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Title: TMC1874 Mathematics For Computing tutorial 4 solution
Description: Tutorial Solution Chapter 4: Real Vector Spaces
Description: Tutorial Solution Chapter 4: Real Vector Spaces
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TMC1874 Mathematics For Computing
Tutorial Solution
Chapter 4: Real Vector Spaces
1
...
;
Solution
(a)
u+v =
3 + (−3)
2+4
=
0
6
u−v =
3 − (−3)
2−4
=
6
−2
2u = 2
2u − 3v =
3
2
6
4
=
,
,
2(3) − 3(−3)
2(2) − 3(4)
15
−8
...
=
(b)
u+v =
0+4
2+2
=
4
4
u−v =
0−4
2−2
=
−4
0
2u = 2
2u − 3v =
W
...
Tiong
0
2
=
0
4
2(0) − 3(4)
2(2) − 3(2)
,
,
,
=
Page 1 of 40
TMC1874 Mathematics For Computing
(c)
u+v =
2+3
6+2
=
5
8
u−v =
2−3
6−2
=
−1
4
2u = 2
2u − 3v =
2
6
=
4
12
2(2) − 3(3)
2(6) − 3(2)
,
,
,
=
−5
6
...
Let u = 2 , v = 3 , w = 2 , c = −2 and d = 3
...
Let x = −3 , y = 1 , z = −1 , u = t
...
K
...
3
s−3
4
...
−2
3
3
3
Solution
1
−3
−1
−3c 1 + c 2 − c 3
2
c 1 + 2c 2 + 4c 3 = −2
...
−3
=
=
=
The augmented matrix is
−3
1
−2
1
2
3
−1
4
3
2
−2
−3
Performing row operations to transform the augmented matrix:
−3
1
−2
W
...
Tiong
1
2
3
−1
4
3
2
1
R 2 ↔R 3
− − − −3
−2
− −→
−2
−3
2
1
3
4
−1
3
−2
3 R 1 + R 2 →R 2
2 −− − − −
− − − −→
2 R 1 + R 3 →R 3
−3
Page 3 of 40
TMC1874 Mathematics For Computing
1
0
0
2
7
7
4
11
11
1
−2
− R 2 + R 3 ↔R 3
−4 − − − − − 0
− − − −→
−7
0
2
7
0
4
11
0
−2
−4
...
e
...
Thus, there is no such scalars
c 1 , c 2, c 3 exist
...
Determine whether the subsets of all vectors of the following forms of R3 are subspaces
...
c
a
(c) a
c
a
(d) b , where 2a − b + c = 1
...
a1
a2
(a) Let u = b 1 and v = b 2
...
0
0
0
So u + v is in the subset
...
0
0
a
ru is also in the subset
...
2
(b) If we take the vector u = 3 in the subset and let the scalar r to be −2
...
K
...
Therefore, the subset
a
of all vectors of the form b where a > 0 is not a subspace of R3
...
c1
c2
a1
a2
a1 + a2
u + v = a1 + a2 = a1 + a2
...
Next,
a1
ra 1
ru = r a 1 = ra 1
...
Therefore, the subset of all vectors of the form a of
c
3
R is a subspace
...
1
2
1
1
2
u + v = 2 + 3 = 5
...
Therefore, the subset of all vectors of the form b where
c
3
2a − b + c = 1 is not a subspace of R
...
Then
2
the scalar multiple
1
−2
ru = −2 3 = −6
2
−4
W
...
Tiong
Page 5 of 40
TMC1874 Mathematics For Computing
is not in the subset since the third entry does not satisfies the formula 2a − b + c = 1
...
6
...
Let W be
the set of all vectors whose tail is at the origin and whose head is a point inside or on the circle
...
Solution
Note that the radius of the circle is 1
...
Take a vector u =
x
y
whose head lies outside
in W and let a scalar r to be 2
...
Therefore, W is not a subspace
...
Determine whether the subsets of all matrices of the following forms of M23 are subspaces
...
K
...
We can rewrite u, v as
a2
d2
a2 + c2
0
c2
0
...
Hence, we can conclude that u + v is
in the subset
...
ru is also in the subset
...
(b) One possible solution:
Take u =
1
3
2
0
−1
0
and r = −1
...
Therefore, the subset
a b c
of all matrices of the form
, where c < 0 is not a subspace
...
Then
d1
0
0
ru = r
a1
d1
b1
0
c1
0
=
ra 1
rd 1
rb 1
0
r c1
0
...
Hence ru need not be in the
a b c
subset
...
W
...
Tiong
Page 7 of 40
TMC1874 Mathematics For Computing
(c) Let u =
a1
d1
u=
b1
e1
−3c 1
d1
c1
f1
b1
e1
and v =
c1
3e 1 + d 1
a2
d2
b2
e2
c2
f2
and v =
...
(2)
Then
u+v =
=
−3c 1
d1
b1
e1
c1
3e 1 + d 1
−3(c 1 + c 2)
d1 + d2
+
b2 + b2
e1 + e2
−3c 2
d2
b2
e2
c2
3e 2 + d 2
c1 + c2
3(e 1 + e 2 ) + (d 1 + d 2 )
Note that u + v has the same form as in (2)
...
Next,
ru =
−3(r c 1)
rd 1
rb 1
r e1
r c1
3(r e 1) + rd 1
...
Therefore, the subset of all matrices of the form
a
d
b
e
c
f
,
where a = −3c and f = 3e + d is a subspace
...
Determine whether the subsets of all polynomials of the following forms of P2 are subspaces
...
Solution
Let u = a 2 t2 + a 1 t + a 0 and v = b 2 t2 + b 1 t + b 0
...
u + v = (a 2 t2 + a 1 t + a 0 ) + (b 2 t2 + b 1 t + b 0 ),
=
(a 2 + b 2 )t2 + (a 1 + b 1 )t + (a 0 + b 0 )
...
ru =
=
r(a 2 t2 + a 1 t + a 0),
ra 2 t2 + ra 1 t + ra 0
...
Therefore the subset of all polynomials of the form a 2 t2 + a 1 t + a 0 , where a 0 = 0 is a subspace
...
W
...
Tiong
Page 8 of 40
TMC1874 Mathematics For Computing
Solution
Let u = a 2 t2 + a 1 t + a 0 and v = b 2 t2 + b 1 t + b 0
...
u + v = (a 2 t2 + a 1 t + a 0 ) + (b 2 t2 + b 1 t + b 0 ),
=
(a 2 + b 2 )t2 + (a 1 + b 1 )t + (a 0 + b 0 )
...
Therefore the subset of
all polynomials of the form a 2 t2 + a 1 t + a 0, where a 0 = 0 is not a subspace
...
Solution
Let
u = a 2 t2 + a 1 t + a 0
and v = b 2 t2 + b 1 t + b 0
...
(3)
u + v = (−a 1 t2 + a 1 t + a 0 ) + (− b 1 t2 + b 1 t + b 0 ),
(4)
So
2
= −(a 1 + b 1 )t + (a 1 + b 1 )t + (a 0 + b 0 )
...
Next,
ru =
r(−a 1 t2 + a 1 t + a 0),
2
= −(ra 1)t + ra 1 t + ra 0
...
Then ru is in the subset
...
9
...
(a) p(t) = t2 + t + 2
(c) p(t) = − t2 + t − 4
(b) p(t) = 2t2 + 2t + 3
(d) p(t) = −2t2 + 3t + 1
Solution
If a vector p(t) = at2 + b t + c belongs to span { p 1 (t), p 2(t), p 3 (t)}, then we are able to find
scalars a 1 , a 2, a 3 such that a 1 p 1 (t) + a 2 p 2 (t) + a 3 p 3 (t) = p(t)
...
K
...
Hence, we have
a 1 p 1 (t) + a 2 p 2 (t) + a 3 p 3 (t) =
p(t) = at2 + b t + c,
a 1 (t2 + 2t + 1) + a 2 (t2 + 3) + a 3 (t − 1) =
at2 + b t + c,
(a 1 + a 2 )t2 + (2a 1 + a 3 )t + (a 1 + 3a 2 − a 3 ) =
at2 + b t + c
...
Thus, the augmented matrix for the
linear system (8) is
1
2
1
1
0
3
0
1
−1
1
1
...
e
...
This means we are able to find scalars
a 1 , a 2, a 3
...
(b) Given p(t) = 2t2 + 2t + 3, then a = 2, b = 2, c = 3
...
3
Performing row operations:
1
2
1
1
0
0
1
0
3
0
1
−1
1
1
2
0
1
−2
−1
2
−2 R 1 + R 2 → R 2
2 −− − − −→
−−−−−
− R 1 + R 3 →R 3
3
2
−2 R 2 + R 3 → R 3
1 −− − − −→
−−−−−
1
1
0
0
1
−2
2
1
0
0
1
1
0
0
1
−1
0
−1
2
0
2
1
− 2 R 2 →R 2
−2 − − − −
− − −→
1
2
1
−1
The system is inconsistent, i
...
we are unable to find scalars a 1 , a 2, a 3
...
W
...
Tiong
Page 10 of 40
TMC1874 Mathematics For Computing
(c) Given p(t) = − t2 + t − 4, then a = −1, b = 1, c = −4
...
−4
Performing row operations:
1
2
1
1
0
0
1
0
3
0
1
−1
1
1
2
0
1
−2
−1
−1
−2 R 1 + R 2 → R 2
1 −− − − −→
−−−−−
− R 1 + R 3 →R 3
−4
−1
−2 R 2 + R 3 → R 3
−3 −− − − −→
−−−−−
2
−3
1
0
0
1
−2
2
0
1
−1
1
0
0
1
1
0
0
1
−2
0
−1
1
− 2 R 2 →R 2
3 −− − −
− − −→
−3
−1
−3
2
0
The system is consistent
...
(d) Given p(t) = −2t2 + 3t + 1, then a = −2, b = 3, c = 1
...
1
Performing row operations:
1
2
1
1
0
0
1
0
3
0
1
−1
1
1
2
0
1
−2
−1
−2
−2 R 1 + R 2 → R 2
3 −− − − −→
−−−−−
− R 1 + R 3 →R 3
1
−2
−2 R 2 + R 3 → R 3
−−−−−
−7 −− − − −→
2
3
1
0
0
1
−2
2
0
1
−1
1
0
0
1
1
0
0
1
−2
0
−2
1
− 2 R 2 →R 2
7 −− − −
− − −→
3
−2
−7
2
10
The system is inconsistent, i
...
we are unable to find scalars a 1 , a 2, a 3
...
10
...
1
...
K
...
2
...
If this system is consistent, then the set spans R2
...
First, we let [ a
b ] be an arbitrary vector in R2
...
The corresponding linear system is
2a 1
a1
a2
a2
−
+
a
...
2
1
−1
1
1
− 3 R 2 →R 2
−− − −
− − −→
a
b
1
0
1
2
R 1 ↔R 2
−− −
− −→
1
1
1
−1
b
a
1
0
−2 R 1 + R 2 → R 2
−− − − −→
−−−−−
1
−3
b
2b − a
b
−
− 2b3 a
Clearly, the system is consistent
...
(b)
a 1 [0
0] + a 2 [1
1] + a 3 [−2
− 2] = [a
b]
...
b
1
0
−2
0
Form the augmented matrix and row reduce it
...
Hence the vectors does not span R2
...
The corresponding linear system is
2a 1
4a 1
W
...
Tiong
−
+
a2
2a 2
=
=
a
...
2
4
−1
2
1
R →R 2
4 2
−− −→
−−−
a
b
1
0
1
R →R 1
2 1
−1
2
1
−1
2
2
1
4
−− −→
−−−
a
2
b
1
−2
4
1
0
−4 R 1 + R 2 → R 2
−− − − −→
−−−−−
a
2
b − 2a
a
2
b −2 a
4
The system is consistent
...
11
...
So we have
0
1] + a 2 [0
1 0
0] + a 3 [1
1 1
1] +
a 4 [1
= [a
1
b
1 0]
c
d]
...
1
1 0 1 1 a
0 1 1 1 b − R 1 + R 4 →R 4 0
− − − −→
−− − − −
0
0 0 1 1 c
1 0 1 0 d
0
1 0 1 1
a
b
0 1 1 1
...
c
d
=
=
=
=
1
1
1
0
1
1
1
−1
a
b
c
d
− R 4 →R 4
−−−
−− −→
The system is consistent
...
(b) [1
−2
3 0], [1
Solution
Let [a b
a 1 [1
W
...
Tiong
c
−2
2
−1
0], [0 0
0
3]
d] be an arbitrary vector in R4
...
Page 13 of 40
TMC1874 Mathematics For Computing
The corresponding linear system is
a1
−2a 1
3a 1
+
+
−
a2
2a 2
a2
=
=
=
=
3a 3
Form the augmented matrix and row reduce it
...
c
d
a
b + 2a
c − 3a
d
1 R →R
4 2 2
−−−
−− −→
a
b +2 a
4
1 R 4 →R 4
3
−−−
−− −→
b+c−a
d
The system is consistent only when b + c − a = 0
...
(c) [1
1 0
0], [1 2
Solution
Let [a b
a 1 [1
1 0
c
−1
1], [0 0
1 1], [2 1
2
1]
d] be an arbitrary vector in R4
...
The corresponding linear system is
a1
a1
+
+
−
a2
2a 2
a2
a2
+
+
a3
a3
+
+
+
+
Form the augmented matrix and row reduce it
...
K
...
c
d
a
b−a
c
d
R 2 + R 3 →R 3
− − − −→
−− − − −
− R 2 + R 4 →R 4
Page 14 of 40
TMC1874 Mathematics For Computing
1
0
0
0
1
1
0
0
0
0
1
1
2
−1
1
2
a
b−a
c+b−a
d+a−b
− R 3 + R 4 →R 4
− − − −→
−− − − −
1
0
0
0
1
1
0
0
0
0
1
0
2
−1
1
1
a
b−a
c+b−a
2a − 2b − c + d
The system is consistent
...
1
2
12
...
Solution
Form the equation
a1
1
2
1
−1
+ a2
4
3
1
0
+ a3
2
0
1
3
Form the augmented matrix and row reduce it
1 4 2 0
1
2 3 0 0 −2 R 1 + R 2 → R 2 0
−−−−−
−− − − −→
1 1 1 0 − R 1 + R 3 →R 3 0
R 1 + R 4 →R 4
−1 0 3 0
0
1
1
4
2 0
0
4
1
0 3 R 2 + R 3 →R 3 0
5
−−−−−
−− − − −→
0 −3 −1 0 −4R2 +R4 →R4 0
−4 R 2 + R 1 → R 1
0
4
5 0
0
6
1
1 0 −5 0
− 9 R 3 + R 4 →R 4 0
4
0 5
0 1
5
−−−−−
−− − − −→
4
0 0
1 0 − 5 R 3 + R 2 →R 2 0
9
6
R + R 1 →R 1
0
0 0
0
5 3
5
=
0
0
0
0
4
−5
−3
4
2
−4
−1
5
0
0
0
0
0
1
0
0
−6
5
0
0
0
0
0
1
0
0
4
5
7
5
9
5
0
0
1
0
0
0
0
0
− 1 R 2 →R 2
5
− − −→
−− − −
5 R 3 →R 3
7
−−−
−− −→
The only solution is a 1 = a 2 = a 3 = 0
...
13
...
The augmented matrix derived from the formula a 1 v1 + a 2 v2 +
a 3 v3 + a 4 v4 = 0 is given by
3 1 4 1 0
1 2 3 1 0
...
K
...
Therefore, S is not linearly independent
...
Which of the given vectors in R3 are linearly dependent? For those which are, express one vector
as a linear combination of the rest
...
The corresponding linear system is
2a 1
−a 1
+
3a 2
2a 2
+
+
+
2a 3
4a 3
3a 3
+
+
+
3a 4
6a 4
6a 4
=
=
=
0
0
...
K
...
2
2
So the vectors are linearly dependent
...
4 2]
Solution
Form the equation:
a 1 [1 1
0] + a 2 [3
4 2] = [0
0 0]
The corresponding linear system is
a1
a1
+
+
3a 2
4a 2
2a 2
=
=
=
0
0
...
Therefore the vectors are linearly independent
...
e
...
Which of the given vectors in P2 are linearly dependent? For those which are, express one vector
as a linear combination of the rest
...
K
...
Equating coefficients of like powers of t, we obtain the system
a1
a1
−
a2
2a 2
+
+
0
0
...
Therefore the vectors are linearly independent
...
Equating coefficients of like powers of t, we obtain the system
2a 1
a1
+
a2
+
0
0
...
K
...
Therefore the vectors are linearly independent
...
Equating coefficients of like powers of t, we obtain the system
2a 1
a1
a1
+
+
−
3a 2
a2
5a 2
+
+
a3
13a 3
=
=
=
0
0
...
Therefore the vectors is linearly dependent
...
16
...
1
0
1
(a) 0 , 1 , 2
0
1
−1
W
...
Tiong
Page 19 of 40
TMC1874 Mathematics For Computing
Solution
One possible solution is to follow the same procedure as in Questions 14 & 15
...
Form the matrix A whose columns are the given vectors
1
0
0
0
1
1
1
2
...
Since det(A) = −3 = 0, then the vectors are linearly independent
...
−3
1
−2
1
The linear system obtained is
2a 1
−2a 1
−3a 1
+
+
3a 2
a2
+
+
+
a3
a3
a3
+
+
−
2a 4
a4
2a 4
=
=
=
0
0
...
K
...
Therefore the vectors is linearly
dependent
...
Form the matrix A whose columns are the given vectors
1
0
0
2
1
1
−1
2
...
Since det(A) = −1 = 0, then the vectors are linearly independent
...
Which of the following sets of vectors are bases for R2 ?
(a)
2
5
1
−1
,
(b)
0
0
,
3
2
,
5
4
Solution
(a) Let V = R2 and S =
2
5
,
1
−1
i
...
K
...
To check if S spans V , we form the
Page 21 of 40
TMC1874 Mathematics For Computing
linear combination
2
5
a1
1
−1
+ a2
=
a
b
Form the augmented matrix and row reduce it
2
5
1
−1
2
− 7 R 2 →R 2
a
b
−− − −
− − −→
1
0
1
R →R 1
2 1
1
5
−− −→
−−−
1
2
1
1
2
a
2
−1
b
−5 R 1 + R 2 → R 2
−− − − −→
−−−−−
1
0
1
2
7
−2
a
2
5 a −2 b
2
a
2
2 b −5 a
7
The system is consistent
...
ii
...
So the augmented matrix in ref is
1
0
1
2
1
0
0
...
Hence S is linearly independent
...
However in this question, note that S contains the zero
vector of R2
...
Hence S is
not a basis for R2
...
Which of the following sets of vectors are bases for P3 ?
(a) { t3 + 2t2 + 3t, 2t3 + 1, 6t3 + 8t2 + 6t + 4, t3 + 2t2 + t + 1}
Solution
Let V = P3 and S = t3 + 2t2 + 3t, 2t3 + 1, 6t3 + 8t2 + 6t + 4, t3 + 2t2 + t + 1
i
...
To check if S spans V , we
form the linear combination
a 1 (t3 + 2t2 + 3t) + a 2 (2t3 + 1) + a 3 (6t3 + 8t2 + 6t + 4)
+a 4 (t3 + 2t2 + t + 1) = at3 + bt2 + ct + d
...
K
...
c
d
Page 22 of 40
TMC1874 Mathematics For Computing
Form the augmented matrix and row reduce it
a
1
2
6
1
1 2 6 1 a
−4
0 b − 2a
2 0 8 2 b −2R1 +R2 →R2 0 −4
−−−−−
−− − − −→
3 0 6 1 c −3R1 +R3 →R3 0 −6 −12 −2 c − 3a
0 1 4 1 d
0
1
4
1
d
a
1
2
6
1
1
4
1
d
R 4 ↔R 2 0
6 R 2 + R 3 →R 3
−− −
− −→
−−−−−
−− − − −→
0 −6 −12 −2 c − 3ac 4R2 +R4 →R4
−2 R 2 + R 1 → R 1
0 −4
−4
0 b − 2a
1 2
6 1
a
a
1 2
6 1
1 R 3 →R 3 0 1
0 1
d
4 1
4 1
d
12
− − −→
−− − −
c −3 a +6 d
0 0 12 4 c − 3a + 6d
0 0
1 1
3
12
0 0 12 4 b − 2a + 4d
0 0 12 4 b − 2a + 4d
1 2 6 1
a
d
1 4 1
−12R3 +R4 →R4 0
−− − − − −
− − − − −→
1
c −3 a +6 d
0 0 1 3
12
0 0 0 0 a + b − c − 2d
The system is inconsistent
...
Therefore S is not a
basis for V
...
To show S span V
Let at3 + bt2 + ct + d be an arbitrary vector in V
...
The corresponding linear system is
a1
a1
a1
a1
W
...
Tiong
+
+
+
+
a2
2a 2
a2
3a 2
+
+
+
+
2a 3
a3
3a 3
2a 3
+
+
+
+
a4
a4
2a 4
2a 4
=
=
=
=
a
b
...
Hence S spans V
...
To show S is linearly independent:
Let a = b = c = d = 0 and perform row reduction
1 0
1 1
2
1 0
0 1 −1
0 0 − R 2 + R 1 →R 1 0 1
− − − −→
−− − − −
0 0
0 0
1 −1 0
0 0
0
1 0
0 0
1 0
1 0 0
4 0
0 1 0 −1 0
R 4 + R 3 →R 3 0 1
−−−−−
−− − − −→
0 0 1 −1 0 R4 +R2 →R2 0 0
−4 R 4 + R 1 → R 1
0 0 0
1 0
0 0
3
−1
1
0
0
0
1
0
1
0
−1
1
0
0
0
1
0
0
0
0
0
0
0
0
R 3 + R 2 →R 2
−−−−−
−− − − −→
−3 R 3 + R 1 → R 1
The only solution is a 1 = a 2 = a 3 = a 4 = 0
...
Therefore, S is a basis for P3
...
Determine which of the given subsets forms a basis for R3
...
1
1
1
(a) −1 , 2 , 1
1
3
0
W
...
Tiong
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TMC1874 Mathematics For Computing
Solution
1
1
1
Let V = R3 and S = −1 , 2 , 1
...
e
...
1
3
0
c
Form the augmented matrix and row reduce it
...
Hence S spans V
...
1 1 1 0
1
− 2 R 3 + R 2 →R 2
0 1 2 0 −3 − − − − 0
− − − −→
−
3
− R 3 + R 1 →R 1
0 0 1 0
0
1 0 0 0
0 1 0 0
0
0
1
0
− R 2 + R 1 →R 1
0 −− − − −
− − − −→
0
0
The only solution is a 1 = a 2 = a 3 = 0
...
Therefore S is a basis for V
...
From our result, we have
3
c
1
0
0
W
...
Tiong
1
1
0
1
a
2
3
b+a
3
−3 c +2 b +5 a
7
1
1
= 0
0
1
1
0
1
2
3
1
2
1
...
Note that S contains a zero vector
...
Therefore, S is not a basis for R3
...
Find a basis for the subspace W of R3 spanned by
What is the dimension of W?
1
2
−3
6
−2 , 1 , −4 , −7
...
Then we have
1
2
−3
6
0
a 1 −2 + a 2 1 + a 3 −4 + a 4 −7 = 0
1
0
1
4
0
W
...
Tiong
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Form the augmented matrix and row reduce it
1 2
−2 1
1 0
1
2
0
1
0 −2
−3
−4
1
6
−7
4
−3
−2
4
6
1
−2
0
1
2
−3
6 0
1
R →R 2
2 R 1 + R 2 →R 2
5 2
5 −10
5 0 −− −→
0 −− − − − 0
− − − −→
−−−
− R 1 + R 3 →R 3
0
0 −2
4 −2 0
0
1 0
1 4 0
2 R 2 + R 3 →R 3
0 − − − − − → 0 1 −2 1 0
−−−−−
−2 R 2 + R 1 → R 1
0 0
0 0 0
0
The leading 1’s appear in columns 1 and 2
...
1
0
Hence dimW = 2
...
Let W be the subspace of P3 spanned by
{ t3 + t2 − 2t + 1, t2 + 1, t3 − 2t, 2t3 + 3t2 − 4t + 3}
...
What is the dimension of W?
Solution
Form the equation
a 1 (t3 + t2 − 2t + 1) + a 2 (t2 + 1) + a 3 (t3 − 2t)
+a 4 (2t3 + 3t2 − 4t + 3) = 0t3 + 0t2 + 0t + 0
...
1 0
1
2 0
1 1
0
3 0 − R 1 + R 2 →R 2
− − − −→
−− − − −
−2 0 −2 −4 0 R1 +R3 →R3
− R 1 + R 4 →R 4
1 1
0
3 0
1
1 0
1 2 0
0 1 −1 1 0
0
− R 2 + R 3 →R 3
− − − −→
−− − − −
0
0 1 −1 1 0
0 0
0 0 0
0
W
...
Tiong
2a 4
3a 4
4a 4
3a 4
+
+
−
+
1
0
0
0
0
1
0
0
0
1
0
1
1
−1
0
0
=
=
=
=
1
−1
0
−1
2
1
0
0
0
0
...
Thus the basis for W is
{ t3 + t2 − 2t + 1, t2 + 1}
...
22
...
Solution
Let S = {[a2
0 1], [0
a
2], [1 0 1]}
...
a 1 [a2
0 1] + a 2 [0
a
2] + a 3 [1
0
1] = [0
0 0]
...
In this case S is not a basis for R3
...
Next, form the augmented matrix and row reduce it
...
Thus we obtain the reduced row echelon form [I 3 | 0], which implies that
the three vectors are linearly independent and hence form a basis for R3
...
23
...
Solution
Let W be the subspace of P2 and let at2 + bt + c be an arbitrary element in W
...
W
...
Tiong
Page 28 of 40
TMC1874 Mathematics For Computing
Thus, the set S = { t2 + 2, t − 3} spans the subspace W
...
Form the augmented matrix and row reduce it
...
Thus, S is linearly independent
...
24
...
c
Solution
Let W be the subspace of R4 and let [a
Since d = a + b, then we have
[a
b
c
d] = [a
a[1 0
=
Let S = {[1
0 0
b
1], [0 1
c
b
c
d] be an arbitrary element in W
...
0]}
...
To determine whether S is a linearly independent set, we form
a[1 0 0
1] + b[0
1 0
1] + c[0 0
Form the augmented matrix and row reduce it
1
1 0 0 0
0 1 0 0
− R 1 + R 4 →R 4 0
− − − −→
−− − − −
0
0 0 1 0
1 1 0 0
0
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 0
0
1
0
1
1 0] = [0 0
0
0
1
0
0
0
0
0
0 0]
...
Hence, S is linearly independent
...
(b) All vectors of the form [a
W
...
Tiong
b
c
d], where c = a − b and d = a + b
...
a−b
a[1 0
=
Let S = {[1
b
a + b]
1 1] + b[0
1
1 1]
...
Then S spans W
...
− R 2 + R 3 →R 3
− − − −→
−− − − −
− R 2 + R 4 →R 4
The only solution is a = b = 0
...
Thus, the basis
for W is S and therefore dimW = 3
...
Let A = 2
−3
3
6
−9
9
18
...
Solution
Form the augmented matrix and row reduce it:
1
2
−3
3
6
−9
9
18
−27
0
1
−2 R 1 + R 2 → R 2
0 −− − − −→ 0
−−−−−
3 R 1 + R 3 →R 3
0
0
3
0
0
9
0
0
0
0
0
The solution is
x3
t,
x2
=
s,
x1
W
...
Tiong
=
= −9t − 3s,
Page 30 of 40
TMC1874 Mathematics For Computing
−9t − 3s
where s, t are any real numbers
...
Solution
−9t − 3s
−9
−3
= t 0 + s 1
...
Solution
26
...
(a)
x1
3x1
+
+
x2
x2
+
−
x3
2x3
+
+
x4
2x4
=
=
0
...
1
3
1
−2
1
0
W
...
Tiong
1
1
1
2
1
1
1
1
5
2
1
2
0
0
0
0
1
0
−3 R 1 + R 2 → R 2
−− − − −→
−−−−−
− R 2 + R 1 →R 1
−− − − −
− − − −→
1
0
1
−2
0
1
1
−5
3
−2
5
2
1
2
1
2
1
−1
0
0
1
− 2 R 2 →R 2
−− − −
− − −→
0
0
Page 31 of 40
TMC1874 Mathematics For Computing
The solution is
x4
=
s,
x3
=
r,
1
5
= − r − s,
2
2
3
1
=
r − s
...
and the dimen
1
−2
1
−2
0
1
,
(b)
x1
x1
+
+
3x2
2x2
x2
−
−
+
2x3
x3
3x3
3x4
2x4
3x4
+
+
+
0
0
...
1
1
0
1
0
0
1
0
0
3
2
1
−2
−1
3
3
2
3
3
1
1
−2
−1
3
3
1
3
0
1
0
1
−1
1
0
1
1
2
1
0
− R 1 + R 2 →R 2
−− − − − 0
0
− − − −→
0
0
0
1
−R + 3
3
− −2−R−→R→ 0
0
−−−−−
−
−3 R 2 + R 1 → R 1
0
0
1
0
R 3 + R 2 →R 2
0 −− − − − 0
− − − −→
− R 3 + R 1 →R 1
0
0
3
−1
1
−2
1
3
3
−1
3
0
1
0
1
−1
4
0
1
2
0
1
0
0
0
1
−1
2
3
2
1
2
0
− R 2 →R 2
0 −− −→
−−−
0
0
1
R →R 3
4 3
0 −− −→
−−−
0
0
0
0
The solution is
x4
x2
W
...
Tiong
1
x3 = − s,
2
1
3
x1 = s
...
27
...
Solution
The null space of A is the solution space of the linear system Ax = 0
...
K
...
= s
−3
2
0
1
+r
5
−4
1
0
...
28
...
3
1
(a) A =
2
2
Solution
Form the (λ I 2 − A)
λ−3
−1
−2
λ−2
...
Thus we have
(λ − 3)(λ − 2) − 2 = 0,
λ2 − 5λ + 4
= 0,
(λ − 4)(λ − 1) = 0,
⇒ λ = 1, 4
...
λ
Next, find the determinant
...
K
...
Find a basis for the subspace V of R3 spanned by
S=
2
−5
1
4
6
2 , 0 , −4 ,
2
8
−13
−3
and write each of the following vectors in terms of the basis vectors, if possible:
2
(a) 4
−6
−2
(b) 8
16
4
(c) −8
16
Solution
Form the equation:
1
4
2
−5
0
6 = 0
...
1 4
2
−5 0
−2 R 1 + R 2 → R 2
2 0 −4
6 0 −− − − −→
−−−−−
3 R 1 + R 3 →R 3
−3 2
8 −13 0
1 4 2 −5 0
1
− R 2 + R 3 →R 3
0 1 1 −2 0 − − − − − → 0
−−−−−
−4 R 2 + R 1 → R 1
0 1 1 −2 0
0
1
0
0
4
−8
14
2
−8
14
0
1
0
−2
1
0
3
−2
0
−5
16
−28
0
0
0
0
1
− 8 R 2 →R 2
0 −− − −
− − −→
1
R →R 3
14 3
0
The leading 1’s is at columns 1 and 2
...
2
−3
a
Next, pick an arbitrary vector in the subspace b and form the equation
c
4
1
a
a1 2 + a2 0 = b
...
K
...
Then
1 0
0 1
0 0
b
2
2a− b
8
c +3 a
b −2 a
14 + 8
0
1
0
b
2
2a− b
8
c +3 a
b −2 a
14 + 8
1
0
=
0
0
1
0
2
0
...
Hence
2
1
4 = 2 2
−6
−3
(b) Take a = −2, b = 8, c = 16
...
2
− 11
14
−2
The system is inconsistent
...
16
(c) Take a = 4, b = −8, c = 16
...
2
Thus the solution is a 1 = −4, a 2 = 2, a 3 = 0
...
Find a basis for the subspace of M22 spanned by
S=
2
1
−1
1
,
2
0
1
1
,
0
1
2
2
,
2
1
−5
−1
...
a1
W
...
Tiong
2
1
−1
1
+ a2
2
0
1
1
+ a3
0
1
2
2
+ a4
2
1
−5
−1
=
0
0
0
0
Page 36 of 40
TMC1874 Mathematics For Computing
Next, form the augmented matrix and row reduce it
...
Thus the basis for the subspace is
2
1
−1
1
,
2
0
1
1
,
0
1
2
2
...
Find a basis for the row space of A consisting of vectors that (a) are not necessarily row vectors
of A; and (b) are row vectors of A
...
1
3
−2
2 −4
3 −2 R 1 + R 2 → R 2
−−−−−
−− − − −→
−4 18 −13 4R1 +R3 →R3
−3 R 1 + R 3 → R 3
3 −1
1
1 3
−2
1
7
0 1 − 10 −3R2 +R1 →R1 0
−−−−−
−− − − −→
0
0 0
0
0 0
0
0
1
0
0
0
0
1
0
0
3
−10
30
−10
1
10
7
− 10
−2
7
−21
7
− 1 R 2 →R 2
10
−−−−−
−− − − −→
3 R 2 + R 3 →R 3
−3 R 2 + R 4 → R 4
0
0
(a) The basis for the row space of A consisting of vectors that are not necessarily row
7
1
vectors of A is {[1 0 10 ], [0 1 − 10 ]}
...
Hence the basis for the row space of A consisting of vectors that are row vectors
W
...
Tiong
Page 37 of 40
TMC1874 Mathematics For Computing
of A is {[1
3
− 2], [2
−4
3]}
...
Find a basis for the column space of A consisting of vectors that (a) are not necessarily column
vectors of A; and (b) are column vectors of A
...
3
2 −3 5
2
1 −1 3
Solution
We could perform column reduction to transform the
we transpose matrix A and perform row reduction
...
However, here
3
2
8
5
−24 −15
5
3
0
1
R
R
5 − 3 ↔−4
−
− −→
0
3 R 2 + R 3 →R 3
− − − −→
−− − − −
− R 2 + R 1 →R 1
1
R →R 4
5 4
3
5
(a) The basis for the row space of A consisting of vectors that are not necessarily row
1
vectors of A is {[1 0 0 0], [0 1 0 3 ], [0 0 1 3 ]}
...
Find the row and column ranks of the given matrices
...
K
...
In some cases we need not go as far as
the reduced row echelon form
...
Therefore row rank =row
column=3
...
K
...
W
...
Tiong
Page 40 of 40
Title: TMC1874 Mathematics For Computing tutorial 4 solution
Description: Tutorial Solution Chapter 4: Real Vector Spaces
Description: Tutorial Solution Chapter 4: Real Vector Spaces