Sec. 2.1
x
17. P
=
y
z
x
Q
=
y
z
0 1 0
x
0 0 1
y
1 0 0
z
0 0 1
y
1 0 0
z
0 1 0
y
=
,
z
x
x
=
y
x
z
You might notice that Q = P . Does this work in general for a permutation matrix? Why?
T
x
19. E
=
y
z
x
E
−1
=
y
z
1 0 0
x
0 1 0
y
1 0 1
z
1
0 0
x
0
1 0
y
−1 0 1
z
3
E −1 E
3
= E −1
4
5
4
x
=
,
y
z+x
x
=
y
,
z−x
3
=
4
8
5
Note, of course, that E −1 E = EE −1 = I
Sec. 2.2
2 x + 3y + z = 8
12. 4x + 7y + 5z = 20
→
2 x + 3y + z = 8
1 y + 3z = 4
Eqn2=Eqn2−2Eqn1
−2y + 2z = 0
→
Eqn3=Eqn3+2Eqn2
2 x + 3y + z = 8
1 y + 3z = 4
−2y + 2z = 0
8z = 8
Back substitution: 8z = −8 , z = 1; y + 31 = 4 , y = 1; 2x + 31 + 1 = 8 , x = 2
x, y, z = 2, 1, 1
2 x − 3y
=3
13. 4x − 5y + z = 7 →
2x − y − 3z = 5
2 x − 3y
=3
1y + z = 1 →
2y − 3z = 2
2 x − 3y
=3
1y + z = 1
-5 z = 0
z = 0 , y = 1 , x = 3 by back substitution
Operations:
Subtract 2 times row 1 from row 2; Subtract 1 times row 1 from row 3;
Subtract 2 times row 2 from row 3;
1
14.
A row interchange is required here if d = 10.
2x + 5y + z = 0
2x
4x + dy + z = 2 →
+
5y + z = 0
d − 10y − z = 2
y−z = 3
2x + 5y + z = 0
→
y−z = 3
d=10
y−z = 3
−z = 2
If d = 11 we obtain a singular system.
2x + 5y + z = 0
2x
4x + dy + z = 2 →
+
5y + z = 0
d − 10y − z = 2
y−z = 3
2x + 5y + z = 0
→
y−z = 2
d=11
y−z = 3
0 = −1
15.
Row interchange:
x + by
=0
x + by
=0
x − 2y − z = 0 → −2 + by − z = 0
y+z = 0
x − 2y
→
=0
y+z = 0
b=−2
y+z = 0
−z = 0
Singular:
x + by
=0
x + by
=0
x − 2y − z = 0 → −2 + by − z = 0
y+z = 0
x−y
→
b=−1
=0
−y − z = 0
y+z = 0
0=0
z can be anything, e.g. if z = 1 we obtain y = −1 , x = −1
16. a) Two row interchanges:
y−z = 0
z=3 →
x+y+z = 1
x+y+z = 1
z=3 →
y−z = 0
x+y+z = 1
y−z = 0
z=3
z = 3, y = 3, x = −5
b) A row interchange, but ultimate breakdown
y−z = 0
y−z = 3
x+y+z = 1
2
19.
x + 4y − 2z = 1
x + 4y − 2z = 1
x + 7y − 6z = 6 →
3y − 4z = 5
3y + qz = t
3y + qz = t
→
x + 4y − 2z = 1
3y − 4z = 5
q=−4
0 = t−5
So the system is singular if q = −4, but if t = 5 there are infinitely many solutions. With
z = 1, we obtain y = 3 , x = −9.
21.
2x + y
=0
x + 2y + z
=0
y + 2z + t = 0
→
2x + y
=0
3/2y + z
=0
4/3z + t = 0
z + 2t = 5
5/4t = 5
(You can do this all in one shot)
Note that the pivots are 2/1 , 3/2 , 4/3 , 5/4 and the solution is
t = 4, z = −3, y = 2, x = −1
2x − y
−x + 2y − z
=0
=0
− y + 2z − t = 0
2x − y
→
=0
3/2y − z
=0
4/3z − t = 0
− z + 2t = 5
5/4t = 5
The pivots are 2/1 , 3/2 , 4/3 , 5/4 and the solution is t = 4, z = 3, y = 2, x = 1
25.
a 2 3
a a 4
a a a
a
→
2
0 a−2
3
1
0 a−2 a−3
a
→
2
0 a−2
0
0
3
1
a−4
Three pivots will not exist if a = 0, 2, 4
3