Homework 8 Solutions
(All questions carry 0.5 mark except otherwise stated)
Sec. 3.2, p.199: 11
(1 mark)
2
5
-3
-1
3
0
1
-3
-6
0
-4
9
4
10
-4
-1
2
5
-3
-1
3
0
1
-3
-6
0
-4
9
0
0
2
1
3
1
-3
0
-2
3
0
2
1
R4-2R1
(-5)
=
3
1
-3
-6
-4
9
0
2
1
3
=
R2+2R1
(-5)
=
(-5)(3) -2
2
(-5)(3)(-8)=120
1
Sec. 3.2, p.199: 13
(1 mark)
2
5
4
1
4
7
6
6
-2
-6
7
R2-2R1=
2
5
4
1
2
0
-3
-2
-4
0
6
-2
7
0
-6
7
=
0
-3
-2
0
6
-2
-4
-4
0
-6
7
7
7
0
R3+R2
0
-3
-2
6
-2
-4
0
5
3
=
(6)
-3
-2
5
3
a
b
c
2d
2e
2f
g
h
I
=(6)(-9+10)=6
Sec. 3.2, p.199:17
a
b
c
g
h
i
d
e
f
=
(-7)
Sec. 3.2, p.199:19
a
b
c
2d+a
2e+b
2f+c
g
h
I
=
=
a
b
c
2 d
e
f
g
h
i
=14
Sec. 3.2, p.199:29
1
0
1
1
1
2
1
2
1
detB = a11detA11 - a12detA12 + a13detA13
detB = (1) det 1
2
2
1
- (1) det 1
1
1 + (1) det 1
1
1
0
1
= (-2). Hence (detB)5= (-32)
Sec. 3.2, p.199:34
det(PAP-1) = (detP)(detA)(detP-1)
= (detP)(detA)(detP)-1
..by theorem6
..exercise31
=(detA)
Sec. 6.1, p.382: 14
We have, dist. between two vectors is given as:
||u-z||2=[0-(-4)]2 + [-5-(-1)]2 + [2-8]2 = 16+16+36 = 68 = 2√
Sec. 6.1, p.382: 16
u.v = 24-9-15 =0. Hence, Orthogonal.
Sec. 6.1, p.382: 17
u.v=-12+2+10+0=0. Hence, Orthogonal.
Sec. 6.1, p.382: 20(d)
True. By Pythogorean theorem.
Sec. 6.1, p.382: 23
u.v=0. ||u||2=30, ||v||2=101. ||u+v||2 = (-5)2+(-9)2+(5)2=131=30+101=||u||2+||v||2
Sec. 6.1, p.382: 31
If x is in W|, then x is orthogonal to every vector in W(including x). This is possible only when x=0(since
for orthogonal vectors, x.x=0).
Sec. 6.2, p.392: 9
(1 mark)
We have, u1.u2 = -1+1=0, u2.u3 = -2+4-2=0, u1.u3 = 2-2=0. Hence, {u1,u2,u3} form an orthogonal set. By
theorem4, set is linearly independent in R3 and form a basis. By theorem5 we have,
c1=x.u1/u1.u1 = (8-3)/(1+1) = 5/2
c2=x.u2/u2.u2=(-8-16-3)/(1+16+1) = -3/2
c3=x.u3/u3.u3=(16-4+6)/(4+1+4) =2
Hence, x=(5/2)u1-(3/2)u2+2u3
Sec. 6.2, p.392: 13
(1 mark)
The orthogonal projection of y onto u is: y^ =(y.u/u.u)u =(-13/15)u = -4/5
7/5
The component of y orthogonal to u is given by y-y^ =
14/5
8/5
Therefore, y = -4/5
7/5
14/5
8/5
Sec. 6.2, p.392: 20
(1 mark)
We have, ||u||2 =4/9+1/9+4/9 = 1
||v||2 = 1/9+4/9 = 5/9.
Hence not orthogonal although u.v=0.
We need to normalize vector v as:
v/||v|| = 1/√
= 3/√
Thus, orthogonal set is -2/3
1/√
1/3
2/√
2/3
0