile system ha,; no solution.
(e)
If two of tl1e pla1les intersect in a line that lies in tile third plaile, tllen each of tile
infinitely n1any different points (x ,y ,=) of tl1i,; li1le provides a solution of tile system.
(f)
If two oftl1e platles intersect in a litle that intersects tl1e third plaile in a single
point, tllen this point (x ,)', =) provides tile w1ique solution of tl1e system.
SECTION 3.2
MATRICES AND GAUSSIAN ELIMINATION
Because. tile litleai· systen1s in Problen1s 1- 10 are already in echelon fonn , ,ve tleed only sta11 at tile
end of tile li.c;t of w1known.,; and work backwai·d,;.
1.
Starting witl1 xJ = 2 from tile third equation, tile second equation gives x, =0, and then
tile fi rst equation gives -'i = I.
2.
Starting witl1 xJ = -3 from tile third equation, tile second equation gives x, = I, and
tllen tile first equation gives x1 =5.
3.
If we set x 3 = 1 tllen tl1e second equation gives x: =2 + 51, and next tile fi rst equation
gives .1j= l3+111.
4.
If we set xJ = 1 tllen tl1e second equation gives x: = 5 + 11, and tlext tile fi rst equation
.
'5 + .,.,,
" .
gives
,\j =.,
5.
lfweset x~=I tllentl1e thirdequationgives x3 =5+31, nextthesecondequationgives
x: = 6+,. ai1d finally t11e fi rst equation gives x, = I3 +41.
6.
If we set x3 = I and x~ =-4 from tl1e third equation, tllen tl1e second equation gives
x: = I 1+31, and next tile first equation gives x, = 17+1.
136
Chapter 3
=
=,.
=
=,.
7.
If we set x3 s and X 4
then the second equation give$ x, = 7 + 2s- 11, and next
thefirst equationgive$ .\j = 3-8s+l91.
8.
If we set x: s and X 4
then the second equation give$ x3 = 10 - 3t. and next the
first equation give$ x, = - 25 + !Os + 221.
9.
Starting with X4 = 6 from the fow1h equation, the third equation gives x3 = -5, next the
second equation gives x: = 3, and finally the.fi rst equation gives x, = I.
10.
If we set x3 = s and x5 = I, then the third equation give$ X 4 =51, next the second
equation gives x: = I3s- 81. and finally the first equation give$ .\j = 63s - 16/.
In each of Proble11L'i 11- 22, we give ju.lit the first two or three steps in the reduction. Then we
display a re$ulting echelon fonn E of the augmented coefficient ,natrix A of the given linear
system, and fi nally list the resulting solution (if any). The student should tmderstand tl1at tl1e
echelon matrix E i'i not w1ique, so a different sequence of ele,nentary row operations ,nay
produce a different echelon matrix.
11.
Begin by interchanging rmvs I and 2 of A. Then subtract twice row I both from row 2
and from row 3.
I 3 2
E = 0 I O
[
0 0 I
12.
Begin by subtracting row 2 of A from row I. Then subtract twice row I both from row
2 and from row 3.
I
E = 0
[
0
13.
'~]·
-{i
-4
I
O
-,
0
I
2
'
Begin by subtracting twice row I of A both from row 2 and from row 3. Then add row
2 torow3 .
I
,
E = 0 :
[
32 13]
3
~
0 0 0 0
14.
Begin by interchanging rows I and 3 of A. Then subb·act twice row I from row 2, and
three ti tne$ row I from row 3.
Section 3.2
137
R2-2RI
2.
~
I 2
[O I
3.
4.
-IJ
3
7
[5 2 8
N2~NI
[32 -57 -IJ
9
R2-NI
~
NI-Rl
~
12
[~
[~ 0 ~]
-29
( - l) R2
Rl-2R2
~
~
~
:J
R2-4RI
6.
7.
~
[ I 42 ']
I
2 I 9
(1 / 2) R2
~
R2-RI
~
[i -,
~
[ 3I -9 -5]
3
I -2 3
R2-R3
~
142
[~ -2
[ 0I 22
2 I
2
'
8.
~
[ 0I
-5]
10
8
[~
~
~3
-5]
18
I -2 3
R,-2R2
~
2
2
-,' ~2]
N3-2RI
[i
2
I
0 ~ I]
IB-RI
~
Rl-2R2
I
0
Chapter 3
[~
~
[ 0I
~3
-5]
0
2
8
~
[i
:J
··~·' [~ 0
9
~I]
R2-3RI
~
I
2
:2
' ]
R3+3R2
~
[i
~
-I
R2-2RI
5.
.>
Rl-12R2
12 - 10]
-5]
10
- 12
0
I
0 ~ I]
18
( - 1112) /B
~
'" -+
[i
0
I
0
~]