# IA Algebra Prac sol

```Mathematics IA Algebra Practice Questions
Questions marked with a * are more challenging, or contain further applications which are not examinable.
1. Matrices and Linear Equations
Matrices


1
1
8 4 2
5 −4
0 , B =
1. Let A =  2
, C =
, D = 3 5 7
2 4 6
0 0
−1 −4
−2
and E =
.
−1
Evaluate each of the following matrices if they exist:
(i) At (ii) AB (iii) BAC (iv) A + C (v) A + B t (vi) ED
(vii) AEDB t C t (viii) C 3
1 2 −1
t
Solution:
(i) A =
1 0 −4
70 −56
(iii) BAC =
20 −16


10
8
8
8
4 
(ii) AB =  16
−16 −20 −26


9 3
(iv) A + C is not defined. (v) A + B t = 6 4
1 2
−6 −10 −14
(vi) ED =
−3 −5 −7


−54 0
125 −100
.
(vii) AEDB t C t = −72 0. (viii) C 3 =
0
0
108 0
a b c
. Find expressions for AAt and At A.
2. Let A =
d e f
a2 + b2 + c2 ad + be + cf
Solution:
=
.
ad + be + cf d2 + e2 + f 2
 2

a + d2 ab + de ac + df
At A = ab + de b2 + e2 bc + ef .
ac + df bc + ef c2 + f 2
AAt
3. Let A, B, C and D be matrices which satisfy the following conditions:
AB t C 2 Dt is defined and is a 3 × 3 matrix, A has 7 columns, and B
has 5 rows. Find the sizes of each of the matrices A, B, C and D.
1
Solution:
Note that for C 2 to be defined then C must be square,
let its size be n × n. Then for C 2 Dt to be defined then Dt must be
n × 3 (since the overall product is 3 × 3. Also, since B has 5 rows, then
B t has 5 columns, so for B t C 2 to be defined m = 5. B t must have 7
rows since AB t is defined and A has 7 columns. Also, A must have 3
rows since the overall product is 3 × 3. Putting all of this together we
have that A is 3 × 7, B is 5 × 7, C is 5 × 5 and D is 3 × 5.
4. Let A be an m × n matrix, and suppose there exist n × m matrices C
and D such that CA = In and AD = Im . Prove that C = D.
Solution:
Consider the product CAD = C(AD) = CIm = C (by
question 4), but also CAD = (CA)D = In D = D. Thus C = D.
5. A square matrix A is said to be symmetric if A = At . Let A be an
m × n matrix. Prove that AAt and At A are symmetric matrices.
Solution:
Note that both AAt and At A are square matrices. To
show that AAt is symmetric, we need to show that (AAt )t = AAt . We
have (AAt )t = (At )t At = AAt , since (At )t = A. Similarly we have
(At A)t = At (At )t = At A.
3 8
1
2
. Find all possible matrices C
and B =
6. Let A =
1 −2
−1 −2
such that both AB = AC and BA = CA.
Solution:
Obviously
are there any oth-
C = B is one solution but
x + 2z
y + 2w
x y
5
4
. AB =
= AC =
.
ers? Let C =
−x − 2z −y − 2w
z w
−5 −4
Equating the matrix entries gives a set of 4 equations, but note that it
is actually two equivalent pairs, so if z = s and w = t then x = 5 − 2s
and y = 4 − 2t. Doing the same thing with BA = CA gives two more
equations x = −5 + y and z = 3 + w. The second says that we can
replace s with 3 + t which gives x = −1 − 2t. Note that the equation
x = −5 + y is satisfied in terms of t so this equation
gives no further
−1 − 2t 4 − 2t
information, thus the matrix C can be of the form
3+t
t
for any choice of t ∈ R.
2
7.* Two square matrices A and B are said to commute if AB = BA.
Find all 2 × 2 matrices A with commute with every 2 × 2 matrix.
a b
(Hint: let A =
; choose some very simple matrices B — for
c d
instance matrices with almost every entry equal to 0 — and investigate
what conditions must be satisfied by a, b, c and d in order to have
AB = BA.)
a b
Solution:
Suppose that A =
commutes with every 2 × 2
c d
1 0
matrix. In particular A commutes with
. Therefore
0 0
a b 1
c d 0
a
=⇒
c
0
1
=
0
0
0
a
=
0
0
0
0
a b
c d
b
0
=⇒ b = c = 0
0 1
. Therefore
Similarly A commutes with
1 0
a 0 0
0 d 1
0
=⇒
d
0
1
=
1
0
0
a
=
a
0
a 0
0 d
d
0
1
0
=⇒ a = d
a 0
. Note that such a matrix commutes with every 2 × 2
0 a
matrix (because such an A is a scalar multiple of the identity matrix
I2 ). Therefore A commutes
with every 2 × 2 matrix precisely when A
a 0
.
is diagonal, i.e. A =
0 a
So A =
Linear Equations
8. Solve the following systems of linear equations:
(a)
2x1 + 6x2
4x1 − 3x2
= 12
= 19
(b)
x1 − 3x2 + 2x3
2x2 − x3
x1 + x3
= -10
=6
=0
Solution: These systems of equations may be solved by the matrix
method, or just by substitution.
(a) The first equation can be rewritten as x1 = 6 − 3x2 , substituting
into the second then gives 24 − 15x2 = 19 which gives the solution
(x1 , x2 ) = (5, 1/3).
3
(b) The second and third equations give x2 = (x3 +6)/2 and x1 = −x3 ,
so we can substitute into the first equation to get −x3 − 32 (x3 + 6) +
2x3 = −10 which gives the solution (x1 , x2 , x3 ) = (−2, 4, 2).
9. Solve the following systems of linear equations:
(a)
2x1 + 6x2
x1 + 3x2
=3
=1
(b)
x1 + 2x2
x2 − x3
=4
=2
Solution:
(a) Multiplying the second equation by 2 gives 2x1 +
6x2 = 2 which clearly contradicts the first equation so there are no
solutions.
Alternatively,
in matrix form we have the augmented matrix
2 6 | 12
and halving the first row, and then subtracting it from
1 3 | 1
1 3 | 6
the second row gives
, so the last row says that 0 = −5,
0 0 | −5
so there are no solutions.
(b) This system has more variables than equations, so there will not
be a unique solution. Let x3 = t then x2 = 2 + t, and x1 = 4 − 2x2 =
4 − 2(2 + t) = −2t, so the solutions are (x1 , x2 , x3 ) = (−2t, 2 + t, t) for
any t ∈ R.
10. Consider the system of linear equations,
x1 − 2x2 = 1
3x1 + ax2 = 4
(a) If this system has a unique solution, (x1 , x2 ) = (5, 2), then what
is a?
(b) If this system has no solutions then what is a?
Solution: (a) The second equation is 15 + 2a = 4, so a = −11/2.
(b) The first equation says that x1 = 1 + 2x2 . Substituting into the
second we get 3 + 6x2 + ax2 = 4 or (6 + a)x2 = 1. This will give a
solution for x2 (and hence also for x1 ) unless 6 + a = 0,that is, a = −6,
in which case there are no solutions.
Gauss-Jordan Elimination
11. For each of the following matrices in reduced row echelon form, give
the set of solutions to the corresponding system of equations (you can
call the variables x1 , x2 , . . . ).




1 0 | 0
1 0 0 | 0
(a) 0 1 | 1 (b) 0 1 0 | 1
0 0 | 0
0 0 0 | 0
4


1 2 0 3 0 | 4
(c) 0 0 1 2 0 | 3 (d) 1 4 8 3 | −1
0 0 0 0 1 | 2
Solution: (a) (x1 , x2 ) = (0, 1)
(b) Let x3 = t. Then (x1 , x2 , x3 ) = (0, 1, t) = (0, 1, 0) + t(0, 0, 1).
(c) Let x2 = s and x4 = t. Then we have x1 + 2x2 + 3x4 = 4, so
x1 = 4 − 2s − 3t. Also x3 + 2x4 = 3 so x3 = 3 − 2t and x5 = 2. Putting
these together we get
(x1 , x2 , x3 , x4 , x5 ) = (4−2s−3t, s, 3−2t, t, 2) = (4, 0, 3, 0, 2)+s(−2, 1, 0, 0, 0)+
t(−3, 0, −2, 1, 0).
(d) Let x2 = r, x3 = s and x4 = t. Then
(x1 , x2 , x3 , x4 ) = (−1−4r−8s−3t, r, s, t) = (−1, 0, 0, 0)+r(−4, 1, 0, 0)+
s(−8, 0, 1, 0) + t(−3, 0, 0, 1).
12. For each matrix below, use
reduced row echelon form:

0 1 |

(a) 0 0 |
1 0 |
row operations to put the matrix into



0
1 0 1 | 1
1 (b) 0 2 0 | 2
0
1 1 0 | 1
(c)
1 2 3 | 4
for some α 6= 0
α 0 1 | 2
Solution:
 (a) Swap R1 and R3 , and then swap R2 and R3 to get

1 0 | 0
0 1 | 0.
0 0 | 1


1 0 1 | 1
(b) Do R3 → R3 − R1 and R2 → 1/2 R2 ; giving 0 1 0 | 1.
0 1 −1 | 0


1 0 1 | 1
Next do R3 → R3 − R2 to get 0 1 0 | 1  then R3 → −R3 :
0 0 −1 | −1




1 0 1 | 1
1 0 0 | 0
0 1 0 | 1 and R1 → R1 − R3 gives the rref 0 1 0 | 1
0 0 1 | 1
0 0 1 | 1
1
2
3
|
4
(c) R2 → R2 − αR1 gives
and then
0 −2α 1 − 3α | 2 − 4α
1 2
3
|
4
R2 → − 1/2α R2 gives
and finally
0 1 − 1/2α + 3/2 | − 1/α + 2
1/α
2/α
1 0
|
R1 → R1 − 2R2 gives
−
1
3
−
1
0 1
/2α + /2 |
/α + 2
5
13. Find the solution space of the following systems of linear equations:
(a) 4x + 2y + 5z = 8
(b) 4x + 2y + 4z = 8
2x + 4y + 4z = 1
x + 4y + 4z
−2y − z = 2
Solution:
=1
2x − 6y − 3z = 6
(a) The corresponding augmented matrix is


4 2
5 | 8
2 4
4 | 1
0 −2 −1 | 2


1 0 1 | 5/2
which row reduces to 0 1 1/2 | −1.
0 0 0 | 0
Let z = t. Then x = 5/2 − t and y = −1 − 1/2t
so the solutions are (x, y, z) = ( 25 − t, −1 − 1/2t, t) = (5/2, −1, 0) +
t(−1, −1/2, 1) for t ∈ R.


4 2
4 | 8
4 | 1 which
(b) The corresponding augmented matrix is 1 4
2 −6 −3 | 6


15
1 0 0 |
/7
row reduces to 0 1 0 | −2/7. Thus there is a unique solution,
0 0 1 |
0
15
2
(x, y, z) = ( /7, − /7, 0).
14. If a 5 × 5 augmented matrix for a system of 5 linear equations in 4
unknowns is row reduced:
(a) How many pivots must there be for a unique solution?
(b) How many zero rows must there be for a unique solution?
Solution:
(a) 4 (one for each unknown). (b) 1, there are 5 rows
and 4 pivots so there must be exactly one row with no pivot.
15. For which values of c
have
(i) no solutions, (ii) a
When solutions exist,
c).
(a) x + cy
cy
does each of the following system of equations
unique solution, (iii) infinitely many solutions?
write down what they are (possibly in terms of
=0
(b)
=1
x + cz = c
x + cy + cz = 0
x + cz + z = 1
6
1 c | 0
Solution:
(a) In matrix form this is
. If c = 0 then
0 c | 1
1 0 | −1
there are no solutions. If c 6= 0 then we can row reduce it to
,
0 1 | 1/c
so there is a unique solution. Thus there are no values of c which give
infinitely many solutions. 

1 0
c
| c
c
| 0. We can subtract the
(b) In matrix form this is 1 c
1 0 c+1 | 1


1 0 c |
c
first row from each of the second and third rows to get 0 c 0 | −c .
0 0 1 | 1−c


1 0 0 | 0
If c = 0 then the reduced row echelon form is 0 0 1 | 1 which
0 0 0 | 0
has infinitely many solutions (x1 , x2 , x3 ) = (0, 0, 1) +t(0, 1, 0).

1 0 c |
c
If c 6= 0 then we can divide the second row by c to get 0 1 0 | −1 
0 0 1 | 1−c
and thensubtract c times row
3
from
row
1
to
get
the
reduced
row eche
1 0 0 |
c2
lon form 0 1 0 | −1  which gives a unique solution (x1 , x2 , x3 ) =
0 0 1 | 1−c
(c2 , −1, 1 − c).
16. Suppose 3 foods are used to make a meal. The foods have the following
units per gram of vitamin B and vitamin C respectively.
Food 1
Food 2
Food 3
Vitamin B Vitamin C
12
18
8
10
4
2
By setting up a system of linear equations, determine if it is possible
to combine these foods into a meal of 500 grams with precisely 2000
units of Vitamin B and 1000 units of Vitamin C. If this is possible,
how many ways can it be done?
Solution:
Let x1 equal the number of grams of Food 1, x2 equal
the number of grams of Food 2 and x3 equal the number of grams
of Food 3 in the meal. Since the meal is of weight 500 grams, we
must have x1 + x2 + x3 = 500. On the other hand, from the table,
the meal will contain 12x1 + 8x2 + 4x3 units of Vitamin B, therefore
12x1 +8x2 +4x3 = 2000. Similarly 18x1 +10x2 +2x3 = 1000. Therefore
x1 , x2 and x3 satisfy the linear system
x1 + x2 + x3 = 500
12x1 + 8x2 + 4x3 = 2000
18x1 + 10x2 + 2x3 = 1000
7
The augmented matrix of the system is


1 1 1 | 500
12 8 4 | 2000
18 10 2 | 1000
Do the row operations R2 → R2 − 12R1 , R3 → R3 − 18R1 to obtain
the augmented matrix


1 1
1
|
500
0 −4 −8 | −4000
0 −8 −16 | −8000
Now do the row operation R2 → −1/4R2 to obtain


1 1
1
|
500
0 1
2
| 1000 
0 −8 −16 | −8000
The row operations R1 → R1 − R2 and R3 → R3 + 8R2 then put the
augmented matrix into reduced row echelon form:


1 0 −1 | −500
0 1 2 | 1000 
0 0 0 |
0
There are infinitely many solutions of this system: x1 = t − 500,
x2 = 1000 − 2t, x3 = t. However, we seek a solution with x1 ≥ 0,
x2 ≥ 0 and x3 ≥ 0 (you can’t eat negative amounts of food). Therefore
we must have t − 500 ≥ 0 and 1000 − 2t ≥ 0. Hence t = 500 and so
there is only one way to make the meal: just take a meal with 500
grams of Food 3.
17.* Let A be an m × n matrix. If the system of equations Ax = b has
a solution for every b, show that there is an n × m matrix D so that
AD = Im .
 
   
1
0
0
0 1
0
 
   
 ,  .  , . . .  .  ∈ Rm consecutively
.
Solution:
Choose b = 
 
   
.
. .
0
1
0
and let x1 , x2 , . . . xm be the corresponding solutions of Ax = b for
these choices. Then the required matrix is D = [x1 |x2 | . . . |xm ].
18.* Let A be an m × n matrix and b be a vector in Rm . Prove that there
cannot be exactly two solutions to Ax = b.
8
Solution:
Let x1 and x2 satisfy Ax1 = b and Ax2 = b, with
x1 6= x2 . Then A(2x1 − x2 ) = 2Ax1 − Ax2 = 2b − b = b. Note also
that 2x1 − x2 cannot be equal to either x1 or x2 as this would imply
that x1 = x2 .
More generally note that A(x2 − x1 ) = 0 and so At(x2 − x1 ) = 0 for
any t ∈ R, thus we have A(x1 + t(x2 − x1 )) = b, so (1 − t)x1 + tx2 is a
solution to Ax = b for any t ∈ R. Hence given two distinct solutions
we can then find infinitely many solutions.
Linear Combinations
19. Decide whether each of the following vectors is in the span of
{(1, 0, 1, 2, 0), (2, 1, 1, 0, 1), (0, 2, 2, 1, 1)}
and where possible write the vector as a linear combination of the
vectors in the span:
(a) (7, −2, 1, 4, 0) (b) (4, 3, 5, 5, 1) (c) (−1, −2, −1/2, 5/2, −3/2)
(d) (−1/2, 13, 17/2, −7/2, 15/2).
Solution: To write a vector v as a linear combination of the vectors
in
 we need to looks for solutions to Ax = v where A =
 the span
1 2 0
0 1 2


1 1 2. We can do all of the calculations at once by row reducing


2 0 1
0 1 1


1 2 0 | 7 4 −1 − 1/2
0 1 2 | −2 3 −2
13 


1 1 2 | 1 5 − 1/2 17/2 .


2 0 1 | 4 5 5/2
− 7/2
0 1 1 | 0 1 − 3/2 15/2


1 0 0 | 3 0 0 − 9/2
0 1 0 | 2 0 0
2 


11/2 .
This row reduces to 

0 0 1 | −2 0 0
0 0 0 | 0 1 0
0 
0 0 0 | 0 0 1
0
From this we can see that the vectors in (b) and (c) are not in the
span as there are no solutions. For (a) we have (7, −2, 1, 4, 0) =
3(1, 0, 1, 2, 0) + 2(2, 1, 1, 0, 1) − 2(0, 2, 2, 1, 1) and for (d) we have
( − 1/2, 13, 17/2, − 7/2, 15/2) = − 9/2(1, 0, 1, 2, 0)+2(2, 1, 1, 0, 1)+11/2(0, 2, 2, 1, 1).

1 5

20. Let A = −1 9
1 3
of the columns of

2
5 , and let W be the set of all linear combinations
1
A.
9
(a) Show that (4, 10, 2)t is in W (you should be able to do this by
inspection, that is, without any row operations).
(b) Solve the homogeneous system Ax = 0.
(c) Given that (1, 2, −1)t satisfies Ax = b, what is the vector b?
(d) Write down the general solution of Ax = b, without directly
solving this system.
 
4

Solution: (a) By inspection we see that 10 is given by the follow2
 
 
 
1
5
2
ing linear combination of the columns: 0 −1 + 0 9 + 2 5.
1
3
1
(b) Row reducing gives:
1
5
2
−1 9
5
1
3
1
1
5
2
0 14 7
0 −2 −1
1
5
2
1
0
1
2
0
0
0
1
0 − 12
1
0
1
2
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0

1
2

Let x3 = t, then x1 = 12 t and x2 = − 12 t, so (x1 , x2 , x3 ) = t − 21 , for
1
any t ∈ R.
 
 
9
1



(c) Multiplying A 2 gives b = 12.
6
−1
(d) The general solution of Ax = b is given by adding together the
general solution of Ax= 0 anda particular
solution of Ax = b, in

1
1
2
this case we have x =  2  + t − 12 , t ∈ R.
−1
1


 
1
3 −4
b1
2 −6 and b = b2 .
21. Let A =  3
−5 −1 8
b3
(a) Show that the equation Ax = b is not consistent for all b ∈ R3 .
(b) Find an equation for the set of all b for which Ax = b is consistent. Describe the set geometrically.
10
(c)* Use your answer to part (b) to deduce that
     
   
3
−4 
0 
 1
 1
span  3  ,  2  , −6 = span 0 ,  1  .




−5
−1
8
1
−2
Solution:
We row reduce the corresponding augmented matrix:
b1
1
3
−4
3
2
−6
b2
−5 −1
8
b3
1
3
−4
b1
0 −7
6
b2 − 3b1
0 14 −12
b3 + 5b1
1
3
−4
b1
0
1 −6/7
−(b2 − 3b1 )/7
0
0
0
b3 + 5b1 + 2(b2 − 3b1 )
(there is no need to reduce further if we are just checking for consistency). This system is not consistent for all b, as it is not always
true 
that
 b3 + 5b1 + 2(b2 − 3b1 ) = −b1 + 2b2 + b3 = 0, for example let
1
b = 0, then the last row says that 0 = −1.
0
(b) For this system to be consistent we require −b1 + 2b2 + b3 = 0.
This equation defines a plane in R3 .
(c) Observe that span{(1, 3, −5), (3, 2, −1), (−4, −6, 8)} is precisely
the set of vectors b such that the system Ax = b is consistent. In
part (b) we observed that the set of vectors b = (b1 , b2 , b3 ) such that
the system Ax = b is consistent is precisely the set {(b1 , b2 , b3 ) ∈
R3 | b3 = b1 − 2b2 }. In other words, the set of vectors b for which the
system Ax = b is consistent is the set {(b1 , b2 , b1 − 2b2 )| b1 , b2 ∈ R}.
Since (b1 , b2 , b1 − 2b2 ) = b1 (1, 0, 1) + b2 (0, 1, −2) we see that this last
set is precisely span{(1, 0, 1), (0, 1, −2)}.
22.* Let u, v and w be vectors in Rn . Prove that w ∈ span{u, u − v, v −
w}. (In other words, you need to prove that w is a linear combination
of u − v, v − w and u.)
Solution: One to do this is by trial and error. A more systematic
way is as follows. We want to find scalars x, y and z such that
xu + y(u − v) + z(v − w) = w.
In other words, we want the vector equation
(x + y)u + (z − y)v − zw = w
11
to be satisfied. This is clearly satisfied if we choose x, y and z so that
x + y = 0, z − y = 0 and −z = 1. Therefore take z = −1, y = −1 and
x = 1. As a check we see that indeed
u − (u − v) − (v − w) = w.
23.* Let v1 , v2 , . . . , vk be a sequence of vectors in Rn . Suppose that
vk+1 ∈ span{v1 , . . . , vk }. Prove that
span{v1 , . . . , vk } = span{v1 , . . . , vk , vk+1 }.
(In other words you need to show that every linear combination of the
vectors v1 , . . . , vk is also a linear combination of the vectors v1 , . . . ,
vk , vk+1 and vice versa).
Solution: Let w ∈ span{v1 , . . . , vk } so that w = x1 v1 + · · · + xk vk
for some scalars x1 , . . . , xk . But then w is also a linear combination
of v1 , . . . , vk , vk+1 since w = x1 v1 + · · · + xk vk + 0vk+1 .
Conversely, suppose that w ∈ span{v1 , . . . , vk , vk+1 }, so that there
exist scalars x1 , . . . , xk+1 such that w = x1 v1 + · · · + xk vk + xk+1 vk+1 .
Since vk+1 ∈ span{v1 , . . . , vk } we may write vk+1 = y1 v1 + · · · + yk vk
for some scalars y1 , . . . , yk . But then
w = x1 v1 + · · · + xk vk + xk+1 vk+1
= x1 v1 + · · · + xk vk + xk+1 (y1 v1 + · · · yk vk )
= (x1 + xk+1 y1 )v1 + · · · + (xk + xk+1 yk )vk
and hence w is a linear combination of v1 , . . . , vk .
Homogeneous Equations
24. (a) Use Gauss-Jordan elimination to solve the following homogeneous
system
x1 − x2 − 4x3 = 0,
x1 + x2 + 4x3 = 0,
2x2 + x3 = 0.
(b) Find by inspection a particular solution of the system
x1 − x2 − 4x3 = −4,
x1 + x2 + 4x3 =
6,
2x2 + x3 =
3,
and hence write down the general solution, using the results of
(b).
12
Solution:
(a) We row reduce the corresponding augmented matrix:
1 −1 −4 0
1 1
4 0
0 2
1 0
1 −1 −4 0
0 2
8 0
0 2
1 0
1 0
0 0
0 1
4 0
0 0 −7 0
1 0
0 0
0 1
0 0
0 0
1 0
   
x1
0
which shows that there is a unique solution x2  = 0.
x3
0
(b) By inspection we see that substituting in x1= 
x2 =x3= 1 gives

x1
0
1
the desired result, hence the general solution is x2  = 0 + 1 =
0
1
x3
 
1
1.
1
Elementary Matrices and the Inverse Matrix
25. Let A,B and C be invertible n × n matrices. Find expressions in terms
of A, B, C and their inverses for the inverses of the following matrices:
(i) ABC (ii) AB −1 A (iii) 3At BC 2 (iv) −BA−1 C t A
Solution: (i) C −1 B −1 A−1 . (ii) A−1 BA−1 (iii) 31 (C −1 )2 B −1 (A−1 )t
(iv) −A−1 (C −1 )t AB −1 .
26. Suppose that A and B are invertible matrices. Is (A + B)−1 = A−1 +
B −1 ? Give reasons to justify your answer.
Solution: Let A = B. Then A+B = 2A, so (A+B)−1 = 12 A−1 . On
the other hand A−1 + B −1 = 2A−1 . Thus it is not true in general that
(A + B)−1 = A−1 + B −1 . (There are various other counterexamples
which could be used).
13
27. Which of the following are elementary matrices?
1 0 0
0 1
1 100
0 −1
1 0
(i)
(ii)
(iii)
(iv)
(v)
0 −1 0
1 0
0 1
1 0
0 0
Solution: (i) No, it is not square.
(ii) Yes, it corresponds to swapping the two rows.
(iii) Yes, it corresponds to adding 100 times row 2 to row 1.
(iv) No, this would correspond to two separate row operations (a swap
and multiplying a row by -1).
(v) No, this corresponds to multiplying the second row by 0 which is
not an elementary row operation.

0
28. Row reduce A = 2
1
down the elementary
Solution:

1 0
2 2 to reduced row echelon form and write
0 2
matrix corresponding to each row operation.


1 0 2
First we swap rows 1 and 3 to get 2 2 2, so E1 =
0 1 0


0 0 1
0 1 0.
1 0 0




1 0 2
1 0 0
Next subtract 2R1 from R2 to get 0 2 −2 so E2 = −2 1 0.
0 1 0
0 0 1




1 0 0
1 0 2
Now multiply R2 by 1/2 to get, 0 1 −1, so E3 = 0 21 0.
0 0 1
0 1 0




1 0 2
1 0 0
Now subtract R2 from R3, 0 1 −1, with E4 = 0 1 0.
0 0 1
0 −1 1
Now do R1=R1-2R3 and R2=R2+R3 which reduces
the
matrix to I3 ,


1 0 −2
and the last elementary matrices are E5 = 0 1 0  and E6 =
0 0 1


1 0 0
0 1 1 .
0 0 1
29. Give the inverse of each of the following elementary matrices:






1 0 0
0 0 1
1 0 0
(i) 0 34 0 (ii) 0 1 0 (iii) 0 1 0
0 0 1
1 0 0
0 −6 1
14
Solution:

0
(ii) 0
1

1

(iii) 0
0
0
1
0
0
1
6


1 0 0
(i) 0 43 0.
0 0 1

1
0.
0

0
0
1
30. Determine E a product of elementary matrices which when premultiplying A performs Gauss-Jordan pivoting on the (3, 3) entry of


1 0 −3 5
A = 0 1 3 4 .
0 0 2 6
Check E by computing EA.
Solution:
1
The required row operations
1 + 3R3 and
 are R3 →
 /2R3 , R1 → R
1 0 0
1 0 3
R2 = R2 − 3R3 , so E1 = 0 1 0 , E2 = 0 1 0 and E3 =
0 0 1/2
0 0 1


1 0 0
0 1 −3. (Note that the order of E2 and E3 could be swapped.).
0 0 1




1 0 3/2
1 0 0 14
Then E = E3 E2 E1 = 0 1 −3/2. Then EA = 0 1 0 −5.
0 0 1/2
0 0 1 3
31. For each of the following matrices find the inverse by using elementary
row operations. Write down the elementary matrix corresponding to
each row operation that you use, and hence in each case express both
the matrix and the inverse as products
matrices.
 of elementary 
1 0 0
0 0 1
1 3
(a)
(b) 0 0 1
(c) 1 1 1.
4 3
1 2 3
0 1 2
Solution:
(a)
1 3
4 3
1 3
0 −9
1 3
0 1
1 0
0 1
1
0
1
−4
1
4/9
− 1/3
4/9
0
1
0
1
0
E1 =
E2 =
− 1/9
1/3
− 1/9
15
E3 =
1
−4
1
0
1
0
0
1
0
− 1/9
−3
1
So
A−1
=
− 1/3
1/3
4/9
− 1/9
.
1 −3 1
Now E3 E2 E1 A = I, so
= E3 E2 E1 =
0 1
0
1 0 1 0
1 3
and A = E1−1 E2−1 E3−1 =
.
4 1 0 −9 0 1
A−1
0
1 0
,
−4 1
− 1/9
(b)
1
0
1
1
0
0
1
0
0
1
0
0
1
0
0
0
0
2
0
0
2
0
2
0
0
1
0
0
1
0

0
1
3
0
1
3
0
3
1
0
3/2
1
0
0
1
1
0
0
1
0
−1
1
−1
0
1
− 1/2
0
1
− 1/2
0
0
1
0
0
1
0
0
0
1
0
0
1
0
− 3/2
1
0
0
1
0
0
1
0
1
0
0
1/2
0
0
1/2
0
E1 =
E2 =
E3 =
E4 =
1
0
−1
1
0
0
1
0
0
1
0
0

0
so A−1 =  − 1/2 − 3/2 1/2 .
0
1
0



1
1 0 0
1 0
0
A−1 = E4 E3 E2 E1 = 0 1 − 3/2 0 1/2 0 0
0
0 0 1
0 0
1
and



1 0 0
1 0 0
1
A = E1−1 E2−1 E3−1 E4−1 = 0 1 0 0 0 1 0
1 0 1
0 1 0
0
1
0
1
0
0
0
1
0
1/2
0
0
1
0
0
0
1
0
1
0
0
0
1
0
− 3/2
1
0
16


1 0 0
0 0
0 1  0 1 0
1 0 −1 0 1


0 0
1 0 0
2 0 0 1 3/2.
0 1
0 0 1
(c)
0
1
0
1
0
0
1
0
0
1
0
0
1
0
0
1
0
0

0 1
1 0 0
1 1
0 1 0
0 0 1
1 2
1 1
0 1 0
1 0 0
0 1
1 2
0 0 1
1 1
0 1 0
1 2
0 0 1
0 1
1 0 0
0 −1 0 1 −1
1 2
0 0 1
0 1
1 0 0
0 0
1 1 −1
1 2
0 0 1
1 0 0
0 1
0 0
1 1 −1
1 0 −2 0 1
0 1
1 0 0

1 1 −1
so A−1 = −2 0 1 .
1 0 0
Then we have
E1 =
E2 =
E3 =
E4 =
E5 =
0 1
0
1 0
0
0 0
1
1 0
0
0 0
1
0 1
0
1 −1 0
0 1
0
0 0
1
1 0
1
0 1
0
0 0
1
1 0
0
0 1 −2
0 0
1
A−1 = E5 E4 E3 E2 E1






1 0 0
1 0 1
1 −1 0
1 0 0 0 1 0
= 0 1 −2 0 1 0 0 1 0 0 0 1 1 0 0
0 0 1
0 0 1
0 0 1
0 1 0 0 0 1
and
A = E1−1 E2−1 E3−1 E4−1 E5−1






1 0 −1 1 0 0
1 1 0
0 1 0
1 0 0
= 1 0 0 0 0 1 0 1 0 0 1 0  0 1 2 .
0 0 1
0 0 1
0 0 1
0 0 1
0 1 0
32. Find all values of α for which the following matrix is not invertible.


1
4 −3
A = −2 −7 6  .
−1 3
α
Solution:


1 0 −3
0 . To further row reduce to the
Row reducing A gives 0 1
0 0 α−3
identity we need α − 3 6= 0, so α 6= 3. Thus A is not invertible if α = 3.
17
33. Let A and B be matrices of the same size and suppose that B is
invertible.
(a) Prove that A is invertible if and only if AB is invertible.
(b)* Hence prove that either A + B and I + AB −1 are both invertible
or else both not invertible.
Solution:
(a) This is an ‘if and only if’ statement. Therefore,
we need to prove two things (with the assumption given, that B is
invertible): (i) if A is invertible then AB is invertible, and (ii) if AB
is invertible, then A is invertible.
The key property of invertible matrices that we will use is the following
fact: if X and Y are invertible matrices of the same size then XY is
also invertible (in fact we saw in class that (XY )−1 = Y −1 X −1 ).
We prove (i). We are given that A is invertible and B is invertible.
Therefore, AB is invertible by the key property above.
We prove (ii). Now we are given that B is invertible and AB is invertible. Observe that (AB)B −1 = A(BB −1 ) = AI = A. But this shows
that A is the product of two invertible matrices, namely AB and B −1 ,
and hence is invertible by the key property above (recall that we saw
in lecture that B −1 is invertible with (B −1 )−1 = B).
(b) Since B is invertible we may write A + B = (AB −1 + I)B. By part
(a), A + B is invertible if and only if AB −1 + I is invertible, which is
equivalent to the statement that A + B and AB −1 + I are either both
invertible or both not invertible.
34. Let a,b, c be fixed non-zero constants.


a 0 0
(a) Find the inverse of 0 b 0 in terms of a, b, c.
0 0 c


a a 0
(b) Find the inverse of 0 a a.
a 0 a
Solution:

1/a
0
(a)  0 1/b
0
0

0
0 .
1/c
(b) We can write this

1
1
so the inverse is a 0
1
(after using the usual


1 1 0
as a 0 1 1,
1 0 1
−1 
1/2a
− 1/2a
1 0


1
1/2a
1 1
/2a
=
− 1/2a
1/2a
0 1
method to find the inverse).
18
1/2a

− 1/2a
1/2a
35. True or False? Examine each of the following statements carefully and
decide whether they are true or false. Give a short reason for your
decision in each case.
(a) Let A and B be square matrices such that AB = O. If A is
invertible then B = O.
(b) Let A and B be invertible matrices of the same size. Then (A +
B)−1 = A−1 + B −1 .
(c) If A is an m × n matrix, then there is an invertible matrix E such
that EA is in reduced row echelon form.
(d) The matrix


1 −1 2
3
7 0


0
1 5
0 1 −2


2 0
4
−3 1 8 


A=

1
−1
2
3
7
0




4 8 11 −21 0 −7
3 5 −6
2
1 4
is invertible.
Solution:
(a) True. Multiply both sides of AB = O by A−1 to get B = A−1 AB =
A−1 O = O.
(b) False. For instance A = I2 and B = −I2 are both invertible, but
A + B = O is not even invertible.
(c) True. If R is the reduced row echelon form of A, then there is a
sequence of elementary matrices E1 , . . . , Ek such that E1 · · · Ek A = R.
The matrix E = E1 · · · Ek is invertible since it is a product of invertible
matrices.
(d) False. Since Row 1 = Row 4, the reduced row echelon form of A
has a zero row. Therefore A is not row equivalent to I6 , and hence A
is not invertible.
36. Prove that if A and B are row equivalent matrices (i.e. you can get
from one to the other by a series of elementary row operations) then
the equations Ax = 0 and Bx = 0 have the same solutions. (Hint:
the proof should use elementary matrices)
Solution:
If A and B are row equivalent then B = EA where E is a product
of elementary matrices, and thus is invertible. Let x satisfy Bx = 0.
Then EAx = 0 so Ax = E −1 0 = 0. On the other hand, suppose y
satisfies Ay = 0. Then EAy = E0 = 0 so By = 0.
19
37. Let A be an n × n matrix with two identical columns. Explain why A
is not invertible.
Solution:
Method 1: At has two identical rows, it was shown in lectures that
this means that At is not invertible (since the rref must have a zero
row), so A must not be invertible since if it was then At would have
inverse (A−1 )t .
Method 2: Suppose A is invertible. Then there is a pivot in each
column of rref(A), which occurs in the corresponding row. Suppose
the ith and jth columns are identical. Then carrying out the same row
operations on the entries in these columns must give the same results,
however the ith column should row reduce to give a pivot in the ith
row, and the jth column should give a pivot in the jth row, hence we
have a contradiction and A cannot be invertible.
38. Suppose that A is an invertible n×n matrix satisfying A3 −3A+2I = 0.
Find an expression for A−1 in terms of A and I.
Solution:
A(A2 − 3I) = −2I so A
3I).
− 1/2(A2
− 3I) = I thus A−1 =
− 1/2(A2
−
2. Determinants
Basic Definitions
39. Calculate the following determinants (answers should be formulas in
terms of a and b):
1 2 a 2
0
a
b
a 1 1
1 0 a a
a
0
.
(c)
(b) b
(a) 1 a 0
0 0 0 1
−a −b −a
a 1 0
1 a b b
Solution:
1 a
(a) Expanding along the 3rd column we have 1
= 1 − a2 .
a 1
b
0
b
a
(b) Expanding along the first row we have 0−a
+b
=
−a −a
−a −b
−a(−ab − 0) + b(−b2 + a2 ) = a2 b − b3 + a2 b = 2a2 b − b3 = b(2a2 − b2 ).
20
1 2
(c) Expanding along the 3rd row gives −1 1 0
1 a
determinant along the 2nd row, giving (−1) −1
a
a . We expand this
b
2 a
1 2
−a
=
a b
1 a
(2b − a2 ) + a(a − 2) = 2b − a2 + a2 − 2a = 2(b − a).
a 5 3 b
0 6 0 0
a b
= 4, evaluate
40. Given that
0 7 −1 0
c d
c 8 4 d
a number).
Solution:
a 3 b
Expanding along the 2nd row we get 6 0 −1 0 , and then expanding
c 4 d
a b
along the second row of this matrix we have 6 −1
= −24.
c d
41. Let Jn be the n × n matrix all of whose entries
example,

1
1 1
, J3 = 1
J1 = [1], J2 =
1 1
1
are equal to 1. For

1 1
1 1 .
1 1
Prove that if n > 1, then the matrix In − Jn is invertible with inverse
(In − Jn )−1 = In −
1
Jn
n−1
(here In is the n × n identity matrix).
1
Solution:
We have to check that (In − Jn )(In − n−1
Jn ) = In =
1
(In − n−1 Jn )(In − Jn ). In fact, we proved a fact in lectures which
allows us to cut out a bit of work. By Theorem 1.4, it is sufficient to
prove that
1
(In − Jn )(In −
Jn ) = In .
n−1
We expand the left hand side:
(In − Jn )(In −
1
1
1
Jn ) = In − Jn −
Jn +
J2
n−1
n−1
n−1 n
n
1
= In −
Jn +
J 2.
n−1
n−1 n
21
We need to think about the matrix power Jn2 . The (i, j) entry of this
matrix is
 
1
1
 
[1 1 · · · 1]  .  = n
 .. 
1
and so we see that Jn2 = nJn . Therefore
(In − Jn )(In −
1
n
1
Jn ) = In −
Jn +
J2
n−1
n−1
n−1 n
n
n
= In −
Jn +
Jn
n−1
n−1
= In .
42. Show that all of the determinants D2 , D3 , D4 , . . . Dn , . . . are equal to
1:
1 1 1 1
1 1 1
1 1
1 2 2 2
D3 = 1 2 2
D2 =
D4 =
...
1 2
1 2 3 3
1 2 3
1 2 3 4
1 1 1 1 ... 1
1 2 2 2 ... 2
1 2 3 3 ... 3
Dn =
1 2 3 4 ... 4
. . . . ... .
1 2 3 4 ... n
Solution:
1 1
D2 =
=2−1=1
1 2
1 1 1 1
1
0
1 2 2 2
D4 =
=
1 2 3 3
0
0
1 2 3 4
In general,for n ≥ 2, Dn
1 1 1
1 1 1
1 1
D3 = 1 2 2 = 0 1 1 =
=1
1 2
1 2 3
0 1 2
1 1 1
1 1 1
1 1 1
= 1 2 2 = 1.
1 2 2
1 2 3
1 2 3
= Dn−1 = . . . = D4 = D3 = D2 and D2 = 1.
Row Operations On Determinants
a b c
43. Given that d e f = 5, find
g h i
22
d e f
(i) g h i
(ii)
a b c
a − 4g d g
(iii) b − 4h e h .
c − 4i f i
d
e
f
2a + d 2b + e 2c + f
g
h
i
Solution:
Given that
a
d
g
f
i
c
b c
e f = 5,
h i
d e
g h
= (−1)(−1) × 5 = 5 (R1 ↔ R2 , R2 ↔ R3 )
a b
d
e
f
(ii) 2a + d 2b + e 2c + f = (−1)(2) × 5 = −10
g
h
i
a − 4g d g
(iii) b − 4h e h = 5
c − 4i f i
where the row operations are:
(ii) R1 ↔ R2 ; R2 → 2R2 and R2 → R2 + R1 ;
(iii) R1 → R1 − 4R3 ; then take the transpose which does not affect the
determinant.
(i)
 2

a a 1
44. Let A =  b2 b 1.
c2 c 1
(a) Use row operations to show that
det A = (b − a)(c − a)(b − c).
(b) Calculate the adjoint and inverse of A when a, b, c are distinct.
(c) Use (b) to find α, β, γ so that y = f (x) = αx2 + βx + γ has
f (1) = 1, f (2) = 0, f (3) = 2.
Solution:
(a)
a2 a 1
b2 b 1
c2 c 1
a2
a
1
− a2 b − a 0
=
c2 − a2 c − a 0
b+a 1
= (b − a)(c − a)
c+a 1
b2
23
=
b2 − a2 b − a
c2 − a2 c − a
= (b − a)(c − a)(b − c)
.

b − c b2 − c2 bc(b − c)
(b) [Aij ] = a − c a2 − c2 ac(a − c) ;
a − b a2 − b2 ab(a − b)


b−c
c−a
a−b
a2 − c2
b2 − a2 .
adj A =  c2 − b2
bc(b − c) ac(c − a) ab(a − b)
1
A−1 = (b−a)(c−a)(b−c)

(c) y1 = f (a), y2 = f (b), y3
 2
a
 b2
c2
= f (c) implies
   
a 1 α
y1
b 1 β  = y2 
c 1
γ
y3
Using f (1) = 1, f (2) = 0, f (3) = 2 and (b),


 2
−1
−1 2 −1
1 1 1
1
 5 −8 3 
22 2 1 =
(2−1)(3−1)(2−3)
2
−6 6 −2
3 3 1


1/2
1
−1
/2
=  − 5/2 4 − 3/2
3
−3
1
α = y1 /2 − y2 + y3 /2 = 3/2;
β = −5y1 /2 + 4y2 − 3y3 /2 =
− 11/2;
γ = 3y1 − 3y2 + y3 = 5
y = 3x2 /2 − 11x/2 + 5.
45.* The numbers 22253, 62951, 54043, 70499 and 19635 are all divisible
by 17. Use this fact to show that if
2
6
D= 5
7
1
2
2
4
0
9
2
9
0
4
6
5
5
4
9
3
3
1
3
9
5
then the number D is divisible by 17, without evaluating the determinant.
Solution: Multiply the first column by 104 , the second column by
103 , the third column by 102 , and the fourth column by 10. Then
20000
60000
10
10 D = 50000
70000
10000
24
2000
2000
4000
0
9000
200
900
0
400
600
50
50
40
90
30
3
1
3
9
5
Now add column 2 to column 1, then column 3 to column 1, then
column 4 to column 1 and finally column 5 to column 1. We get
22253
62951
10
10 D = 54043
70499
19635
2000
2000
4000
0
9000
200
900
0
400
600
50
50
40
90
30
3
1
3
9
5
Since each of the numbers in the first column are divisible by 17,
we can take out a factor of 17 and deduce that the determinant is
divisible by 17. But this means that 1010 D is divisible by 17 (clearly
the determinant is an integer in this case). Since 17 is prime the only
way that this can happen is if D is divisible by 17.
General Properties of Determinants
46. Suppose A and B are n × n matrices with det A = 4 and det B = −3.
Find each of the determinants below:
(a) det(AB)
(b) det(A2 )
(c) det(B −1 A)
(d) det(2A)
(e) det(3B t )
(f) det(AAt )
Solution:
(a) det(A) det(B) = −12.
(b) det(A) det(A) = 16.
(c)
(d)
1
det(B) det(A) = −4/3.
2n det(A) = 4.2n = 2n+2 .
(e) 3n det(B) = −3n+1 .
(f) det(A) det(At ) = det(A)2 = 16.
47. What can you say about det(A) if a square matric A satisfies
(a) A2 = A (such a matrix is called idempotent).
(b) Am = 0 for some m > 1 (such a matrix is called nilpotent).
Solution:
25
(a) We have det(A)2 = det(A), so det(A)(det(A) − 1) = 0 and either
det(A) = 0 or det(A) = 1.
(b) det(A)m = 0 so we must have det(A) = 0.
48.*
(a) Let A be a 2 × 3 matrix. Then AT A is a 3 × 3 matrix. Show that
det(AT A) = 0.
(b) Suppose again that A is a 2 × 3 matrix; is it necessarily true that
det(AAT ) = 0?
Solution:
(a) Since A is a 2 × 3 matrix, there exists a vector x 6= 0 such that
Ax = 0 (less equations than variables). Thus there exists x 6= 0
with AT (Ax) = 0 or (AT A)x = 0. This means that the square
matrix AT A is not invertible; that is, det(AT A) = 0.
(b) There are 2 × 3 matrices A with det(AAT ) 6= 0. For example,


1 1 1
3 2
1 1 1
, det(AAT ) 6= 0; AT A = 1 2 2
AAT =
A=
2 2
0 1 1
1 2 2
T
but det(A A) = 0.


1 2 3
49. Consider the matrix A = 4 5 6 .
7 8 10
(a) Find the matrix [Aij ] of minors of A.
(b) Find the matrix [Cij ] of cofactors of A.
(c) Find the adjoint, adj(A), of A.
(d) Find det(A).
(e) Hence write down the inverse of A.
Solution:


2
−2 −3
(a) [Aij ] = −4 −11 −6
−3 −6 −3


2
2
−3
(b) [Cij ] =  4 −11 6 
−3
6
−3
26


2
4
−3
(c) adj A = [Cij ]T =  2 −11 6 
−3
6
−3
(d) |A| = 1A11 − 2A12 + 3A13 = −3

− 2/3 − 4/3
1
1
adj A =  − 2/3 11/3 −2
(e) A−1 = |A|
1
−2
1
50.* Let A = [aij ] be an n × n matrix. If det(A) = 1 and each entry of A
is an integer (i.e. aij ∈ Z for all i, j), prove that all of the entries A−1
are also integers.
Solution: We have (det(A))A−1 = adj(A) and hence A−1 = adj(A).
Therefore to show that every entry of A−1 is an integer, we need to
show that every cofactor Cij of A is an integer, and hence that every
(i, j) minor Aij of A is an integer. But Aij is the determinant of the
(n − 1) × (n − 1) matrix obtained by removing the i-th row and j-th
column from A; since every entry of A is an integer also every entry
of this (n − 1) × (n − 1) matrix is an integer. Therefore all we need to
do is to show that the determinant of a matrix all of whose entries are
integers, is itself an integer. But the determinant of a n × n matrix A
is a sum of terms of the form ±a1i1 a2i2 · · · anin ; since each entry aij is
an integer, so must all of the terms in this sum be integers, therefore
det(A) ∈ Z.
3. Optimisation and Convex Sets
Convex Sets
51. (a) Write down the definition of a convex set.
(b) Define a vertex of a convex set.
Solution:
(a) A convex set C is a set of points (or region) in Rn such that for any
two points in C, the line segment joining the two points lies completely
in C.
−→
(b) A point P with position vector OP = v is a vertex of the convex
set C if v can not be written in the form
(1 − t)u + tw
0≤t≤1
for any (distinct) points u, w ∈ C except when v = u (t = 0) or v = w
( t = 1).
27
52. Sketch the following sets in R2 and determine the vertices.
(a)
{(x1 , x2 ) | x1 + 3x2 ≤ 6, 2x1 − x2 ≥ 4, x1 ≥ 0, x2 ≥ 0}
(b)
{(x1 , x2 ) | x1 + x2 ≥ 2, x1 − x2 ≤ 4, x1 ≥ 1, x2 ≥ 0}
(c)
{(x1 , x2 ) | −x1 + 4x2 ≤ 5, x1 − 4x2 ≤ 2, x1 + 2x2 ≥ 2}
Solution:
Graphs:
(a)
(b)
(c)
Vertices: (a): (2,0), (6,0), (18/7,8/7)
(b) (1,1), (2,0), (4,0)
(c) (2,0), (-1/3,7/6)
53. Sketch the following sets in R2 and hence determine whether they are
bounded or unbounded, and whether they are convex or not. In the
case where the set is not convex, give an example of a line segment
where the definition of convexity breaks down.
(a) {(x, y) | x + y ≥ 1, x − y ≤ 2, x ≥ 0}
(b) {(x, y) | 1 ≤ |x| ≤ 2, |y − 3| ≤ 2}
(c) {(x1 , x2 ) | x21 + x22 ≤ 4 and x1 + x2 ≤ 1}
(d) {(x1 , x2 ) | x21 + x22 ≥ 1 and x1 + x2 ≤ 4}
28
Solution:
(a)
Unbounded, convex
b)
bounded, not convex.
Consider the line segment from (-1,1) to (1,1).
(c)
Bounded, convex.
29
(d)
unbounded, not convex.
Consider the line segment from (-1,0) to (1,0).
54. Using the definition of convexity, show that the set
C = {(x1 , x2 ) ∈ R2 | (5, −1) · (x1 , x2 ) ≤ 7}
is convex.
Solution:
Let u and v be any two points in the set C, and consider any point x
on the line segment between them, so x = (1 − t)u + tv for 0 ≤ t ≤ 1.
We need to show that (x1 , x2 ) is in C, that is x · (5, −1) ≤ 7. Consider
x · (5, −1) = [(1 − t)u + tv] · (5, −1) = (1 − t)u · (5, −1) + tv · (5, −1)
where we have used the basic properties of the inner product. Since
each of u and v is in C, u · (5, −1) ≤ 7 and v · (5, −1) ≤ 7, thus
(1 − t)u · (5, −1) + tv · (5, −1) ≤ 7(1 − t) + 7t = 7
as required. Note that the final inequality depends on both t ≥ 0 and
(1 − t) ≥ 0 which is true since 0 ≤ t ≤ 1.
55. Let u and v be the points (−2, 1, 0) and (1, −5, 6) in R3 .
(a) Let x be the point (−1, −1, 2). Find t such that
x = (1 − t)u + tv.
Is x on the line segment joining u and v?
(b) Is the point y = (−4, 5, −4) on the line segment joining u and v?
Is y on the line through u and v?
30
(c) Is the point z = (2, −4, 4) on the line segment joining u and v?
Is z on the line through u and v?
Solution:
(a) We need t such that (−1, −1, 2) = (1 − t)(−2, 1, 0) + t(1, −5, 6)
or (−1, −1, 2) = (−2 + 3t, 1 − 6t, 6t), so t = 1/3. Since 0 ≤ t ≤ 1
then this means that x is on the line segment joining u and v.
(b) We need to solve (−4, 5, −4) = (−2 + 3t, 1 − 6t, 6t). This is
satisfied when t = −2/3 but since t < 0 we can say that y is on
the line through u and v but it is not on the line segment joining
u and v.
(c) This time we need to solve (2, −4, 4) = (−2 + 3t, 1 − 6t, 6t). The
third entry requires that t = 2/3, however this is not consistent
with the other entries so there is no solution. Thus z is not on
the line through u and v, which also means that it is not on the
line segment joining the two points.
56.* Let C = {(x1 , x2 ) ∈ R2 | x1 + x2 ≤ 1, x1 ≥ 0, x2 ≥ 0}. Consider
v = (1, 0). Suppose there exist u = (u1 , u2 ) and w = (w1 , w2 ) in C,
and t ∈ (0, 1), such that v = (1 − t)u + tw.
(a) Show that if u1 < 1 or w1 < 1, then (1 − t)u1 + tw1 < 1. This
means u1 = w1 = 1. Why?
(b) Hence or otherwise show that v is a vertex.
Solution:
(a) Note that the requirement that u and w are in C means that
0 ≤ u1 ≤ 1 and 0 ≤ w1 ≤ 1, so that (1 − t)u1 ≤ 1 − t and tw1 ≤ t.
Thus (1−t)u1 +tw1 ≤ (1−t)+t = 1. Now if either u1 < 1 or w1 <
1 then the corresponding inequalities become strict inequalities
(i.e. (1 − t)u1 < 1 − t or tw1 < t, thus (1 − t)u1 + tw1 < 1).
Thus we must have u1 = w1 = 1 since (1, 0) = v = (1 − t)u + tw
implies that we must have (1 − t)u1 + tw1 = 1.
(b) We have shown that if 0 < t < 1 then there are no two distinct
points u and w such that v = (1 − t)u + tw. Thus if 0 ≤ t ≤ 1
then the the only possible ways of writing v = (1 − t)u + tw
would be t = 0 or t = 1, hence v is a vertex.
31
57.* Let S = {u1 , . . . , un } be a set of vectors in Rn . A vector w is said
to be a convex combination of the vectors u1 , . . . , un if w is a linear
combination w = x1 u1 + · · · + xn un where the scalars x1 , . . . , xn are
non-negative real numbers and x1 + · · · + xn = 1. Let Conv(S) be the
set of all convex combinations of the vectors in S. Use the definition
of convexity to show that Conv(S) is a convex set.
Solution: Let u and v be points of Conv(S). We have to prove that
for any 0 ≤ t ≤ 1, the vector (1 − t)u + tv belongs to Conv(S). Since
u and v belong to Conv(S), there exist non-negative real numbers
x1 , . . . , xn and y1 , . . . , yn such that x1 + · · · + xn = 1, y1 + · · · + yn = 1,
u = x1 u1 + · · · + xn un , and v = y1 u1 + · · · + yn un . Therefore
(1 − t)u + tv
=(1 − t)(x1 u1 + · · · + xn un ) + t(y1 u1 + · · · + yn un )
=((1 − t)x1 + ty1 )u1 + · · · + ((1 − t)xn + tyn )un
Since
((1 − t)x1 + ty1 ) + · · · + ((1 − t)xn + tyn )
= (1 − t)(x1 + · · · + xn ) + t(y1 + · · · + yn )
= (1 − t) + t = 1
and each (1 − t)xi + tyi ≥ 0, we see that (1 − t)u + tv ∈ Conv(S), as
required.
Methods for Solving Optimisation Problems
58. We wish to find the minimum value of g(x, y) = 5x/2 + 5y subject to
the constraints
3x + 2y ≥ 36
3x + 5y ≥ 45
x≥0
y≥0
Sketch the feasible region, identify its vertices, and find the vertex or
vertices at which the smallest value of g(x, y) is taken.
Solution:
The feasible region is shown in the diagram below:
32
The vertices are (15,0), (0,18) and (10,3). Substituting each of these
into g(x, y) gives the values 37 1/2, 90 and 40 respectively, so the minimum value is 37 1/2 at the point (15,0).
59. Consider the convex set given by
2x1 + x2 ≤ 8
2x1 + 3x2 ≥ 12
x1 ≥ 0
x2 ≥ 0
(a) Add slack variables to write this in standard form
C = {x ∈ Rn | Ax = b, x ≥ 0}.
(b) Find the basic solutions of Ax = b and indicate those which are
feasible.
(c) Check your answer to the previous point by sketching the feasible
region.
(d) Find the maximum value of f (x) = x1 − x2 over C and write
down the corresponding optimal choice for x.
Solution:
(a) Multiplying the second inequality by -1 and adding slack variables
x3 , x4 ≥ 0 gives
2x1 + x2
+x3
−2x1 − 3x2
for x1 , x2 , x3 , x4 ≥ 0.
(b) The basic solutions are as follows:
33
=8
+ x4 = − 12
Pivots
Matrix
Basic Solution
x3 , x4
2
1 1 0 |
8
−2 −3 0 1 | −12
x2 , x4
x1 , x4
x1 , x2
x1 , x3
Feasible?
Vertex
(0, 0, 8, −12)
no
2 1 1 0 | 8
4 0 3 1 | 12
(0, 8, 0, 12)
yes
1 1/2 1/2 0 | 4
0 −2
1 1 | −4
(4, 0, 0, −4)
no
(3, 2, 0, 0)
yes
(6, 0, −4, 0)
no
1 0 3/4
1/4 | 3
0 1 −1/2 −1/2 | 2
1 3/2 0 −1/2 | 6
0 −2 1
1
| −4
2/3 1 0 −1/3 | 4
(0, 4, 4, 0)
x2 , x3
4/3 0 1 1/3 | 4
Therefore the vertices are (0, 8),(3, 2) and (0, 4).
(0, 8)
(3, 2)
yes
(c) The feasible region and its vertices are shown below:
(d) Evaluating f at each vertex gives f (0, 8) = −8, f (0, 4) = −4 and
f (3, 2) = 1, so the maximum is 1 at (3, 2).
34
(0, 4)
60. Consider the region given by
x≤4
x+y ≥1
x−y−z ≤8
x, y, z ≥ 0
(a) Formulate this problem using slack variables s1 , s2 and s3 . Write
down the basic solution obtained by choosing the slack variables
as pivots and state whether it is feasible.
(b) Find the basic solution corresponding to pivots z, s1 and s2 and
state whether is is feasible.
(c) Find the basic solution corresponding to pivots y,s1 and s3 and
state whether it is feasible.
Solution:
(a) Adding slack variables gives the following system of equations:
x
+ s1
−x − y
=4
=−1
+s2
x − y −z
+ s3 =8
with x, y, z, s1 , s2 , s3 ≥ 0. In matrix form

1
0
0 1 0 0
−1 −1 0 0 1 0
1 −1 −1 0 0 1
we have

| 4
| −1
| 8
The resulting basic solution is (0, 0, 0, 4, −1, 8) which is not feasible.
(b) We carry out the operation Pivot(3,3) to

1
0 0 1 0 0
−1 −1 0 0 1 0
−1 1 1 0 0 −1
get

| 4
| −1
| −8
The resulting basic solution is (0, 0, −8, 4, −1, 0) which is not feasible.
(c) We carry out the operation Pivot(2,2) on the original matrix to
get


1 0 0 1 0 0 | 4
1 1 0 0 −1 0 | 1
2 0 −1 0 −1 1 | 9
which gives the basic solution (0, 1, 0, 4, 0, 9) which is feasible.
35
Formulation of Optimisation Problems
61. A pet food manufacturer produces two types of food: regular and
premium. A 20kg bag of regular food requires 2 hours to prepare
and 3 hours to cook; a 20kg bag of premium food requires 2 hours to
prepare and 5 hours to cook. The materials used to prepare the food
are available 8 hours per day and the oven used to cook the food is
available 15 hours per day. The profit on a 20kg bag of regular food
is \$42 and on a 20kg bag of premium food it is \$50. Determine how
many bags of each type of food should be made to maximise the profit
(fractional bags can be produced – they just sell them in smaller bags).
Solution:
Let x1 = number of bags of regular food, x2 = number of bags of
premium food. We wish to maximise profit, P = 42x1 + 50x2 such
that 2x1 + 2x2 ≤ 8, 3x1 + 5x2 ≤ 15, x1 , x2 ≥ 0. The feasible region is
5
4
3
2
( 52 , 32 )
1
1
0
1
2
3
4
5
6
7
-
1 (0, 0) = 0, P (4, 0) = 168, P ( 5 , 3 ) = 180
P (0, 3) = 150, P
2 2
So maximum is \$180 when (x1 , x2 ) = ( 25 , 32 ). This means we want
bags of regular food, and 32 bags of premium food.
5
2
62. A trucking company owns two types of trucks. Type A has 20 cubic
metres of refrigerated space and 30 cubic metres of non-refrigerated
space. Type B has 20 cubic metres of refrigerated space and 10 cubic
metres of non- refrigerated space. A customer wants to haul some
produce a certain distance and will require 160 cubic metres of refrigerated space and 120 cubic metres of non-refrigerated space. The
trucking company figures that it will take 300 litres of fuel for the type
36
8
A truck to make the trip and 200 litres of fuel for the type B truck.
Find the number of trucks of each type that the company should allow
for the job in order to minimise fuel consumption
(a) graphically, and
(b) algebraically (i.e. by using slack variables).
Solution:
Let x1 = number of trucks of type A, x2 = number of trucks of type
B (x1 , x2 ≥ 0).
We wish to minimise f = 300x1 + 200x2 subject to 20x1 + 20x2 ≥
160, 30x1 + 10x2 ≥ 120.
(a) The feasible region is
12.5
10
7.5
(2, 6)
5
2.5
- .5
2
0
2.5
5
7.5
10
12.5
15
Vertices: (0, 12), (2, 6), (8, 0)
f (0, 12) = 2400, f (2, 6) = 1800, f (8, 0) = 2400
∴ Min is 1800L for 2 type As and 6 type Bs.
(b) Add slack variables x3 , x4 ≥ 0 :
−20x1 − 20x2 +x3
= −160
−30x1 − 10x2
+ x4 = −120
Number of possible basic solutions is 42 = 6. The first colum
gives the choices of pivots.
x1 x2 x3 x4
Basic soln
Feasible? Vertex Value of f
X X (0, 0, −160, −120)
N
−
−
X
X
(0, 8, 0, −40)
N
−
−
X
X
(8, 0, 0, 120)
Y
(8, 0)
2400
X
X
(4, 0, −80, 0)
N
−
−
X X
(2, 6, 0, 0)
Y
(2, 6)
1800
X X
(0, 12, 80, 0)
Y
(0, 12)
2400
So again min is 1800 at (2,6).
37
63. Two oil refineries produce three grades of gasoline, A, B and C. At
each refinery the various grades of gasoline are produced in a single
operation so that they are in fixed proportions. Assume that one
operation at Refinery 1 produces 1 units of A, 1 unit of B, and 3 units
of C. One operation at Refinery 2 produces 1 unit of A, 4 unit of
B, and 2 units of C. Refinery 1 charges \$400 for one operation, and
Refinery 2 charges \$300 for one operation. A consumer needs 170 units
of A, 200 units of B, and 360 units of C. How should the orders be
placed if the consumer’s needs are to be met most economically? (Use
graphical means to solve).
Solution:
Let x be the number of operations at refinery 1, and y be the number of
operations at refinery 2. Then we wish to maximise the cost, C(x, y) =
400x + 300y (in dollars), subject to the constraints:
x + y ≥ 170
(units of A)
x + 4y ≥ 200
(units of B)
3x + 2y ≥ 360
(units of C)
x, y ≥ 0
(non-negativity).
The feasible region is:
The vertices are (0, 180), (200, 0), (160, 10), (20, 150). Evaluating
the cost at each vertex gives C(0, 180) = 54000, C(200, 0) = 80000,
C(160, 10) = 67000, and C(20, 150) = 53000. Therefore the cheapest
option is to carry out 20 operations at refinery 1 and 150 operations
at refinery 2.
38
64. Formulate as an optimisation problem, but do not solve:
A manufacturer of a particular product is planning the production
schedule for the months of October, November, and December. The
demands for the product are for 70 units in October, 90 units in
November, and 120 units in December. The industrial plant has the
capacity to produce at most 100 units per month. It costs \$150 to
produce each unit in October, and \$160 in each of November and December. As well, each unit held over in storage from one month to the
next, attracts a storage fee (inventory cost). There is no storage cost
in the month of production, but there is a cost of \$30 in any following
month. (Thus the storage cost for one unit is found by multiplying
the number of items still in storage on the first day of the month by
\$30.) The manufacturer plans to have no items in storage at the beginning of October, and none in storage at the end of December. The
manufacturer wishes to determine the number of units which should
be produced in each month in order to minimise costs. (Fractional
units can be produced.)
Solution:
Let xi be the number of items made in month i (i = 1 corresponds to
October, etc). Then to satisfy demand, we need x1 ≥ 70. At the end
of October, we will have x1 − 70 items left, which will incur a storage
charge of \$30 per item in November. However, we will then only need
to ensure (x1 − 70) + x2 ≥ 90, the demand for November. That is,
x1 +x2 ≥ 160. There will then be (x1 −70)+x2 −90 = x1 +x2 −160 items
in storage at the end of November. Proceed similarly for December.
The objective is then to minimise costs of 150x1 + 160(x2 + x3 ) +
30(x1 − 70) + 30(x1 + x2 − 160) = 210x1 + 190x2 + 160x3 − 6900 dollars.
Thus the LP is:
Minimise z = 210x1 + 190x2 + 160x3 − 6900 (cost in dollars)
Subject to
x1
x1 +x2
x1 +x2 +x3
x1
x2
x3
≥ 70
(demand for October)
≥ 160
(demand for November)
= 280 (demand for December; none left)
≤ 100
(monthly plant capacity)
≤ 100
(monthly plant capacity)
≤ 100
(monthly plant capacity)
and x1 , x2 , x3 ≥ 0 (nonnegativity).
4. The Vector Space Rn
Linear Dependence and Independence
65. Let u = (2, 1, 4, 3), v = (3, 4, 1, 2), w = (1, 2, −1, 0).
39
(a) Are the vectors {u, v, w} linearly independent? Justify your
(b) Are the vectors {v, w} linearly independent? Justify your answer.
(c) Are the vectors {v, w, 0} linearly independent? Justify your
(d) Are the vectors {u, w, 5u − 3w} linearly independent? Justify
Solution:
(a) No, as 2u + (−3) v + 5w = 0.
(b) Yes, as v is not a multiple of w.
(c) No, as 0 is in the list.
(d) No, as 5.u + (−3).w + (−1).(5u − 3w) = 0.
66. Which of the following lists of vectors are linearly independent:
(a) {(3, 4), (4, 3)}
(b) {(2, 1, −3, 6), (5, 3, 7, 8), (1, 1, 13, −4)}
(c) {(2, 1, 3), (2, −2, −5), (7, 3, 9)}
(d) {e1 , e1 + 2e2 , e1 + 2e2 + 3e3 , . . . , e1 + 2e2 + 3e3 + · · · + nen }
where ei ∈ Rn is the vector which has 1 in the i-th place and 0
everywhere else.
Solution:
(a) Linearly independent: the 2 vectors are not multiples of each
other.




2 5 1
1 0 −2
1 3 1
0 1 1 



(b) Linearly dependent: as 
−3 7 13  row reduces to 0 0 0 
6 8 −4
0 0 0
so 2u − v + w = 0.




2 2 7
1 0 0
(c) Linearly independent: as 1 −2 3 row reduces to 0 1 0.
3 −5 9
0 0 1


1 1 1 ... 1
0 2 2 . . . 2 


0 0 3 · · · 3 
(d) Linearly independent: as A = 
 row reduces to
 .. .. .. . .
.
. . .
. .. 
0 0 0 ... n
the identity n×n matrix. (In fact, the determinant of this matrix
is 1 × 2 × . . . × n 6= 0 so it’s invertible which means that there is
a unique solution to Ax = b for any b.)
40
67. For what value(s) of d are the following sets of vectors linearly dependent? Justify your answers.
(a) {(1, −1, 4)), (3, −5, 5), (−1, 5, d)}
(b) {(3, 7, −2), (−6, d, 4), (9, 1, −4)}
Solution:
(a) Linear dependent
only if d = 10 as 1 3 −1R − 1 −5 5R4 5 d
rowreduces to 1 0 5R0 1 −2R0 0 −10 + d .
(b) Linear dependent
only if d = −14 as 3 −6 9R7 d 1R − 2 4 −4
row reduces to 1 −2 0R0 d + 14 0R0 0 1 .
68. Determine, with reasons, if the following statements are (A) always
true, (B) always false, or (C) might be true or false.
(a) If v1 , ..., v5 ∈ R5 and v1 = v2 + v3 then the set {v1 , ..., v5 } is
linearly dependent.
(b) If v1 and v2 ∈ R2 lie on the same straight line through the origin,
then the set {v1 , v2 } is linearly dependent.
(c) If {v1 , v2 , v3 } is linearly dependent, then so is {v1 , v2 }.
(d) If {v1 , v2 } is linearly independent, then so is {v1 + v2 , v1 − v2 }.
(e) There is a set {v1 , . . . , v5 } of linearly independent vectors in R4 .
Solution:
(a) (A) Always true, since one vector is a linear combination of the
others, or v1 − v2 − v3 + 0v4 + 0v5 = 0.
(b) (A) Always true, since one vector is a multiple of the other.
(c) (C) Neither e.g. result is false for v1 = (1, 0), v2 = (0, 1), v3 =
(1, 1), but true for v1 = (1, 0), v2 = (2, 0), v3 = (1, 1).
(d) (A) Always true. Suppose a(v1 + v2 ) + b(v1 − v2 ) = 0. Then
(a + b)v1 + (a − b)v2 = 0. Since v1 , v2 are linearly independent,
this gives a + b = 0, a − b = 0 giving a = b = 0 only.
(e) (B) Always false. The maximum number of linearly independent
vectors in R4 is four.
41
69. What is the definition of a subspace of Rn ?
Solution:
W ⊆ Rn is a subspace of Rn if it satisfies the following three conditions:
• 0 ∈ W;
• u ∈ W ⇒ c u ∈ W for any c ∈ R;
• u, v ∈ W ⇒ u + v ∈ W .
Subspaces
70. Which of the following are subspaces of Rn for some n? Give a reason
(a) W1 = {(x, y, z) ∈ R3 | 2x − y = 3z + x = 0}
(b) W2 = {(x, y, z, w) ∈ R4 | x + y = zw}
(c) W3 = {(α, β, γ) ∈ R3 | γ = 0}
(d) W4 = {(x, y, z) ∈ R3 | x + y + z = 1}
(e) W5 = {(x, y) ∈ R2 | x2 + y 2 = 1} ∪ {(0, 0)}
(f) W6 = {(x, y) ∈ R2 | x2 + y 2 = 0}.
Solution:
(a) W1 is a subspace as it is the solution set for two homogeneous
linear equations.
(b) W2 is not a subspace as (1, 1, 1, 2) ∈ W2 , but 2(1, 1, 1, 2) =
(2, 2, 2, 4) 6∈ W2 , so W2 is not closed under scalar multiplication.
(c) W3 is a subspace of R3 since it is the solution set of a homogenous
linear equation.
(d) W4 is not a subspace, since (0, 0, 0) = 0 6∈ W4 .
(e) W5 is not a subspace, since even though 0 ∈ W5 , it is not closed
under addition or scalar multiplication, for example (1, 0) ∈ W5
since 12 + 02 = 1 but 2(1, 0) = (2, 0) ∈
/ W5 since 22 + 02 6= 1
(f) W6 is a subspace, since W6 = {(0, 0)} ⊂ R2 .
71. Let A be an m × n matrix. Show that
(a) R = {y ∈ Rm | Ax = y for some x ∈ Rn } is a subspace of Rm
(b) K = {x ∈ Rn | Ax = 0} is a subspace of Rn .
42
Solution:
(a) Note that 0 ∈ R since A0 = 0. If y1 , y2 ∈ R then ∃ x1 , x2 such
that Ax1 = y1 and Ax2 = y2 . Then A(x1 + x2 ) = Ax1 + Ax2 =
y1 + y2 so y1 + y2 ∈ R. Also, A(cx) = cAx = cy so cy ∈ R.
Therefore R is a subspace of Rm .
(b) Obviously 0 ∈ K. Note that if x1 , x2 ∈ K then A(x1 + x2 ) =
Ax1 + Ax2 = 0 + 0 = 0 so x1 + x2 ∈ K. Also, if Ax = 0 then
A(cx) = c(Ax) = c0 = 0 so cx ∈ K. Therefore K is a subspace
of Rn .
72. Suppose that U, V ⊂ Rn are subspaces of Rn . Prove that the intersection U ∩ V is a subspace of Rn .
Solution: Since U and V are subspaces of Rn we have 0 ∈ U and
0 ∈ V . Therefore 0 ∈ U ∩ V . Suppose that x ∈ U ∩ V and y ∈ U ∩ V .
Then x, y ∈ U and x, y ∈ V . Therefore x + y ∈ U and x + y ∈ V
since U and V are subspaces. Hence x + y ∈ U ∩ V . Finally, suppose
that x ∈ U ∩ V and c ∈ R. Then cx ∈ U and cx ∈ V since U and V
are subspaces. Therefore cx ∈ U ∩ V . Hence U ∩ V is a subspace.
73. Consider the following two subsets of R4 :
U = {(x, x + 1, y + z, 2x)| x, y, z ∈ R}
V = {(x, x + y + 1, z, 2x)| x, y, z ∈ R}
Although these sets look similar, one is a subspace and one is not.
(a) Explain why U is not a subspace of R4 .
(b) Show that V = span{v1 , v2 , v3 }, where v1 = (1, 1, 0, 2), v2 =
(0, 1, 0, 0) and v3 = (0, 0, 1, 0). Deduce that V is a subspace of
R4 .
Solution: (a) U is not a subspace because it does not contain the
zero vector 0: the system of equations x = 0, x + 1 = 0, y + z = 0,
2x = 0 is not consistent, therefore 0 ∈
/ U.
(b) We have
(x, x + y + 1, z, 2x) = x(1, 1, 0, 2) + (y + 1)(0, 1, 0, 0) + z(0, 0, 1, 0).
Therefore any vector in V is a linear combination of v1 , v2 and v3 .
On the other hand, as x, y and z range over all real numbers, and
so (x, x + y + 1, z, 2x) ranges over all the vectors in V , we get every
such linear combination (in more detail, r(1, 1, 0, 2) + s(0, 1, 0, 0) +
43
t(0, 0, 1, 0) = (r, r + (s − 1) + 1, t, 2r) ∈ V ). Therefore V is the set
of all linear combinations of v1 , v2 and v3 , i.e. V = span{v1 , v2 , v3 }.
Hence V is a subspace since the span of a set of vectors is always a
subspace.
Basis
74. What constraints must be imposed on a, b ∈ R, so that
a
0
,
is
0
b
a basis for R2 .
Solution:
We require that the two vectors are linearly
independent. Clearly this
x
will be the case if a, b 6= 0. Any vector
∈ R2 can now be formed
y
from the linear combination
y 0
x a
x
,
+
=
y
a 0
b b
which is well defined if a, b 6= 0 — the same restriction.
 

x




 

y


75. Find a basis for W =   x − 2y + w = 0, 3x − z + 3w = 0 .
z






w
Solution:
W is the solution space of the two linear homogenous equations x −
2y + w = 0 and 3x − z + 3w = 0. (Boxes in the following designate
where we will pivot.)
0
1 −2
3
0
−1
1 −2
0
0 6
−1
1
0 −1/3
0
1 −1/6
1
3
1
0
1
0
0
0
0
0
0
0
The solution is of the form
  

 
 
s/3 − t
1/3
−1
x
 y   s/6 
1/6
0
 =

 
 
z   s  = s  1  + t  0 
w
t
0
1
44
The expression above shows that the two vectors span W , and the two
vectors are linearly independent
they
 are not multiples of each
 because
  
1
−1
/
3




1   
0
/
6
,  .
other. So a basis for W is 
 1   0 





1
0
76. Find a basis for the subspace U = {x ∈ R4 | Ax = 0} where


1 1 1 1
A = 2 0 2 0 .
3 1 3 1
Solution:


1 0 1 0
Row reducing A gives the matrix 0 1 0 1. Let x3 = s and x4 = t
0 0 0 0
be the free variables. Then x1 = −s and x2 = −t so the solutions are
of the form
   
 
 
x1
−s
−1
0
x2   −t 
0
−1
  
 
 
x=
x3  =  s  = s  1  + t  0 
x4
t
0
1
for s, t ∈ R. Thus we see that U is spanned by
   
−1
0 


   

0
−1
 ,  .
 1   0 





0
1
Since these two vectors are not multiples of each other they are linearly
independent, hence they are a basis for U .
77. Show that if {u, v} is a basis for a subspace W , then so is {u + v, v}.
Solution: Let x ∈ W , then x = c1 u + c2 v for some c1 , c2 ∈ R, since
{u, v} spans W . We have
x = c1 u + c2 v = c1 (u + v) + (c2 − c1 )v
so x ∈ span{u + v, v}. That is, still spans W .
To see that {u+v, v} is linearly independent, consider c1 (u+v)+c2 v =
0, so c1 u + (c1 + c2 )v = 0, but since {u, v} is linearly independent
this implies that c1 = 0 and c1 + c2 = 0, so c1 = c2 = 0 and hence
{u + v, v} is linearly independent.
45
78.* Let v1 , . . . , vr be a sequence of vectors in a subspace W of Rn with
the property that every vector in W can be written uniquely as a linear
combination of the vectors v1 , . . . , vr . Prove that the vectors v1 , . . . ,
vr form a basis for W .
Solution:
To show that the sequence of vectors v1 , . . . , vr is a
basis for W we need to show that (i) it spans W and (ii) it is linearly independent. We are told that every vector in W is a linear
combination of v1 , . . . , vr , therefore the sequence of vectors spans
W . We need to show that it is linearly independent. Suppose that
c1 v1 + · · · + cr vr = 0. We need to show that c1 = · · · = cr = 0.
We’re told that every vector in W can be written uniquely as a linear
combination of the vectors v1 , . . . , vr . Well, 0 is a vector in W (W is
a subspace) and certainly 0 = 0v1 + · · · + 0vr . But this must be the
only way in which we can write 0 as a linear combination of v1 , . . . ,
vr . Therefore we must have c1 = 0, . . . , cr = 0. Hence the vectors v1 ,
. . . , vr are linearly independent and form a basis for W .
79. Explain why the following are not bases for the indicated subspaces.
x
−6
6
x + 6y = 0 ;
for V =
,
(a)
y
1
−1

 
 

 x
 4 
(b) 1 for W = y  x − 2y − 2z = 0 ;




z
1
   
2 
 4
(c) 1 ,  0  for W as in part (b).


1
−1
     
5 
4
 1
(d) 4 , 2 , 6 for R3 .


1
3
2
Solution:
(a)
(b)
(c)
(d)
6
−6
0
The vectors are linearly dependent:
+
=
.
−1
1
0
The equation is clearly that of a plane, and so the space W is 2
dimensional, and hence should have 2 basis vectors.
 
2

The vector 0  doesn’t satisfy the equation, so isn’t even in
−1
W
.
     
5
1
4
6 = 4 + 2 so these vectors are linearly dependent.
3
2
1
46
5. Eigenvalues and Eigenvetors
Eigenvalue Problem
80. Let A be an n × n matrix.
(a) Define what it means for x ∈ Rn to be an eigenvector of A with
eigenvalue λ.
(b) Define the characteristic polynomial of A.
Solution:
(a) An eigenvector of A with eigenvalue λ is a vector x 6= 0 satisfying
Ax = λx.
(b) The characteristic polynomial of A is |λI − A|.
81. Find all eigenvalues and eigenvectors for the following matrices.
0
3
(a) A1 =
6 −3


2 1 0
(b) A2 =  1 3 1 
0 1 2


1 5 5
(c) A3 =  0 3 2 
0 2 3
Solution:
(a) First find the characteristic polynomial of A1 :
|λI − A1 | =
λ
−3
−6 λ + 3
=λ(λ + 3) − 18
=λ2 + 3λ − 18
=(λ + 6)(λ − 3)
So the two eigenvalues of A2 are λ = −6, 3. To compute the
eigenspaces, solve (λI − A1 )x = 0.
λ = −6
-6
-6
1
0
-3
-3
1/2
0
47
λ=3
0
0
0
0
3
-6
1
0
-3
6
-1
0
0
0
0
0
−1/2
= s
s ∈ R and E3 =
1
So the eigenspaces are: E−6
1
s
s∈R ,
1
(b) The characteristic polynomial is:
|λI − A2 | =
λ−2
−1
0
−1 λ − 3
−1
0
−1 λ − 2
−1 −1
λ − 3 −1
+1
0 λ−2
−1 λ − 2
= (λ − 2) [(λ − 3)(λ − 2) − 1] − (λ − 2)
= (λ − 2)
= (λ − 2) [(λ − 3)(λ − 2) − 2]
= (λ − 2) λ2 − 5λ + 4
= (λ − 2)(λ − 1)(λ − 4)
So the three eigenvalues of A2
• λ = 2:
0
−1
0
-1
0
0
1
0
0
1
0
0
are λ = 2, 1, 4.
−1
0
−1 −1
−1
0
−1 −1
−1
0
−1
0
1
1
0
-1
−1
0
0
1
1
0
0
0
  
 −1
So the eigenspace is E2 = s  0 

1
• λ = 1:
−1 −1
0
−1 −2 −1
0 −1 −1
1
0 −1
0
1
1
0
0
0
  
1

So the eigenspace is E1 = s −1

1
• λ = 4:
2 −1
0
−1
1 −1
0 −1
2
1
0 −1
0
1 −2
0
0
0
48
0
0
0
0
0
0
0
0
0
0
0
0


s∈R .

0
0
0
0
0
0


s∈R .

0
0
0
0
0
0
  

 1

So the eigenspace is E4 = s 2 s ∈ R .


1
(c) Using the first column to compute the determinant:
λ−1
−5
−5
0 λ−3
−2
0
−2 λ − 3
|λI − A3 | =
λ−3
−2
−2 λ − 3
= (λ − 1) [(λ − 3)(λ − 3) − 4]
= (λ − 1) λ2 − 6λ + 5
= (λ − 1)
= (λ − 1)(λ − 1)(λ − 5)
So the three eigenvalues of A3 are λ = 1, 1, 5.
• λ = 1:
0 -5 −5 0
0 −2 −2 0
0 −2 −2 0
0
1
1 0
0
0
0 0
0
0
0 0
  

 
0
 1

Eigenspace is E1 = s 0 + t −1 s, t ∈ R .


0
1
• λ = 5:
−5
−5 0
4
0
2
−2 0
0
−2
2 0
1 −5/4 −5/4 0
0
2
−2 0
0
−2
2 0
1
0 −10/4 0
0
1
−1 0
0
0
0 0
  

 10/4

So the eigenspace is E5 = s  1  s ∈ R .


1
82. Show that if A has eigenvalues λ1 , λ2 , . . . , λn then
(a) cA has eigenvalues cλ1 , cλ2 , . . . , cλn for any constant c ∈ R.
(b) If A−1 exists, then A−1 has eigenvalues 1/λ1 , 1/λ2 , . . . , 1/λn
m
m
(c) Am has eigenvalues λm
1 , λ2 , . . . , λn for m = 1, 2, 3, . . ..
Solution:
49
(a) If 0 = |λi I − A| then
0 = cn |λi I − A| = |c(λi I − A)| = |(cλi )I − (cA)|.
Hence cA has eigenvalues cλi , i = 1, 2, . . . , n.
(b) If 0 = |λi I − A| then λi 6= 0 (otherwise det A = 0 and so A−1
doesn’t exist). Hence
0 = |λi I − A| = |A|.|λi .A−1 − I| = (λi )n |A|.|A−1 −
since λi 6= 0. Hence 0 = | λ1i I − A−1 | so
for all i = 1, 2, . . . , n.
1
λi
1
I|
λi
is an eigenvalue of A−1
(c) Ax = λi x so A2 x = A(Ax) = A.λi x = λi (Ax) = λ2i x. Then, if
Ak x = λki x, we get
Ak+1 x = A.Ak x = A.λki x = λki .(Ax) = λki .(λi x) = λk+1
x.
i
So by mathematical induction, if λi is an eigenvalue for A, then
m
m
Am x = λ m
i x i.e. A has eigenvalues λi for all i. Note they have
the same eigenvectors.
83. Show that if 0 < θ < π then the matrix
cos θ − sin θ
A=
sin θ cos θ
has no real eigenvalues. Give a geometric interpretation of this fact,
given that the vector Av is obtained from the vector v by a counterclockwise rotation about 0 through an angle θ.
Solution: The characteristic polynomial of A is given by the determinant |λI − A| and hence is equal to
λ − cos θ
sin θ
− sin θ λ − cos θ
= (λ − cos θ)2 + sin2 θ
= λ2 − 2λ cos θ + cos2 θ + sin2 θ
= λ2 − 2λ cos θ + 1
The discriminant of the quadratic λ2 − 2λ cos θ + 1 is 4 cos2 θ − 4 =
−4 sin2 θ < 0, since 0 < θ < π. Therefore the quadratic has no real
roots and hence A has no real eigenvalues.
Given that Av is obtained from v by a rotation about the origin, it is
clear that the only way that Av can be a multiple of v is if Av = v
(a rotation through 0 radians) or if Av = −v (a rotation through π
radians). Both of these possibilities are excluded by the constraint on
θ.
50
Properties of eigenvalues
84. For each of the following matrices:


5 0 0
A = 0 4 1 ,
0 1 4


3 1 1
B = 1 3 1 ;
1 1 3
(a) determine all eigenspaces, Eλ , and state whether the multiplicity
of λ equals the dimension of Eλ ;
(b) check that the determinant equals the product of the eigenvalues,
and the trace is their sum.
Solution:
(a) Characteristic polynomial:
|λI − A| =
λ−5
0
0
0
λ − 4 −1
0
−1 λ − 4
λ − 4 −1
−1 λ − 4
=(λ − 5) (λ − 4)2 − 1
=(λ − 5) λ2 − 8λ + 15
=(λ − 5)
=(λ − 5)(λ − 3)(λ − 5)
So the roots, and hence the eigenvalues are λ = 3, 5, and the
multiplicities are 1 and 2 respectively.
λ=3
λ=5
−2
0
0 0
0 −1 −1 0
0 −1 −1 0
1
0
0 0
0
1
1 0
0  0 0 0

0


E3 = s−1  s ∈ R


1
0
0
0 0
0
1 −1 0
0 −1
1 0
0
0
0 0
0
1 −1 0
0
0  0 0  


0
 1

E5 = s 0  + t 1  s, t ∈ R


0
1
The dimensions of E3 , and E5 are 1 and 2 respectively, so they
are equal to the multiplicity of the corresponding eigenvalues.
det(A) = 75 = 3 × 5 × 5,
51
tr(A) = 13 = 3 + 5 + 5
(b) Characteristic polynomial:
|λI − B| =
λ − 3 −1
−1
−1 λ − 3 −1
−1
−1 λ − 3
λ − 3 −1
−1 −1
−1 λ − 3
+1
−1
−1 λ − 3
−1 λ − 3
−1 −1
=(λ − 3) (λ − 3)2 − 1 − 2(λ − 2)
=(λ − 3)
=(λ − 3)(λ2 − 6λ + 8) − 2(λ − 2)
=(λ − 3)(λ − 2)(λ − 4) − 2(λ − 2)
=(λ − 2) [(λ − 3)(λ − 4) − 2]
=(λ − 2)(λ2 − 7λ + 10)
=(λ − 2)2 (λ − 5)
So the roots, and hence the eigenvalues are λ = 2, 5 with multiplicity 2 and 1 respectively.
λ=5
λ=2
−1 −1 −1 0
−1 −1 −1 0
−1 −1 −1 0
1
1
1 0
0
0
0 0
0
0
0 0
  
 
−1 
 −1
E2 = s  1  + t  0 


0
1
2
−1
−1 0
−1
2
−1 0
−1
−1
2 0
1 −1/2 −1/2 0
0
3/2 −3/2 0
0 −3/2
3/2 0
1
0
−1 0
0
1
−1 0
0  0  0 0
 1 
E5 = s  1 


1
The dimensions of E2 , and E5 are 2 and 1 respectively, so they
are equal to the multiplicity of the corresponding eigenvalues.
det(B) = 20 = 2 × 2 × 5,
tr(B) = 9 = 2 + 2 + 5
85. Answer the following true or false and give reasons.
(a) λ4 + λ2 − 1 cannot be the characteristic polynomial of a 3 × 3
matrix.
(b) An eigenvalue of multiplicity 7 can have an eigenspace of dimension 3.
(c) A real n × n matrix always has a real eigenvalue when n is odd.
(d) If tr(A) = 0 then A cannot be invertible.
(e) A polynomial with real coefficients must have real roots.
52
(f) If λ is an eigenvalue of an n × n matrix A with eigenvector x,
then for any scalar µ, λ − µ is an eigenvalue of A − µIn with x as
a corresponding eigenvector.
Solution:
(a) True: a 3 × 3 matrix can have at most 3 eigenvalues, whereas a
degree 4 polynomial can have 4 roots. Alternatively, if A is an
n × n matrix then its characteristic polynomial is of degree n,
hence the degree of the polynomial λ4 + λ2 − 1 is too large.
(b) True: the dimension of an eigenspace is equal to or less than its
multiplicity.
(c) True: The characteristic polynomial of an n × n matrix is of
degree n, and a polynomial of odd degree always has a real root
(complex roots of polynomials come in complex conjugate pairs).
0 1
0 1
1 0
(d) False: e.g.
=
.
1 0
1 0
0 1
(e) False: consider λ2 + 1 = 0, which has roots λ = ±i.
(f) True: if Ax = λx, then (A − µIn )x = λx − µx = (λ − µ)x;
therefore x is an eigenvector of A − µIn with eigenvalue λ − µ.
86. (a) Show that λ = 0 is an eigenvalue of A if and only if A is not
invertible.
(b) Without calculation find one eigenvalue and two linearly independent eigenvectors of


1 2 3
A = 1 2 3 .
1 2 3
Solution:
(a) The determinant of A is equal to the product of eigenvalues,
which is equal to zero if and only if one of the eigenvalues equals
zero. The matrix is invertible if and only if the determinant is
not equal to zero. Hence the result.
(b) The determinant of A is zero, so λ = 0 must be an eigenvalue.
The vectors




3
2
 0  and  −1 
−1
0
when pre-multiplied by A, take combinations of columns which
result in zero, and therefore give the zero vector, thus they are
eigenvectors corresponding to the eigenvalue λ = 0.
53
87. (a) Let A be an n × n matrix. Show that A and its transpose At have
the same characteristic polynomial and hence the same eigenvalues.
(b) Beware! Even though A and At have the same eigenvalues, the
eigenspace for A at an eigenvalue λ need not be equal to the
eigenspace for At at λ. Investigate this phenomenon with the
matrix
1 1
A=
0 1
Solution:
(a) The characteristic polynomial of A is det(λI − A).
The characteristic polynomial of At is det(λI − At ). The claim is that
these two polynomials are equal. Why? First we need to recall the
following property of the determinant: the determinant of a square
matrix is equal to the determinant of its transpose, i.e. det(B) =
det(B t ) for an n × n matrix B. Since the matrix I is equal to its own
transpose we have λI − At = (λI − A)t . Therefore det(λI − At ) =
det((λI −A)t ) = det(λI −A). Therefore the characteristic polynomials
of A and At are equal (and so the matrices have exactly the same
eigenvalues).
(b) The given matrix A has characteristic polynomial equal to (λ−1)2 .
Therefore A has one eigenvalue λ = 1 of multiplicity 2. Solving the
homogenous system (λI − A)x = 0 when λ = 1 we get
0
0
0
0
-1
0
1
0
0
0
0
0
1
and so we find that the λ = 1 eigenspace of A is span
. On the
0
other hand At has exactly the same characteristic polynomial, but if
we solve the homogenous system (λI − At )x = 0 when λ = 1 we get
0
-1
1
0
0
0
0
0
0
0
0
0
which shows that the λ = 1 eigenspace of At is span
88. Suppose that for some 3 × 3 matrix A, we have
   
   
1
2
1
2
A 1 = 2 and A 0 = 0
1
2
1
2
54
0
.
1
(a) Give one eigenvalue of A.
 
0

(b) Explain why 1 is an eigenvector for this eigenvalue and hence
0
 
0
find A 1.
0
(c) If det(A)=12, then what is the multiplicity of the eigenvalue from
(a)?
Solution:
(a) 2 is an eigenvalue of A.
(b) Since [1 1 1]t , [1 0 1]t ∈ E2 and E2 is a subspace then [0 1 0]t =
[1 1 1]t − [1 0 1]t ∈ E2 .
Hence A[0 1 0]t = [0 2 0]t .
(c) The multiplicity m of an eigenvalue is greater than or equal to
the dimension of the corresponding eigenspace. As the two eigenvectors are linearly independent, the dimension of E2 is at least
2, and so the multiplicity of the eigenvalue 2 is at least 2. As
A is a 3 × 3 matrix, the multiplicity of the eigenvalue 2 can be
at most 3. The product of the three eigenvalues is equal to the
determinant, so 2×2×λ3 = 12, and so λ3 = 3, and the eigenvalue
2 has multiplicity 2.
Diagonalisation
89. Consider the matrices
−12 7
A1 =
−7 2


4 2 2
A2 =  2 4 2 
2 2 4


−1
3
9
A3 =  0 −7 −18 .
0
2
5
For each matrix Ai above
(a) Determine the eigenvalues and eigenvectors.
(b) For each eigenvalue, state its multiplicity and give the dimension
of its associated eigenspace.
(c) Hence determine whether the matrix Ai is diagonalisable, stating
55
(d) For each diagonalisable matrix Ai , determine a matrix P such
that P −1 Ai P = D, where D is a diagonal matrix. What is D?
(For the purposes of this exercise, order your eigenvalues from
smallest to largest i.e. λ1 ≤ λ2 . . . ≤ λn .)
(e) Determine if each matrix Ai is invertible and, where possible, use
your results from (d) to write the inverse matrix A−1
i in the form
P ∆P −1 , where ∆ is a diagonal matrix. (You do not need to find
A−1 if not possible by this method.)
Solution:
A1 : (a) A1 has eigenvalue −5 with multiplicity 2. Eigenvectors are
x = t(1, 1), t 6= 0.
(b) E1 = span{(1, 1)}, dim(E1 ) = 1.
(c) A1 is not diagonalisable as it does not have two linearly independent eigenvectors.
(d) No answer required as A1 is not diagonalisable.
(e) det(A) = 25 6= 0 so A is invertible though not diagonalisable.
A2 : (a) A2 has eigenvalues 2 and 8 with multiplicities 2 and 1 respectively. For λ = 2 the eigenvectors are x = s(−1, 1, 0) +
t(−1, 0, 1), s and t not both zero. For λ = 8 the eigenvectors
are x = t(1, 1, 1), t 6= 0.
(b) E2 = span{(−1, 1, 0), (−1, 0, 1)}, dim(E2 ) = 2. E8 = span{(1, 1, 1)},
dim(E8 ) = 1.
(c) A2 is diagonalisable as it has 3 linearly independent eigenvectors.




2 0 0
−1 −1 1
0 1 and D = 0 2 0.
(d) P =  1
0 0 8
0
1 1


1/2
0
0
(e) det(A2 ) = 32 6= 0 so A2 is invertible. ∆ =  0 1/2 0 
0
0 1/8
with P as above.
A3 : (a) A3 has eigenvalue -1 with multiplicity 3. The eigenvectors
are x = t(1, 0, 0) + s(0, −3, 1), s and t not both zero.
(b) E3 = span{(1, 0, 0), (0, −3, 1)}, dim(E3 ) = 2.
(c) A3 is not diagonalisable as it does not have three linearly
independent eigenvectors.
(d) No answer required as A3 is not diagonalisable.
(e) det(A3 ) = −1 6= 0 so A3 is invertible though not diagonalisable.
2 3
90. Let A =
.
4 1
56
1
(a) Verify that v1 =
is an eigenvector of A. What is the corre1
sponding eigenvalue λ1 ?
(b) Use the trace of A to find a second eigenvalue λ2 .
(c) Find the eigenspace for λ2 .
(d) Write down an invertible matrix P such that P −1 AP = D, where
D = diag(λ1 , λ2 ).
Solution:
2 3 1
5
1
(a)
=
= 5
. Thus (1, 1)t is an eigenvector of A
4 1 1
5
1
corresponding to the eigenvalue 5.
(b) A is a 2 × 2 matrix so it has two eigenvalues. We know λ1 = 5.
But tr(A) = 2 + 1 = 3 = λ1 + λ2 so λ2 = −2.
−4 −3
(c) −2I − A =
. Solving (−2I − A)x = 0 gives x =
−4 −3
s[−3/4 1]t . Thus E−2 = span{[−3/4 1]t }.
5 0
1 −3/4
(d) D =
and P =
.
0 −2
1
1


3 0 0
91. Show that the matrix A = 0 2 0 is not diagonalizable.
0 1 2
Solution:
The characteristic polynomial of A is
λ−3
0
0
0
λ−2
0
0
−1 λ − 2
= (λ − 3)(λ − 2)2
and so A has eigenvalues λ = 3 (multiplicity 1) and λ = 2 (multiplicity
2). By Theorem 5.6, A is diagonalizable if and only if it has 3 linearly
independent eigenvectors v1 , v2 and v3 . Since λ = 3 has multiplicity
1 the λ = 3 eigenspace has dimension 1 (since the dimension of an
eigenspace is always ≤ to the multiplicity of the corresponding eiegenvalue). So if the linearly independent vectors v1 , v2 and v3 exist then
only one of them could be an eigenvector for λ = 3. So the only way
A can be diagonalizable is if the λ = 2 eigenspace has 2 linearly independent eigenvectors and hence is of dimension 2. To find the λ = 2
eigenspace E2 we solve (λI − A)x = 0 when λ = 2:
-1
0
0
1
0
0
0
0
-1
0
1
0
57
0
0
0
0
0
0
0
0
0
0
0
0
 
0


0 has dimension 1. Therefore A is not diagHence E2 = span
1
onalizable.
92.*
(a) A square matrix A is called nilpotent if Ak = 0 for some integer
k > 1. Show that 0 is the only eigenvalue of a nilpotent matrix.
(b) A square matrix A is called idempotent if A2 = A. Show that
the only possible eigenvalues of A are 0 or 1.
(c) Suppose that A is a diagonalizable matrix such that every eigenvalue of A is 0 or 1. Prove that A is idempotent.
Solution:
(a) Let Ax = λx. Note that, since x is an eigenvector, x 6= 0.
Observe that A2 x = Aλx = λ2 x. More generally (you can prove
this by induction), Ak x = λk x. Since Ak = 0, this implies λk x =
0, and hence (given x 6= 0) we can see that λk = 0 and hence
λ = 0.
(b) Let Ax = λx. As before A2 x = λ2 x, but as A is idempotent
A2 = A, and so A2 x = λx. Thus λ2 x = λx, so (λ2 − λ)x = 0,
and as before x 6= 0 so (λ2 − λ) = 0, which can be rewritten
λ(λ − 1) = 0. Thus the possible values of λ are 0 or 1.
(c) Since A is diagonalizable there is an invertible matrix P such
that P −1 AP = D, where D is a diagonal matrix whose diagonal
entries are all either 0 or 1 (since the eigenvalues of A are all either
equal to 0 or 1). Observe that D2 = D (since 02 = 0 and 12 = 1).
We have (P −1 AP )2 = (P −1 AP )(P −1 AP ) = P −1 A2 P and so
P −1 A2 P = P −1 AP . Therefore A2 = A, i.e. A is idempotent.
93. Let s be a real number, and let

s
−1
s−1
0
1
1
A=
0
1
1
2
2
0 s −s s −s

0
0
.
0
2
(a) For which values of s is A diagonalisable?
(b) For s = 0, find an invertible matrix P such that P −1 AP is a
diagonal matrix.
58
Solution:
(a) First, find the eigenvalues of A. The characteristic polynomial is


λ−s
1
1−s
0
 0
λ−1
−1
0 

det(λI − A) = det 
 0
−1
λ−1
0 
0
s − s2 s − s2 λ − 2


λ−1
−1
0
λ−1
0 
= (λ − s) det  −1
2
2
s−s s−s λ−2
λ − 1 −1
= (λ − s)(λ − 2) det
−1 λ − 1
= (λ − s)(λ − 2) (λ − 1)2 − 1
= (λ − s)(λ − 2)2 λ.
So the eigenvalues of A are 2, 0 and s.
For A to be diagonalisable, there must exist four linear independent eigenvectors of A (Theorem 5.6 in the lecture). Since the
eigenvalue 2 has multiplicity > 1 (2 if s 6= 2 or 3 if s = 2), it is
sensible to first find out which dimension the eigenspace E2 has.
Remark: If dim E2 is less than the multplicity of the eigenvalue 2,
then we know we cannot find four linearly independent eigenvectors, since the combined dimension of the eigenspaces is at most
3.
So compute E2 by solving (2I − A)x = zero. Apply row reductions:



2−s 0
2−s
2−s
1
1−s 0

 0
 0
1
−1
1
−1
0


(2I − A) = 

 0

0
0 2(s − s2 )
−1
1
0
2
2
0
0
0
0
s−s s−s 0
If s 6= 0, 1, 2, then this linear system has only one free parameter,
hence dim E2 = 1. So A is clearly not diagonalisable if s 6= 0, 1, 2.
Now check if A is diagonalisable if s either 0, 1 or 2:
s = 2: Continue the row reduction above for s = 2:




0 0 0 0
0 1 0 0
0 1 −1 0
0 0 1 0




0 0 −4 0
0 0 0 0 .
0 0 0 0
0 0 0 0
So the eigenspace has dimension dim E2 = 2. However, for
s = 2, the multiplicity of the eigenvalue 2 is 3, so dim E2 is
less than the multiplicity, and thus A is not diagonalisable
for s = 2.
s = 1: For s = 1, the row reduction is already complete:


1 0 1 0
0 1 −1 0


0 0 0 0 .
0 0 0 0
59

0
0
.
0
0
The solution is

  
−1
0
n 1  0o
  
E2 = span 
 1  , 0 .
0
1
So the eigenspace E2 has dimension dim E2 = 2, which equals
the multiplicity of the eigenvalue 2. The other eigenvalues
are 0 and s = 1, and both eigenspaces E0 and E1 are at least
one-dimensional, so in total we find four linearly independent
eigenvectors by picking an eigenvector from E0 and E1 each
and combining them with the two basis vectors of E2 .
Hence A is diagonalisable for s = 1.
s = 0: Continue the row reduction above for s = 0:




1 0 1 0
2 0 2 0
0 1 −1 0
0 1 −1 0




0 0 0 0 .
0 0 0 0 
0 0 0 0
0 0 0 0
The solution is, as in the previous case,
   
0
−1
n 1  0o
  
E2 = span 
 1  , 0 .
1
0
So the eigenspace E2 has dimension dim E2 = 2, which equals
the multiplicity of the eigenvalue 2. But now the eigenvalue
0 = s also has multiplicity 2, so we need to determine the
dimension of the eigenspace E0 in order to find out if we can
obtain four linearly indepdenent eigenvectors. Solve (0I −
A)x = zero:




0 1 1 0
0 1
1
0
0 0 0 1
0 −1 −1 0 




0 0 0 0 .
0 −1 −1 0 
0 0 0 0
0 0
0 −2
The solution is,
   
1
0
n0 −1o
  
E0 = span 
0 ,  1  .
0
0
So E0 is indeed of dimension dim E0 = 2, and we can obtain
four linearly independent eigenvectors for A by choosing two
basis vectors from E0 and combining them with two basis
vectors from E2 .
Hence A is diagonalisable for s = 0.
60
(b) By part (a) we know that A is diagonalisable for s = 0, and we
can find a matrix P by choosing its columns to be four linearly
independent eigenvectors of A (Theorem 5.7 in the lecture). Use
the basis vectors for E0 and E2 from part (a):


1 0 −1 0
0 −1 1 0
.
P = v 1 v2 v3 v4 = 
0 1
1 0
0 0
0 1
Then we know that

0

0
P −1 AP = 
0
0
0
0
0
0
0
0
2
0

0
0
 = diag(0, 0, 2, 2).
0
2
Cayley-Hamilton Theorem
94. Use the Cayley-Hamilton theorem to compute the inverse of the matrix


1 2 5
A = 0 2 −1 .
0 0 1
Solution:
Expanding around the bottom row, the characteristic polynomial for
A is
|λI − A| = (λ − 1)
λ − 1 −2
= (λ − 1)2 (λ − 2) = λ3 − 4λ2 + 5λ − 2.
0
λ−2
By the Cayley-Hamilton Theorem
A3 − 4A2 + 5A − 2I = O
and therefore
2A−1 = A2 − 4A + 5I

 
 

1 6 8
4 8 20
5 0 0
= 0 4 −3 − 0 8 −4 + 0 5 0
0 0 1
0 0 4
0 0 5


2 −2 −12

1 .
= 0 1
0 0
2
Finally

A−1

1 −1 −6
= 0 1/2 1/2  .
0 0
1
61


1 1 0
95. Let A = 1 0 1. Use the Cayley-Hamilton theorem to compute
0 1 1
3
4
A and A .
Solution:
|λI − A| =
First we find the characteristic equation.
λ − 1 −1
0
−1
λ
−1
0
−1 λ − 1
= (λ − 1)
λ
−1
−1 −1
+
−1 λ − 1
0 λ−1
= (λ − 1)[λ(λ − 1) − 1] − (λ − 1)
= (λ − 1)[λ(λ − 1) − 2]
= λ3 − 2λ2 − λ + 2
Therefore A3 − 2A2 − A + 2I =
calculating

 
4 2 2
1
2A2 + A − 2I = 2 4 2 + 1
2 2 4
0
0 and hence A3 = 2A2 + A − 2I,
 
 

1 0
2 0 0
3 3 2
0 1 − 0 2 0 = 3 2 3
1 1
0 0 2
2 3 3
and then A4 = A(A3 ) = A(2A2 + A − 2I) = 2A3 + A2 − 2A,

 
 
 
6 6 4
2 1 1
2 2 0
6 5
A4 = 6 4 6 + 1 2 1 − 2 0 2 = 5 6
4 6 6
1 1 2
0 2 2
5 5
so

5
5
6
Dynamical Systems
−5 9
96. Consider the dynamical system with xk+1 = Axk , where A =
.
−3 11
2
Show that one eigenvector of A is (2, 1)T . What is the corresponding
eigenvalue? Hence determine the remaining eigenvalue and its eigenvectors.
What happens to the system over a long period of time?
−5
9
2
−1
2
Solution:
=
= (−1/2)
so
−3 11/2
1
−1/2
1
λ1 = −1/2 for v1 = (2, 1)T . Since λ1 + λ2 =tr(A)=1/2, we get λ2 = 1.
For λ2 = 1, eigenvector s are multiples of v2 = (3, 2)T .
Ak x0 gives
2
2
3
2
3
1 k
k
k
k
+ c2
= c1 λ1
+c2 λ2
= c1 (− 2 )
+
xk = A c1
1
2
1
2
1
62
c2
.1k .
3
2
So xk → c2
3
of
2
3
2
as k → ∞. Hence whatever x0 is, xk is a multiple
97.*
(a) A square matrix B is called nilpotent if B k = 0 for some integer
k > 1. Show that 0 is the only eigenvalue of a nilpotent matrix.
(b) A square matrix C is called idempotent if C 2 = C. What are the
possible eigenvalues of an idempotent matrix?
Solution:
(a) Let Bx = λx. Note that, since x is an eigenvector, x 6= 0.
Observe that B 2 x = Bλx = λ2 x. More generally (you can prove
this by induction), B k x = λk x. Since B k = 0, this implies λk x =
0, and hence (given x 6= 0) we can see that λk = 0 and hence
λ = 0.
(b) Let Cx = λx. As before C 2 x = λ2 x, but as C is idempotent
C 2 = C, and so C 2 x = λx. Thus λ2 x = λx, so (λ2 − λ)x = 0,
and as before x 6= 0 so (λ2 − λ) = 0, which can be rewritten
λ(λ − 1) = 0. Thus the possible values of λ are 0 or 1.
98. Find the steady-state probability vectors for the Markov processes
with transition matrices:


1/2 3/4 1/3
3/4 1/3
(a)
(b) 1/2 1/4 1/3
1/4 2/3
0
0 1/3
1−b
a
(c)
b
1−a
(with 0 < a, b < 1)
Solution:
For each of the probability matrices P we must solve
(I − P )x = 0.
1/4 −1/3 0
1 −4/3 0
4/3
=⇒
=⇒ x = t
.
(a)
−1/4 1/3 0
0
0
0
1
As we need a probability vector we choose t = 1/(4/3 + 1) = 3/7
and q = (4/7, 3/7)0 .
63
1/2 −3/4 −1/3 0
(b) −1/2 3/4 −1/3 0
0
0
2/3 0


3/2
x = t  1 .
0
1 −3/2 −2/3 0
0
0
1
0
0
0
0
0
=⇒
=⇒
 
3/5

Choose t = 1/(3/2 + 1) = 2/5 and q = 2/5.
0
b −a 0
1 −a/b 0
a/b
=⇒ x = t
.
(c)
=⇒
−b a 0
0
0
1
a/(a + b)
Choose t = 1/(a/b + 1) = b/(a + b) and q =
.
b/(a + b)
99. A country is divided into four regions (A, B, C and D). It is found
that each year 18% of the residents in A move to B, 12% move to C
and 20% move to D. Of the residents in B, 30% move to C, 10% move
to D and only 5% move to A. Of the residents in D, 50% move to
A, 25% move to C and 10% move to B; while the residents of C will
either stay where they are or move to A with equal probability. Write
down the transition matrix for this Markov process.
Solution:
Using the order A, B, C, D the transition matrix is


0.50 0.05 0.50 0.50
 0.18 0.55
0 0.10 

P =
 0.12 0.30 0.50 0.25  .
0.20 0.10
0 0.15
Note that the columns all sum to 1.
100.* Fibonacci numbers turn up in a huge variety of applications. They
are determined according to the following rules: F0 = 0; F1 = 1;
Fk+1 = Fk + Fk−1 ∀ k ≥ 1.
1 1
Fk+1
1
, k ≥ 0 and x0 =
Let A =
, xk =
1 0
Fk
0
Explain why the sequence {Fk : k = 0, 1, 2...}can be determined from
xk+1 = Axk . Hence show xk+1 = Ak x0 .
Show that Fk =
Find F1000 .
λk1 −λk2
√
5
Show that as k → ∞,
Solution:
where λ1 , λ2 are eigenvalues of A, (λ1 > λ2 ).
Fk+1
Fk
→
√
1+ 5
2
Eigenvalues are λ1 =
(Golden Ratio) .
√
1+ 5
2 ,
λ2 =
√
1− 5
2 .
The corresponding eigenvectors are multiples of v1 =
64
λ1
, v2 =
1
λ2
.
1
Then x0 =
Hence xk =
1
0
=
Fk+1
Fk
Therefore Fk =
√1 v1
5
√1 (λk
5 1
Fk+1
=
Fk
=
−
√1 v2
5
√1 λk
5 1
λ1
1
so xk =
−
√1 λk
5 2
√1 Ak (v1
5
− v2 ).
λ2
.
1
− λk2 ) and
√1 (λk+1
5 1
√1 (λk
5 1
− λk+1
2 )
− λk2 )
as k → ∞, since |λ2 /λ1 | < 1.
65
=
λ1 − λ2 (λ2 /λ1 )k
→ λ1
1 − (λ2 /λ1 )k
```
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