Homework 1 Solutions:
(0.5 marks each)
Sec. 1.1, p.11: 1:
x1+5x2=7
1
5
-7
-2x1-7x2=-5
-2
-7
-5
By 2R1 + R2 on R2
1
5
7
0
3
9
Therefore, 3x2=9
x2=9/3=3
Now, x1+5x2=7
x1=7-15= -8
Solution is: (x1,x2)=(-8,3)
Sec. 1.1, p.11: 3:
x1+5x2=7
1
5
7
x1-2x2=-2
1
-2
-2
By R1-R2 on R2,
1
5
7
0
7
9
Hence, 7x2=9
x2=9/7 & x1+5x2=7
Therefore, x1=7-5(9/7) = 4/7
Solution is: (4/7, 9/7)
Sec. 1.1, p.11: 12:
x1-3x2+4x3= -4
1
-3
4
-4
3x1-7x2+7x3= -8
3
-7
7
-8
-4x1+6x2-x3= 7
-4
6
-1
7
By 3R1-R2 on R2,
1
-3
4
-4
0
-2
5
-4
-4
6
-1
7
By 4R1+R3 on R3,
1
-3
4
-4
0
-2
5
-4
0
-6
15
-9
By 3R2-R3 on R3,
1
-3
4
-4
0
-2
5
-4
0
0
0
-3
Now, Row3 is 0x1+0x2+0x3 = -3 i.e. 0≠ -3
This shows that the system is inconsistent.
Sec. 1.1, p.11: 17:
Consider the 2 lines:
x1-4x2=1
-x1-3x2=4
1
-4
1
-1
-3
4
By R1+R2 on R2,
1
-4
1
0
-7
5
Therefore, -7x2=5
x2= (-5/7)
x1=1 + 4(-5/7) = (-13/7)
Hence, (x1,x2)=(-13/7, -5/7)
Substituting these values in equation for line 3, we get,
2x1-x2= -3
2(-13/7) – (-5/7) = -3
i.e. -3=-3
Thus (x1,x2)=(-13/7, -5/7) is also on line 3. Thus they have a common point of intersection :
(x1,x2)=(-13/7, -5/7)
Sec. 1.1, p.11: 21:
1
3
-2
-4
h
8
R2= 4R1+R2
1
3
-2
0
12+h
0
Here we are getting a row of the form [0 X 0] where X=12+h
Now irrespective of X be non-zero or zero. Therefore h can be any value.
Sec. 1.1, p.11: 23(a):
Every elementary row operation is reversible.-> True
Sec. 1.2, p.25: 10:
1
-2
-1
3
3
-6
-2
2
R2 = 3R1- R2
1
-2
-1
3
0
0
-1
7
1
-7
0
6
5
x3= -7 & x1=2x2+x4+3 = 2x2-4
Solution:
x1= -4+2x2
x2= Free
x3= -7
Sec. 1.2, p.25: 12:
1
-7
0
6
5
0
0
1
-2
-3
0
0
1
-2
-3
-1
7
-4
2
7
0
0
-4
8
12
R3= 4R2+R3
1
-7
0
6
5
0
0
1
-2
-3
0
0
0
0
0
x1-7x2+6x4=5 & x3-2x4=-3
x1=5+7x2-6x4 & x3=-3+2x4
Solution:
x1=5+7x2-6x4
x2=free
x3=-3+2x4
x4=free
R3=R1+R3
Sec. 1.2, p.25: 14:
1
2
-5
-6
0
-5
0
1
-6
-3
0
2
0
0
0
0
1
0
0
0
0
0
0
0
Here x1,x2 & x5 are basic whereas x3 & x4 are free.
Now, x2=6x3+3x4+2 & x1= -12x3-6x4-4+5x4+6x4-5 = -7x3-9
2(x2)
Solution:
x1=-7x3-9
x2=6x3+3x4+2
x3 free
x4 free
x5=0
Sec. 1.2, p.25: 20:
a). h=9 & k=-6
b). h≠9
c). h=9 & k=6
1
3
2
3
h
k
R2=3R1-R2
1
3
2
0
9-h
6-k
Sec. 1.2, p.25:31:
Yes, it can be consistent.
x1-4x2=1
1
-4
1
2x1-x2=-3
2
-1
-x1-3x2=4
-1
R3=R1+R3
1
-4
1
-3
2
-1
-3
-3
4
0
-7
5
R3=R2-R3
1
-4
1
Now, by R2=2R1-R2
1
-4
1
0
-7
5
0
-7
5
0
-7
5
0
0
0
Hence system is consistent. Also, x2= -5/7 & x1= -13/7
Sec. 1.3, p.37: 9:
x1 0 + x2 1
+ x3
5 =
0
4
6
-1
0
-1
3
-8
0
Sec. 1.3, p.37: 10:
x1 4 + x2 1
+ x3
3 =
9
1
-7
-2
2
8
6
-5
15
Sec. 1.3, p.37: 12:
1
0
2
-5
-2
5
0
11
2
5
8
-7
R2= 2R1+R2
1
0
2
-5
0
5
4
1
2
5
8
-7
R3=2R1-R3
1
0
2
-5
0
5
4
1
0
-5
-4
-3
R3=R2+R3
1
0
2
-5
0
5
4
1
0
0
0
-2
Therefore, it is inconsistent. (Since third row gas the form [0 0 0 X] where X≠0)
Also, b is not linear combination of x1,x2 & x3.
Sec. 1.3, p.37: 14:
Augmented matrix:
1
-2
-6
11
0
3
7
-5
1
-2
5
9
R3=R1-R3
1
-2
-6
11
0
3
7
-5
0
0
-11
2
Therefore, a3= -2/11
a2= -41/33
a1= 243/33
Therefore, b is a linear combination of vectors formed from columns of matrix A.
Sec. 1.3, p.37: 23(c,d):
c). True ->( ½)V1 (=(1/2)V1 + 0V2)
d). True -> See page 34
Sec. 1.3, p.37: 24(c,d):
c). False-> See page 32
d). True -> See page 35
Sec. 1.3, p.37: 26:
a).
2
0
6
10
-1
8
5
1
-2
R1=R1/2
1
0
3
5
3
-1
8
5
3
1
3
1
-2
1
3
1
0
3
5
R2=R1+R2
1
0
3
5
R3=R1-R3
0
8
8
8
0
8
8
8
1
-2
1
3
0
2
2
2
R3=R2-4R3
1
0
3
5
0
8
8
8
0
0
0
0
a3=free
a2=1-a3
a1=5-3a3
Therefore, b is in W
b). From a). we say that W is span (a1,a2,a3). Since it is set of all linear combination
of vectors a1,a2,a3.
Therefore, vectors of the form
a3=0a1+0a2+(1)a3 will be contained in the span of W.