Download Tutorial 1 — Solutions - School of Mathematics and Statistics

T HE U NIVERSITY OF S YDNEY
P URE M ATHEMATICS
Linear Mathematics
2010
Tutorial 1 — Solutions

1.
 
3 0 4
5



1 1 2
−2.
a) Calculate the matrix product
−1 3 5
4
b) Find the unique solution of the linear system of equations which corresponds to the augmented matrix


3 0 4 31
 1 1 2 11  .
9
−1 3 5
Solution

   
3 0 4
5
31
a)  1 1 2 −2 = 11.
−1 3 5
9
4
x1 3 0 4
31
x2
1 1 2 . Then we want to find x =
11 . Notice that
b) Let A = −1
such
that
Ax
=
x3
3 5
9
the matrix A must be invertible because the question says that there is a unique solution.
(It is easy to see that A is invertible because det A = −7 6= 0—check!)
5
The shortest way to solve this question is to observe that −2 is a solution by part (a).
4
5 −2
As det A = −7 the matrix A is invertible, so x =
is the unique solution.
4
For completeness, we also find the solution using Gaussian elimination:






3 0 4 31
1 1 2 11
1
1
2 11
R1 ↔R2
R2 :=R2 −3R1
 1 1 2 11 −
−−−→  3 0 4 31 −−
−−−−−→ 0 −3 −2 −2
−1 3 5 9
5 9
−1 3 5 9 
−1 3
1 1
2 11
1 1 2 11
R3 :=R3 +R1
R2 :=R2 +R1



−−−−−−−→ 0 −3 −2 −2 −−−−−−−→ 0 1 5 18
7 20 
0 4 7 20
0 4
1 1 2 11
1 1 2
11
1
R3 :=− R3
R3 :=R3 −4R2
13



18 −−−−−−−→ 0 1 5 18
−−−−−−−→ 0 1 5
0 0 1 4
0 0 −13 −52
Therefore x3 = 4, x2 = 18−5x
3 = −2, x1 = 11−x2 −2x3 = 5. Using back substitution
5
the unique solution is −2 , as before.
4
2. Let A =
1 −3 2
2 −5 6
−1 0 −8
.
x1 a) Use Gaussian elimination to find all solutions of the equation Ax = 0, where x = xx23
0
and 0 = 0 .
0
b) Hence show that some non–zero linear combination of the columns of A is equal to 0.
Linear Mathematics
c) Let b =
b1 b2
b3
Tutorial 1 — Solutions
Page 2
∈ R3 and suppose that Ax = b has at least one solution. Using (b), show
that the equation Ax = b has infinitely many solutions.
Solution
a) Using Gaussian elimination to reduce A to row echelon form we find
1 −3 2 1 −3 2 1 −3 2 R3 :=R3 +R1
R2 :=R2 −4R1
2 −5 6
0
1
2
−−−−−−−→ 0 1 2
−−−−−−−→
0 −3 −6
−1 0 −8
−1 0 −8
1 −3 2 R3 :=R3 +3R2
−−−−−−−→ 0 1 2
0 0 0
Therefore, x3 is the free variable. Set x3 = t, where t is an arbitrary element of R. Then,
by back substitution, x2 = −2t and x1 = 3x2 − 2x3 = −6t − 2t = −8t. That is, the
general solution of the equation Ax = 0 is
−8 −8t where t ∈ R.
x = −2t = t −2 ,
t
1
b) Taking t = 1 in part (a), we see that x =
−8 −2
1
is a solution of Ax = 0. Therefore,
1 −3 2 0
2
3
−8 −1
− 2 −5 + −8
=0= 0 ,
0
0
is one linear combination of the columns of A which sums up to 0. In fact,
1 −3 2 2
3
−8t −1
− t −5 + t −8
= 0,
0
for any t ∈ R.
x1 c) Suppose that x = xx23 is a solution to the equation Ax = b. Then, by part (b),
A
x1 −8t x2 −2t
x3 +t
= Ax + A
−8t −2t
t
= Ax + 0 = b,
for any t ∈ R. Hence, there are infinitely many solutions.
3. Let V be the set of all ordered pairs of real numbers, and consider the following addition and
scalar multiplication operations on u = (u1 , u2 ) and v = (v1 , v2 ):
u + v = (u1 + v1 , u2 + v2 ),
ku = (ku1 , 0)
a) Compute u + v and ku for u = (−1, 2), v = (3, 4) and k = 3.
b) Explain in words why V is closed under addition and scalar multiplication.
c) Since addition on V is the standard addition operation on R2 , certain vector space axioms
hold for V because they are known to hold for R2 . Which axioms are they?
d) Show that the axioms S2, S3 and S4 hold.
e) Show that V is not a vector space under the given operations.
Solution
a) (−1, 2) + (3, 4) = (2, 6); 3(−1, 2) = (−3, 0).
b) Adding two ordered pairs of real numbers produces another ordered pair of real numbers,
so V is closed under addition. Multiplying an ordered pair of real numbers by a real
number (using the operation defined above) produces an ordered pair of real numbers, so
V is closed under scalar multiplication.
Linear Mathematics Extra questions for outside the tutorial — Solutions
Page 3
c) Axioms A1–5.
d) S2: If u = (u1 , u2 ), v = (v1 , v2 ) ∈ A and k ∈ R then k(u + v) = k(u1 + v1 , u2 + v2 ) =
(ku1 + kv1 , 0) = (ku1 , 0) + (kv1 , 0) = ku + kv.
S3 If u ∈ A and k, l ∈ R then (k + l)u = ((k + l)u1 , 0) = (ku1 , 0) + (lu1 , 0) = ku + lu.
S4 If u ∈ A and k, l ∈ R then (kl)u = ((kl)u1 , 0) = k(lu1 , 0) = k(lu).
e) 1u = (u1 , 0) 6= u. So axiom S5 does not hold, and hence V is not a vector space.
4. Each of the following matrices is the reduced row echelon form of an augmented matrix belonging to a system of linear equations in the variables xi , (i = 1, 2, . . .). (Both the systems
represented here have infinitely many solutions – why?)
For each augmented matrix (i) determine the number of parameters needed to solve the system
and (ii) express the solution
of the system in parametric form.
!
!
a)
1 0 4
0
0 1 −5 −1
0
0 0 0
b)
1
0
0
2
0
0
0
1
0
0 6
0 5
1 −1
Solution
a) i) The variable x3 is a parameter.
ii) Let x3 = t. Then x2 = −1 + 5t and x1 = −4t.
  
  
 
x1
−4t
0
−4
x2  = −1 + 5t = −1 + t  5  , t ∈ R.
x3
t
0
1
b)
i) There is one parameter; namely, x2 .
ii) Let x2 = t. Then x4 = −1, x3 = 5, x1 = 6 − 2t.
  
  
 
x1
6 − 2t
6
−2
x 2   t   0 
1
 =
  
 
x3   5  =  5  + t  0  , t ∈ R.
x4
−1
−1
0
Extra questions for outside the tutorial — Solutions
n x1 o
x2 x1 , x2 , x3 ∈ R
is a vector space. Describe each of the following
x3
subsets of n
R3geometrically
and determineowhether or not each
n xsubset
is a vector space.
o
x1
1
x
x
3
3 2
2
2
1
2
x2
x2
a) A =
∈
R
=
−
=
x
.
b)
A
=
∈
R
x
+
x
+
x
≤
1
.
3
1
2
3
2
3
x3
x3
5. Recall that R3 =
Solution
a) The set A is a line in R3 .
We claim that A is a vector space. To check this we have to verify that each of the vector
space axioms is satisfied.
a
d
b
A1 Suppose that c ∈ A and fe ∈ A. Then a2 = − 3b = c and d2 = − 3e = f , so
a+d a
d
b
e
+ 2 = − 3 − 3 = c + f . Hence, b+e ∈ A. Hence, A is closed under addition.
2
c+f
A2 If u, v, x ∈ R3 then (u + v) + x = u + (v + x) so, in particular, this is true when
u, v, x ∈ A. Hence, addition is associative in A.
A3 If u, v ∈ R3 then u + v = v + u so, in particular, this is true when u, v ∈ A. Hence,
addition is commutative in A.
Linear Mathematics
A4 The vector
Tutorial 1 — Solutions
0
0
0
Page 4
∈ A and
a
b
c
+
0
0
0
a
=
a
b
c
=
0
0
0
+
a
b
c
a
0
∈ R3 ; in particular, this holds if cb ∈ A, so 0 is a zero vector in A.
0
a −a a
A5 If cb ∈ A then − cb = −b ∈ A since −a − (−b) − (−c) = −(a − b − c) = 0.
−c
Also
a −a −a a 0
b
+ −b = 0 = −b + cb ,
c
0
−c
−c
a
a
so − cb is a negative of cb .
a
a ka = − kb
= kc.
S1 Suppose that cb ∈ A and k ∈ R. Then k cb = kb ∈ A since ka
2
3
kc
S2 If u, v ∈ A and k ∈ R then k(u + v) = ku + kv since this is true for all points
in R3 .
S3 If u ∈ A and k, l ∈ R then (k + l)u = ku + lu since this is true in R3 .
S4 If u ∈ A and k, l ∈ R then (kl)u = k(lu) since this is true in R3 .
S5 If u ∈ A then (1)u = u since this is true in R3 .
b) The set B is the unit ball of radius 1 and centre (0, 0, 0) in R3 , including its boundary.
The set B is not a!vector space because it is not closed under scalar multiplication. For
1
2
1
1
example, a = 2 ∈ A but 2a = 1 ∈
/ A.
for all
b
c
1
2
1