CHAPTER
13
SOLUTIONS
TO
EXERCISES
IN
MATRICES
AND
DETERMINANTS
Exercise on 13.2
Write the following system of equations in matrix form Ax = b. In each case, specify
the location of the coefficient – 3 in the matrix A in the form aij
i)
x – 3y = 1
2x + 4y = 4
ii) 3x + 2y – z = 1
–x – 3y + 2z = 1
x+ y+ z=2
Solution
i) For the system of equations
x – 3y = 1
2x + 4y = 4
we introduce the matrices of coefficients, variables and right hand sides in
the form
 1 –3 
A =  2 4 


 1 
b =  4 


 x 
x =  y 


and then the equations can be written in the form
 1
 2

–3   x 
 1 
4   y  =  4 
In this case the – 3 is in the first row and second column, so
a12 = – 3
ii) For the equations
3x + 2y – z = 1
–x – 3y + 2z = 1
x+ y+ z=2
we introduce the matrices
–1–
 3 2 –1 
A =  –1 –3 2 
 1 1 1 
 x 
x =  y 
 z 
 1 
and b =  1 
 2 
for the coefficients, unknowns and right hand sides and then the system can
be written in the form
 3 2 –1   x
Ax =  –1 –3 2   y
 1 1 1   z




 1 
=  1 
 2 
=b
In this case the –3 is in the second row and second column and so
a22 = – 3
Exercises on 13.3
1.
If
3 
X+Y 
 x+y
 2

 = 

1 
 –1 x – y 
 Y–X
find x, y, X, Y.
Solution
Two matrices are equal if and only if their corresponding elements are equal.
In this case this gives four equations
x+y=2
x–y=1
which give (check!)
3
x=2
1
and y = 2
and
X+Y=3
X–Y=1
which give
X = 2 and Y = 1
–2–
2.
For the matrices
 –1
A =  2
 3
evaluate
i)
iv)
0 1
1 0
2 –1
 4
B =  –2
 3




A + B ii) A – B
AB
v) BA
2
4
2
3
1
1
iii)




3A + 2B
vi)
A2
Answers
i)
 –1
A + B =  2
 3
0 1
1 0
2 –1

 4
 +  –2



 3
 3
=  0
 6
2
4
2
3
1
1
2 4
5 1
4 0








on adding corresponding elements.
 –1
ii) A – B =  2
 3
0 1
1 0
2 –1

 4
 –  –2



 3
 –5
=  4
 0
iii)
 –1
3A + 2B = 3  2
 3
 –3
=  6
 9
0 1
1 0
2 –1
2
4
2
3
1
1




–2 –2 
–3 –1 

0 –2 

 4
 + 2  –2



 3
0 3
3 0
6 –3

 8
 +  –2



 6
 5
=  2
 15
4 9
11 2
10 –1
–3–
2
4
2
4
8
4




3
1
1




6
2
2




iv)
 –1
AB =  2
 3
0 1  4
1 0   –2

2 –1   3
2
4
2
3
1
1




 –1
 6

 5
=
0 –2 
8
7 

12 10 
We will give one calculation of the product elements as an example, and you
can then check the rest for yourself. Specifically, we calculate the element 6 in
the second row and first column of the product. This comes from plugging
the second row of the first factor, A, into the first column of the second factor
B to give:
2(4) + 1(– 2) + 0(1) = 8 – 2 + 0 = 6
as stated.
 4
v) BA =  –2
 3
vi)
2
4
2
 –1
A =  2
 3
2
3
1
1




0 1
1 0
2 –1
 –1
 2

 3




 –1
 2

 3
0 1
1 0
2 –1

 9
 =  13


 4

0 1
1 0
2 –1
8 1 
6 –3 

4 2 

 4
 =  0


 –2

2 –2 
1 2 

0 4 
Exercises on 13.4
1.
 3

 –1
Evaluate i)
2 

1 
 3
ii)  1

3 

1 
Solution
i)
 3

 –1
 3
ii)  1

2 

1  = 3  1 – (– 1)  2 = 3 + 2 = 5
3 

1  =31–13=3–3=0
Or, alternatively, simply note that the two rows are proportional, or the two
columns are identical
2.
Expand by i) first row ii) second column
 3
 1

 2
2
0
4
0
7
1




–4–
iii) second row
Write down all minors and all cofactors of the determinant.
Solution
i) By the first row we have
 3
 1

 2
2
0
4
0
7
1

 = 3  0

 4

7 
 1


1  – 2 2
7 
 1


1  +0 2
0 

4 
= 3 (0 – 28) – 2(1 – 14) + 0 = – 84 + 26 = – 58
ii) By the second column we have
 3
 1

 2
2
0
4
0
7
1

 = – 2  1

 2

7 
 3
 + 0
1 
 2
0 
 3
 –4
1 
 1
0 

7 
= – 2(– 13) – 4 (21) = – 58
By the second row we have
 3
 1

 2
2
0
4
0
7
1

 = – 1  2

 4

0 
 3
 + 0
1 
 2
0 
 3
 –7
1 
 2
2 

4 
= –2 – 56 = – 58
Note that these three values are the same, as they should be - the value of a
determinant does not depend on how it is evaluated.
 9
3. Simplify and evaluate  2
 3
10
3
3
10
–1
1




Solution
Referring to the properties of determinants given in this section we have
 9
 2

 3
10
3
3
10
–1
1

 9
 =  2



 3
10
3
3
0
–4
–2

 ( col 3 – col 2: property v)


–5–
 9
=  2
 3
1
1
0
0
–4
–2

 (col 2 – col 1)


 0
=  – 7
 3
1
1
0
0
–4
–2

 (col 1 – 9 col 2)


This is now easily expanded by the first row to give
 –7
=–1 3

–4 

– 2  = –(14 + 12) = – 26
Exercise on 13.5
Solve the system of equations where possible
i)
3x – 2y + z = 1
x + y + z = 0
2x – y + 2z = 1
ii)
x + 2y – 3z = 0
2x – y + z = 1
x – 3y + 4z = 2
Solution
i) For the system
3x – 2y + z = 1
x + y + z = 0
2x – y + 2z = 1
the determinant of coefficients is
 3
 1

 2
–2
1
–1
1
1
2

 = 3  1

 –1

1 
 1


2  +2 2
1 
 1


2  +  2
1 

–1 
= 3(2 + 1) + 0 + (– 3) = 6  0
So the determinant of coefficients is not zero and we can proceed with the
solution, knowing that one exists. Using Cramer's rule we then have
1
1 
x=6  0
 1
–2
1
–1
0
1
2
 1
 =   1
 6  –1

–6–
1 
 –2
+ 
2 
 1
1 

1 
1
= 6 (3 + (– 3)) = 0
3
1 
y=6  1
 2
1
0
1
1
1
2
 1
 = –  1
 6   1

1   3
 
2 –  2
1 

1 
1
1
= 6 (– 1 – 1) ) = – 3
3 –2 1
1 
z=6  1 1 0
 2 –1 1
 1
 =
 6

  –2
– 
  –1
1 
 3
+ 
1 
 2
1 

1 
1
1
= 6 (1 + 1) ) = 3
So the solution is
x=0
1
y=–3
1
z=3
ii) For the system
x + 2y – 3z = 0
2x – y + z = 1
x – 3y + 4z = 2
the determinant of coefficients is
 1 2
 2 –1

 1 –3
–3
1
4

 = 1  – 1

 –3

1 
 2
 –2
4 
 1
1 
 2 –1 
 – 3

4 
 1 –3 
= 1(– 1) – 2(7) – 3(– 5) = 0
So there is no solution in this case - the equations are inconsistent
–7–
Exercise on 13.6
Where possible, solve the systems of Exercise 13.5 using the inverse matrix.
Solution
We know that ii) has no solution, so we concentrate on i). In matrix form the
system
3x – 2y + z = 1
x + y + z = 0
2x – y + 2z = 1
is
 3
Ax =  1
 2
–2
1
–1
1
1
2
 x 
 1
 y  =  0
 

 z 
 1

 =b


So
 3
x = A b =  1
 2
–2
1
–1
–1
3
1 
=6 3
 –3
0
4
–2
3
1 
= 6 0
 –3
0
1 
=6 –2
 2
1
1
2
–3
–1
5
3
4
–1

 =






–3
–2
5







T




0
1
–3
1
3
–1
 1 
 0 
 
 1 
 1 
 0 
 
 1 
 1 
 0 
 
 1 



So the solution is
x=0
1
y=–3
1
z=3
–8–
as before
Exercise on 13.7
Find the eigenvalues and corresponding eigenvectors for the matrix
 4 0 1 
A =  –2 1 0 
 –2 0 1 
Solution
The eigenvalues, , and eigenvectors, u,are determined by the condition that
the following equations should have non-trivial solutions.
 4 0 1   1
(A – I)u = –2 1 0  – 0
 –2 0 1   0
0
1
0
0
0
1

 u = 0


or
 4– 0
 –2 1– 

0
 –2
1
0
1–


u=0

There are non-trivial solutions only if
 4– 0
 –2 1– 

0
 –2
1
0
1–


 =0

Expanding the determinant gives the polynomial equation
3 – 62 + 11 – 6 = 0
Using the factor theorem this can be factorized to give (actually of course
since we are after the roots anyway the factor theorem is somewhat
superfluous!)
 – 1)( – 2)( – 3) = 0
giving eigenvalues
 = 1, 2, 3
–9–
To find the corresponding eigenvectors we must solve the system of
equations
 4 0 1   1
 –2 1 0  – 0

 
 –2 0 1   0
or
 4– 0
 –2 1– 

0
 –2
0
1
0
1
0
1–
0
0
1

 u = 0




u=0

for each of the values of .
 x 
 
For  = 1 we have, with u =  y 
 z 
 4– 0
 –2 1– 

0
 –2
1
0
1–
 x 
 3 0 1   x
 y 
 –2 0 0   y
  = 
 
 –2 0 0   z
 z 

 =0


This gives the system of equations
3x
– 2x
– 2x
+z=0
=0
=0
As we expect for a homogeneous system not all of the equations are
independent and in fact the last two are the same in this case, giving x= 0.
The first equation then gives z = 0. However, y is not determined and is
therefore arbitrary. So the most general solution for  = 1 is
x = 0 y = arbitrary = t and z = 0
We write this as an eigenvector
 0 
u =  t 
 0 
or, choosing t = 1
– 10 –
 0 
u =  1 
 0 
Repeating a similar process you can confirm that the other eigenvectors are
For  = 2
u =
 – 12 
 
 1 
 1 
For  = 3
 –1
u =  1
 1




Remember that each of the eigenvectors could be multiplied by a different arbitrary
constant.
REINFORCEMENT EXERCISES FOR MATRICES AND DETERMINANTS
1.
Express the following in matrix form
i)
iii)
2x – y = – 1
x + 2y = 0
ii)
3x – y + 2z = 1
2x + 2y + 3z = 2
– 3x
– z=–3
a + 2b + 3c = 1
iv)
4u – 2v = 1
a– b – c=3
3u + 3v = 2
u–v=–1
Solution
i) Using the rule for multiplying matrices, the system
2x – y = – 1
x + 2y = 0
can be written in the matrix form (always beware of the signs)
– 11 –
 2 –1   x 
 –1 
 1 2   y  =  0 

 



ii) The system
3x – y +2z = 1
2x + 2y + 3z = 2
– 3x – y = – 3
can be written in the form
 3 –1 2
 2 2 3

 –3 0 –1




 x 
 1
 y  =  2



 z 
 –3




iii) The system
a + 2b + 3c = 1
a–b–c=3
can be written in the form
 a 
 1 2 3   b 
 1 
 1 –1 –1  
 3 
=



 
 c 
iv) The system
4u – 2v = 1
3u + 3v = 2
u–v=–1
can be written in the form
 4 –2 
 1 
 3 3   u  =  2 

  v 


 1 –1 
 –1 
2.
 –1 2 3 
If A =  4 0 2 
 –1 1 2 
i)
 0 1 3 
B =  2 –1 4 
 3 –1 2 
Write down a12, a31, a33, b11, b21, b32
– 12 –
3
ii)
Evaluate
3

a1k bk3,
k=1
iii)
a3k bk2
k=1
Evaluate a) 2A – 3B
Solution
i)

b) A2
c) AB
d) BA
With
 –1 2 3

A=  4 0 2
 –1 1 2
 0 1 3

B =  2 –1 4
 3 –1 2








we have, for example, that a12 is the element in the first row and second
column of A and so
a12 = 2
Similarly we have, by inspection
a31 = –1, a33 = 2, b11= 0, b21 = 2, b32 = – 1
ii)
3

a1k bk3 = a11b13 + a12b23 + a13b33
k=1
= (– 1)(3) + (2)(4) + (3)(2) = – 3 + 8 + 6 = 11
3

a3k bk2 = a31b12 + a32b22 + a33b32
k=1
= (– 1)(1) + (1)(– 1) + (2)(– 1) = – 1 – 1 – 2 = – 4
iii) We have
 –1 2 3
2A – 3B = 2  4 0 2
 –1 1 2

 0 1 3
 – 3 2 –1 4



 3 –1 2
– 13 –




 –2 4 6 
 0


=  8 0 2  +  –6
 –2 2 4 
 –9
 –2 1 – 3
=  2 3 – 8
 – 11 5 – 2
 –1 2 3
A =  4 0 2
 –1 1 2
2
2
 =


–3 –9 
3 – 12 

3
–6 




 –1 2 3
 4 0 2

 –1 1 2




 –1 2 3
 4 0 2

 –1 1 2
1 7 
 6

=  – 6 10 16 
 3 0 3 
 –1 2 3
AB =  4 0 2
 –1 1 2




 0 1 3
 2 –1 4

 3 –1 2




 13 – 6 11 
=  6 2 16 
 8 –4 5 
 0 1 3
BA =  2 –1 4
 3 –1 2
  –1 2 3
 4 0 2

  –1 1 2
3 8 
 1

=  – 10 8 12 
 – 9 8 11 
3.
If
 x –1 y 
 3 a x 
 2 0 3  =  b 0 c 




 z 3 2 
 2 3 d 
evaluate a, b, c, d, x, y, z.
Solution
With
– 14 –








 x –1 y 
 3 a x 
 2 0 3  =  b 0 c 




 z 3 2 
 2 3 d 
equating corresponding elements gives
x = 3, – 1 = a, x = y, b = 2, c = 3, z = 2, 2 = d
or
x = y = 3, z = 2, a = – 1, b = 2, c = 3, d = 2
4.
If
 cos 

 – sin 
sin  
 =
cos  



1
2


3/2

3/2
1
–2
determine  in the first quadrant, ie 0 <  < 90°.
Solution
Equating corresponding elements in
 cos 

 – sin 
sin  
 =
cos  



1
2


3/2

3/2
1
–2
gives
3
cos  = 2
1
and sin  = 2

So, in the range 0 <  < 90° we have  = 6 = 30°
5.
Using the following matrices evaluate every possible sum and product of pairs
of the matrices (repetitions such as A2 allowed)
 2 –1 
A =  3 0 


 3 
B =  –1 
 4 
 0 1 –1 
D =  2 0 3 
 –1 –3 2 
 3 4 
E =  –1 2 
 0 3 
 1 1 1 
F =  2 2 2 


G = [2 1 0]
– 15 –
C = [ –2 0 ]
Solution
Additions
Since no two matrices are the same size the only possible additions of two,
with repetiton allowed, are of the form A + A = 2A, etc which can all be
obtained by scalar multiplication.
 2 –1 
 4 –2 
2A = 2  3 0  =  6 0 




 3 
 6


 –2
–1
2B = 2
 = 
 8
 4 




2C = 2 [ –2 0 ] = [ –4 0 ]
 0 1 –1 
 0 2 –2 


2D = 2  2 0 3  =  4 0 6 
 –2 –6 4 
 –1 –3 2 
 3 4 
 6 8 


2E = 2  –1 2  =  –2 4 
 0 6 
 0 3 
 1 1 1 
 2 2 2 
2F = 2  2 2 2  =  4 4 4 




2G = 2[ 2 1 0 ] = [ 4 2 0 ]
To give you lots of practice there is a long list of conformable products of
two matrices. They are given below.
 2 –1   1 1 1 
 0 0 0 
AF =  3 0   2 2 2  =  3 3 3 



 

 2 –1 
A2 =  3 0 


2
 2 –1   2 –1 
 1 –2 
=  3 0   3 0  =  6 –3 



 

 3 
BG =  –1  [ 2 1 0 ] =
 4 
3 0 
 6
 –2 –1 0 


4 0 
 8
– 16 –
 2 –1 
CA = [ –2 0 ]  3 0  = [ –4 2 ]


 1 1 1 
CF = [ –2 0 ]  2 2 2  = [ – 2 – 2 – 2 ]


 0 1 –1 
D =  2 0 3 
 –1 –3 2 
2
2
 0 1 –1   0 1 –1 
=  2 0 3   2 0 3 
 –1 –3 2   –1 –3 2 
3
 3

=  –3 –7
 –8 –7
1
4
2
 0 1 –1   3
DB =  2 0 3   –1
 –1 –3 2   4
 0 1 –1   3 4
DE =  2 0 3   –1 2
 –1 –3 2   0 3





 –5
 =  18


 8






 –1 –1
 =  6 17


 0 –4





 3 4 
 18 – 3 
2 –1 



EA =  –1 2   3 0  =  4 1 


 9 0 
 0 3 
 3 4 
 11 11 11 
 1 1 1 


 3 3 3 
–1
2
EF = 
  2 2 2  = 

 6 6 6 
 0 3 
 3 
 1 1 1   –1 
 6 
 12 
FB =  2 2 2  
=





 4 
 0 1 –1 
 1 1 1   2 0 3 
 1 –2 4 
FD =  2 2 2  
=  2 –4 8 





 –1 –3 2 
 3 4 
 1 1 1   –1 2 
 2 9 


 4 18 
FE = 2 2 2
=




 
 0 3 
– 17 –
 3 
GB = [ 2 1 0 ]  –1  = 5
 4 
 0 1 –1 
GD = [ 2 1 0 ]  2 0 3  = [ 4 2 1 ]
 –1 –3 2 
 3 4 
GE = [ 2 1 0 ]  –1 2  = [ 5 10 ]
 0 3 
6.
 2
A =  3
 –1
–1
0
1
 3
B =  0
 1




2
1
1
–1
2
1

 4
 C =  3



 0
2
1
–1




 2 1 4 
D =  3 0 –1 


 –2 
u =  1 
 –1 
 1 
v =  2 
 3 
Find, where possible, i) 3A + B
iii) 3D – 2A
iv) 2AD + 3B
vi) 2u + 3v – w
vii) u – 2w + Bv
 –1 
w =  3 
 2 
ii) 4A + 2C
v) 3u – 2v + B
Solution
i) 3A + B is not possible since A and B are of different size.
ii) A and C are the same size, so
 2
4A + 2C = 4  3
 –1
–1
0
1

 4
 +2 3



 0
=
 4
iii) 3C – 2A = 3  3
 0
2
1
–1


 –2




2
1
–1

 8
 =  12



 –4
0
 16
 18
2

 –4 2
2 –1 
3
0 

–1
1 
– 18 –
–4
0
4

 8
 +  6



 0
6
3
–3
  4
 – 6
 
  –2
4
2
–2








 12
=  9
 0
–2
0
2




 8 8
=  3 3
 2 –5




iv) AD is 3 by 3 and therefore the same size as B and so 2AD + 3B is possible
and we have
 2
2AD + 3B = 2 3
 –1
–1
0
1

  2 1 4  + 3
  3 0 –1 

9
 1 2

= 2  6 3 12
 1 –1 –5

 9
 +  0



 3
6
3
3
4
18
 2

12
6
24
= 
 2 – 2 – 10

 9
 +  0



 3
6
3
3
 3
 0

 1
–3
6
3
2
1
1
–1
2
1








–3
6
3




 11 10 15 
=  12 9 30 
 5 1 –7 
v) u, v, B are not all same size, so we cannot form a linear combination of
them.
vi) u, v, w are all the same size and we have
 –2 
 1 
 –1




2u + 3v – w = 2  1  + 3  2  –  3
 –1 
 3 
 2

 –4
 =  2



 –2
  3
 + 6
 
  9

 1
 +  –3



 –2
 0 
=  5 
 5 
vii) Bv is 3 by 1 and therefore the same size as u and w, so we have
 –2 
u – 2w + Bv =  1  – 2
 –1 
 –1 
 3
 3  +  0



 2 
 1
– 19 –
2
1
1
–1
2
1




 1 
 2 
 
 3 




 –2 
 2


=  1  +  – 6
 –1 
 –4
7.

 4 
 4 
 +  8  =  3 



 
 1 

 6 
A, B, C are the matrices
 3
A =  –1

1 

2 
 2
B =  3

–1 0 

1 2 
 1
C =  –2

1
2
2 

3 
Verify that A(B + C) = AB + AC. Is it true that (B + C)A = BA + CA?
Solution
 3
AB =  –1

 3
AC =  –1

1  2

2  3
–1 0 
 9


1 2  = 4
1   1
 
2   –2
1
2
–2 2 

3 4 
2 
 1
 = 
3 
 –5
5
3
9 

4 
So
 9
AB + AC =  4

–2 2 
 1
 + 
3 4 
 –5
 10
=  –1

3
6
5
3
9 

4 
11 
8 
On the other hand
 2
B+C= 3

–1 0 
 1
 +
1 2 
 –2
 3
= 1

0
3
1
2
2 

3 
2 

5 
So
 3
A(B + C) =  –1

1   3
 
2   1
0
3
2 
 10


5  =  –1
= AB + AC
as it should be.
– 20 –
3 11 

6 8 
However, it is not true that (B + C)A = BA + CA. Indeed none of (B + C)A,
BA or CA exist at all.
8.
Evaluate
i)
iii)
 2 3 


 –2 4 
 1 0 2
ii)  3 4 5
 5 6 7
 1 0 6 
 3 4 15 


 5 6 21 
 1 0 0 
iv)  2 3 5 
 4 1 3 
 0 2 3 
v) –2 0 4 
 –3 –4 0 
 cos 
vi) 
 –sin 




sin  

cos  
Solution
 2 3 
i)  –2 4  = (2)(4) – (–2)(3) = 8 + 6 = 14


ii) Expanding by the first row we have
 1 0 2
 3 4 5

 5 6 7

 = 1 4 5  + 0  3 5  + 2 3 4 

 6 7 
 5 7 
 5 6 

= 1(28 – 30) + 2(18 – 20) = – 2 – 4 = – 6
Of course we needn't have bothered to even write down the second
determinant in the expansion.
iii) Note that by taking a 3 out of the last column we have
 1 0 6
 3 4 15

 5 6 21

 1 0 2
 =3 3 4 5



 5 6 7
from ii).
iv) Again, expanding by the first row:
– 21 –

 = 3(– 6) = – 18


 1 0 0
 2 3 5

 4 1 3

 = 1 3 5  = 1(9 – 5) = 4

 1 3 

v) By the first row
 0 2 3 
 –2 0 4  = – 2 – 2


 –3
 –3 –4 0 
 cos 
vi) 
 –sin 
9.
4 
 –2 0 
 + 3
 = – 2(12) + 3(8) = 0
0 
 –3 –4 
sin  
 = cos2  + sin2  = 1
cos  
Write down the following expansions of the determinant
 3 –1 2 
|A| =  0 1 2 
 4 –1 2 
i)
ii)
iii)
iv)
By first row
By second row.
By 3rd column.
By last row.
and check that they all lead to the same result.
Solution
i) By the first row we have
 3 –1 2 
 0 1 2  = 3 1 2  – (– 1)  0 2  + 2 0 1 


 –1 2 
 4 4 
 4 –1 
 4 –1 2 
= 3(4) + (–8) + 2(– 4) = – 4
ii) By the second column (don't forget the order of the signs)
 3 –1 2 
 0 1 2  = – 0 – 1 2  + 1  3 2  – 2 3 – 1 


 –1 2 
 4 2 
 4 –1 
 4 –1 2 
– 22 –
= 0 + (6 – 8) – 2(– 3 + 4) = – 4
iii) By the 3rd column
 3 –1 2 
 0 1 2  = 2 0 1  – 2  3 – 1  + 2 3 – 1 


 4 –1 
 4 –1 
 0 1 
 4 –1 2 
= 2(– 4) – 2(– 3 + 4) + 2(3) = – 4
iv) By the last row we have
 3 –1 2 
 0 1 2  = 4 – 1 2  + 1  3 2  + 2 3 – 1 


 1 2 
 0 2 
 0 1 
 4 –1 2 
= 4(– 2 – 2) + 6 + 2(3) = – 4
10.
Invert the following matrices
i)
 2 3 


 1 4 
iv)
 2 3 4
 4 3 1

 1 2 4
 2 3 1
ii)  1 2 3
 3 1 2




 1
v)  3
 2
2
–1
0




0
4
6
iii)
 1 2 –1 
 –1 1 2 


 2 –1 1 




Solution
Remember that in each case it is as well to evaluate the determinant first, in case it
turns out to be zero, in which case we won't be able to evaluate the inverse of the
matrix
i)
 2 3 

 = 2(4) – 3(1) = 5, so the inverse exists, and
 1 4 
 2 3  –1 1  4 – 1 

 = 

5 –3 2 
 1 4 
T
– 23 –
1 4 –3 
=5 –1 2 


 2 3 1 
ii)  1 2 3  = 2(4 – 3) – 3(2 – 9) + 1(1 – 6) = 2(1) – 3(– 7) – 5 = 18  0, so the
 3 1 2 
inverse exists, and
 2 3 1
 1 2 3

 3 1 2
 –1 1
 =

18

 1 7 –5 T
 –5 1 7  = 1


18
 7 –5 1 
 1 –5 7
 7
1 –5

1
 –5 7
 1 2 –1 
iii) –1 1 2  = 3 – 2(– 5) – (– 1) = 3 + 10 + 1 = 14  0, so
 2 –1 1 
 1 2 –1  – 1 1  3 5 –1  T
 –1 1 2  =
 –1 3 5 



14 
 2 –1 1 
 5 –1 3 
3 –1 5
1 
3 –1
= 14  5
3
 –1 5
 2 3 4
iv)  4 3 1
 1 2 4





 = 2(10) – 3(15) + 4(5) = 20 – 45 = – 5  0


 2 3 4  –1
 10 – 15 5  T
 4 3 1  = 1  –4
4 –1 



–5
 1 2 4 
 – 9 14 – 6 
10 – 4 – 9
1 
–
= – 5  15 4 14
 5 –1 –6
 1
v)  3
 2
2
–1
0
0
4
6
 1
 3

 2
9

 –10 4
 = 1  15 – 4 – 14
 5 
6
 –5 1


 = – 26, and


2
0  –1
– 6 – 10 2
1 

–1 4
– 12
6
4
 = – 26 
0
6 
–4 –7
 8
– 24 –




T







– 6 – 12 8 
1 
6 –4 
= – 26  10

4 –7 
 2
6
12 – 8 
1 
–
10
–
6 4 
= 26 

 –2 –4 7 
11.
Solve the equations below by matrix inversion where possible
i)
x+
y+ z = 4
2x + 5y – 2z = 3
x + 7y – 7z = 5
iii)
ii) 2x + 3y – 4z = – 15
3x – 2y + 3z = 15
5x + 7y + 5z = – 6
2x + 3y – z = 5
x – y + 3z = 8
3x + 4y – 2z = 5
Solution
i) In matrix form the system is
 1 1 1
 2 5 –2

 1 7 –7




 x 
 4
 y  =  3



 z 
 5




with solution, if it exists:
 x 
 1 1 1 
 y  = 2 5 –2 




 z 
 1 7 –7 
–1
 4 
 3 


 5 
But
1 1 
 1
 2 5 – 2  = (– 35 + 14) – (– 14 + 2) = (14 – 5) = 0


 1 7 –7 
and so in this case no solution exists
ii) In matrix form the system is
 2 3 –4   x 
 – 15
 3 – 2 3   y  =  15

 


5   z 
 5 7
 –6
– 25 –




with solution, if it exists:
 x 
 2 3 –4 
 y  = 3 –2 3 




5 
 z 
 5 7
–1
 – 15
 15

 –6




You can entertain yourself by confirming that
 2 3 –4 
 3 – 2 3  = – 186


5 
 5 7
So a solution does exist, and specifically:
0
31
 x 
 – 31
 y  = – 1  – 43 30
1


186
z
1
–
18
–
13



– 31 – 43
1
1 
0
30
–
18
= – 186
1 – 13
 31




 – 15
 15

 –6




 – 15 
 15 


 –6 
T

 – 186
 = – 1  558

186 

 – 372
So the solution is
x = 1, y = – 3, z = 2
iii) In matrix form the system is
 2 3 –1   x 
 5
 1 –1 3   y  =  8

 


 3 4 –2   z 
 5




with solution:
 x 
 2 3 –1 
 y  = 1 –1 3 




 z 
 3 4 –2 
1
= 6

– 10 11
7
2
–1 1
8 –7 –5




– 26 –
T
–1
 5 
 8 


 5 
 5 
 8 


 5 

 1
 =  –3



 2




– 10 2
8 
1
= 6 11 – 1 – 7 
1 –5 
 7
 5 
 1 
 8  =  2 
 
 
 5 
 3 
So the solution is
x = 1, y = 2, z = 3
12.
Solve the equation
 x+1
 1–x

 x –1
6
–5
4
2
x–1
0

 = 0


Expanding by the first row we get
 x+1
 1–x

 x –1
6
–5
4
2
x–1
0

 = (x + 1)(– 4(x – 1)) + 6 (x – 1)2 + 2(4 – 4x + 5x – 5)


= –4(x – 1)(x + 1) + 6 (x – 1)2 + 2(x – 1)
= (x – 1)(– 4x – 4 + 6x – 6 + 2) = (x – 1)(2x – 8) = 0
So
x = 1 or 4
13.
Solve the following systems of equation by Cramer's rule – where possible.
i)
iii)
x+y=1
2x – y = 2
ii)
x+ y+ z=1
iv)
x + 2y + 3z = 0
x– y– z=0
4x – 12y = 3
11x – 2y = 1
x – y + 2z = 1
x + 2y – 3z = 0
2x + y – z = 2
Solution
i) For the system
x+y=1
2x – y = 2
The determinant of coefficents is
– 27 –
 1 1 

 =–1–2=–30
 2 –1 
So this system has a solution which is given by Cramer's rule as



x=



1 1 

2 –1  –1–2
1 1  = –3 = 1

2 –1 
and
 1 1 


 2 2 
y=
=0
–3
So the solution is
x = – 1, y = 0
ii) The system
4x – 12y = 3
11x – 2y = 1
has the solution



x=



3 – 12 

1 –2 
– 6 + 12
6
3
=
=
4 – 12 
– 8 +132
124 = 62

11 – 2 
 4 3 


4 – 33
29
 11 1 
y=
=
=
–
124
124
 4 – 12 


11
–
2


iii) The system
x+y+z=1
x + 2y + 3z = 0
x–y–z=0
has, by Cramer's rule, the solution
– 28 –




x=




1 1
1 
0 2
3 

0 –1 –1 
1
1 1
1  = 2
1 2
3 

1 –1 –1 
1 1
1
1 
3
y=2  1 0
1
–
1
–
1

1 1
1 
z=2 1 2
 1 –1
1
0
0

 = – 1 (– 4) = 2

2


 = 1  1 2  = 1 (– 3) = – 3

2
2 1 –1  2

iv) For the system
x – y + 2z = 1
x + 2y – 3z = 0
2x + y – z = 2
we find that the determinant of coefficients vanishes:
 1 –1 2
 1 2 –3

 2 1 –1

 =0


and so the system is inconsistent. We can see this directly from the equations
– the LHSs of the first two add to give the LHS of the third, but this is not the
case for the corresponding RHSs.
14.
Decide which of the following systems of equations have non-trivial solutions
i)
3x – 2y = 0
x+y =0
ii)
x + 4y = 0
2x + 8y = 0
x+y+z=0
x – y + 2z = 0
2x + y – 3z = 0
iv)
x+y+z=0
x–y–z=0
3x + y + z = 0
v)2x – 2y + z = 0
3x – y + z = 0
vi)
2x – y + z = 0
x+ y– z=0
iii)
x+y
=0
3x + 2y + 2z = 0
– 29 –
Solution
In each case we simply have to evaluate the determinant of coefficients - only if this
is zero will the homogeneous system have a non-trivial solution.
 3 –2 
i)  1 1  = 5  0 and so in this case we only have a trivial solution


x=y=0
 1
ii)  2

4 

8  = 0 and so non-trivial solutions exist.
In fact, the second equation is clearly twice the first and any values of x and y
satisfying x + 2y = 0 will satisfy both equations. So if s is any number then y
= s and x = – 2s will be a solution of the system.
1 
 1 1

1
–
1
2  = 1(3 – 2) – 1(– 3 – 4) + 1(1 + 2) = 11  0
iii) 

 2 1 –3 
so no non-trivial solutions in this case.
1 
 1 1

1
–
1
–
1  = 1(0) – (4) + (4) = 0
iv) 

1 
 3 1
so in this case there are non-trivial solutions
 2 –2
v)  3 – 1
 1 1
1
1
0

 = 0, so nontrival solutions exist


 2 –1 1 
vi)  1 1 – 1  = 2(2 + 2) + 1(2 + 3) + 1(2 – 3) = 8 + 5 – 1  0
2 
 3 2
and so no nontrivial solutions exist in this case
15.
Construct a system of three linear homogeneous equations in three unknowns
that has a non-trivial solution
Solution
– 30 –
We want a system of equations for which the determinant of coefficients
definitely vanishes. We can assure this by arranging it so that one row of the
set of coefficients is a linear combination of the other two (this follows
essentially from property v) of determinants). This we can do by taking our
first two equations to be any linear homogeneous equations in three
variables x, y, z and then taking the third equation to be a linear combination
of these. For example
x+y+z=0
x + 2y + 3z = 0
i)
ii)
2x + 3y + 4z = 0
16.
i) + ii)
Determine the eigenvalues of the following:
i)
iv)
vii)
 1 0 


 0 2 
ii)
 6 –3 


 2 1 


v)


2
0
–1



 0 2 0  viii) 



 –1 0 2 

 8 –4 


 2 2 
–1
0
3
1
2
–1
1
–2
–9
2
1
1
iii)
0
2
6
–1
1
0








 2 3 


 –1 –2 
vi)
 1 0 


 2 1 
ix)
 1 0 0 
 2 1 –2 


1 
 3 2
Solution
 1 0 
A =  0 2  is a diagonal matrix - it only has non-zero elements on the


'leading diagonal', and zeros everywhere else. In general, it is not difficult to
see that the eigenvalues of such a diagonal matricx will always be precisely
the diagonal elements. We will illustrate how this works with this example.
i)
The eigenvalues of A are given by the solutions of |A – I| = 0, ie
0 
 1–

 = (1 – )(2 – ) = 0
0 2– 

which clearly has solutions
 = 1, 2
which are the required eigenvalues.
– 31 –
ii) The eigenvalues of
 6 –3 

 are given by the roots of
 2 1 
 6– –3 

 = (6 – )(1 – ) + 6
2 1– 

= 2 – 7 + 12
= ( – 3)( – 4) = 0
So the eigenvalues are
 = 3, 4
iii)
 2 3 
For  –1 –2  we have


3 
 2–

 = – (2 – )(2 + ) + 3
 –1 –2– 
= 2 – 4 + 3 = 2 – 1
= ( – 1)( + 1) = 0
So the eigenvalues are
 = 1, – 1
iv)
 8 –4 
For  2 2 


 8– –4 

 = (8 – )(2 – ) + 8
2 2– 

= 2 – 10 + 24
= ( – 4)( – 6) = 0
So the eigenvalues are
– 32 –
 = 4, 6
 –1 1 0 
v) ForA =  0 –2 2  exactly the same approach applies, but now of
 3 –9 6 
course the calculations are (much!) more messy
1
0 
 –1–
2 
 –2–

0
–2–
2  = (– 1 – )
 –
|A – I| = 

 –9 6– 
3
–9
6– 

2 
 0

1
3
6
–
 

=– (1 + )(( + 2)( – 6) + 18) – (– 6)
= – ( + 1)(2 – 4 – 12 + 18) + 6
= – (3 – 42 + 6 + 2 – 4 + 6) + 6
= – 3 + 32 – 2 = 0
and so
(2 – 3 + 2) = ( – 1)( – 2) = 0
The eigenvalues in this case are therefore
 = 0, 1, 2
 1 0 
vi) For  2 1  we have


0 
 1–

 = (1 – )2 = 0
2
1
–
 

and so the eigenvalues are
 = 1 (twice)
 2 0 –1 
vii)For A =  0 2 0  we have
 –1 0 2 
– 33 –
0
–1
 2–

0 2–
0
|A – I| = 
0 2–
 –1

 2– –1 



=
(2
–
)

 –1 2– 

by expanding by the second row
= (2 – )[( – 2)2 – 1]
= (2 – )(2 – 4 + 4 – 1)
= (2 – )(2 – 4 + 3)
= (2 – )( – 1)( – 3) = 0
The eigenvalues in this case are therefore
 = 1, 2, 3
 1 2 –1 
viii) ForA =  2 1 1  we have
 –1 1 0 
2
 1–

2 1–
|A – I| = 
1
 –1
1 
 1–
 2
 – 2
= (1 – )
1 – 
 –1

–1
1
–




1 
 2 1– 
 – 1

– 
1 
 –1
= (1 – )( ( – 1) – 1) – 2(– 2 + 1) – (2 + 1 – )
=(1 – )(2 –  – 1) + 4 – 2 – 3 + 
= 2 –  – 1 – 3 + 2 +  + 5 – 5
= ( – 1)(– 2 +  + 1 + 5) = ( – 1)(– 2 +  + 6)
= – ( – 1)(2 –  – 6) = – ( – 1)( – 3)( + 2)
and so the eigenvalues in this case are
– 34 –
 = 1, – 2, 3
 1 0 0 
ix) Nobody said eigenvalues had to be real! In fact, for A = 2 1 –2  we
1 
 3 2
have
0
0
 1–

2 1– –2
|A – I| = 
3
2 1–


 1– –2 


 = (1 – )
2 1– 

= (1 – )(( – 1)2 + 4) = (1 – )(2 – 2 + 5) = 0
So in this case we have  = 1, or
2 – 2 + 5 = 0
Solving this quadratic by formula we get
=
2
4 – 20
2
=
2  4j
= 1  2j
2
So in this case the eigenvalues are
 = 1, 1  2j
17.
Determine the eigenvectors for each of the matrices in Q16.
i)
 1 0 


 0 2 
 –1 1 0
v) 0 –2 2
 3 –9 6




 1 2 –1 
viii)  2 1 1 
 –1 1 0 
ii)
 6 –3 


 2 1 
iii)
 1 0 
vi)  2 1 


 1 0 0 
ix)  2 1 –2 
1 
 3 2
– 35 –
 2 3 


 –1 –2 
iv)
 2 0 –1 
vii)  0 2 0 
 –1 0 2 
 8 –4 


 2 2 
Solution
 1 0 
i) The eigenvalues of  0 2  were found to be 1 and 2 and the


corresponding eigenvectors are given by:
For  = 1
 1 0  x 
 x 


 = 1

 0 2  y 
 y 
which gives
x=x
2y = y
The only solution for y is clearly y = 0, while x can take any value. So the
most general possible solution is
 x 
 x 

 =

 y 
 0 
and we may take the eigenvector corresponding to  = 1 to be
 1 
u=  0 


For  = 2
 1 0  x 
 x 


 = 2

 0 2  y 
 y 
which gives
x = 2x
y=y
The only solution for x is clearly x = 0, while y can take any value. So the
most general possible solution is
 x 
 0 

 =

 y 
 y 
and we may take the eigenvector corresponding to  = 2 to be
– 36 –
 0 
u=  1 


 6 –3 
ii) The eigenvalues of  2 1  were found to be 3 and 4 and the


corresponding eigenvectors are given by:
For  = 3
 6 –3  x 
 x 


 = 3

2
1
y



 y 
which gives
3x – 3y = 0
2x – 2y = 0
with solution x = y. So the most general possible solution is
 x 
 x 

 =

 y 
 x 
and we may take the eigenvector corresponding to  = 3 to be
1  1 
 1 


u =  1  or, normalizing, u =


2  1 
For  = 4
 6 –3  x 
 x 


 = 4

 2 1  y 
 y 
which gives
2x – 3y = 0
2x – 3y = 0
with solution 2x = 3y. So the most general possible solution is
 x 
 x 

 =  2 
 y 
x
 3 
and we may take the eigenvector corresponding to  = 4 to be
– 37 –
 1 
1  3 
 2 
u =  2  or, normalizing, u =


13
3


 2 3 
iii) The eigenvalues of  –1 –2  were found to be 1 and the


corresponding eigenvectors are given by:
For  = 1
 2 3  x 
 x 


 = 1

 –1 –2  y 
 y 
which gives
x + 3y = 0
– x – 3y = 0
with solution x = – 3y. So we may take the eigenvector corresponding to  =
1 to be
1  –3 
 –3 


u =  1  or, normalizing, u =


10  1 
For  = –1
 2 3  x 
 x 


 = – 1

 –1 –2  y 
 y 
which gives
x+y=0
with solution x = – y. So the most general possible solution is
 x 
 x 

 =

y


 –x 
and we may take the eigenvector corresponding to  = – 1 to be
1  1 
 1 


u =  – 1  or, normalizing, u =


2  –1 
– 38 –
 8 –4 

 were found to be 4 and 6 and the
 2 2 
corresponding eigenvectors are given by:
iv) The eigenvalues of
For  = 4
 8 –4  x 
 x 


 = 4

 2 2  y 
 y 
which gives
x–y=0
with solution x = y. So we may take the eigenvector corresponding to  = 4
to be
1  1 
 1 


u =  1  or, normalizing, u =


2  1 
For  = 6
 8 –4  x 
 x 


 = 6

 2 2  y 
 y 
which gives
x – 2y = 0
with solution x = 2y. So we may take the eigenvector corresponding to  = 6
to be
1  2 
 2 


u =  1  or, normalizing, u =


5  1 
 –1 1 0
v) The eigenvalues of  0 –2 2
 3 –9 6
eigenvectors given by:

 are  = 0, 1, 2, with corresponding


=0
 –1 1 0 
 0 –2 2 


 3 –9 6 
 x 
 x 
 0 
 y  =0 y  =  0 






 z 
 z 
 0 
– 39 –
This gives the result x = y = z, and we can take for a normalized eigenvector
1
u=
3
 1 
 1 


 1 
=1
 –1 1 0
 0 –2 2

 3 –9 6




 x 
 x 
 y  =1 y 




 z 
 z 
3
This gives the result y = 2x and z = 2 y = 3x, and the general form of the
eigenvector is
 x 
u = 2x 
 3x 
so we may take a normalized eigenvector as
1
u=
14
 1 
 2 


 3 
=2
 –1 1 0
 0 –2 2

 3 –9 6




 x 
 y  =2


 z 
 x 
 y 


 z 
This gives the result y = 3x, z = 2y = 6x. The general form of the eigenvector
is then
 x 
u =  3x 
 6x 
and a normalized eigenvector may be taken to be
1
u=
46
 1 
 3 


 6 
– 40 –
 1 0 
vi) We only found one distinct eigenvalue for  2 1  ,  = 1. When we have


a repeated eigenvalue in this way, there may be one or more independent
eigenvectors corresponding to the single eigenvalue, depending on the
matrix. In this case there is in fact only one eigenvector (up to a
normalization). It is given by the solution to
 1 0  x 
 x 


 = 1

 2 1  y 
 y 
which gives x = 0 and y = arbitrary. We can therefore take the normalized
eigenvector to be
 0 
u=  1 


 2 0 –1 
vii) The eigenvalues of  0 2 0  are  = 1, 2, 3, with corresponding
 –1 0 2 
eigenvectors given by:
=1
 2 0 –1   x
 0 2 0  y


 –1 0 2   z

 x 
 =1 y 




 z 
This gives the result x = z, y = 0 and we can take for a normalized
eigenvector
1
u=
2
 1 
 0 


 1 
=2
 2 0 –1   x
 0 2 0   y

 
 –1 0 2   z

 x 
 =2 y 




 z 
This gives the result x = z = 0 and y = arbitrary, and so we may take a
normalized eigenvector as
– 41 –
 0 
u =  1 
 0 
=3
 2 0 –1   x
 0 2 0   y

 
 –1 0 2   z

 x 
 =3 y 




 z 
This gives the result x = – z, y = 0, and a normalized eigenvector may be
taken to be
1
u=
2
 1 
 0 


 –1 
 1 2 –1 
viii) The eigenvalues of  2 1 1  are  =1, –2, 3, with corresponding
 –1 1 0 
eigenvectors given by:
=1
 1 2 –1   x 
 x 
 2 1 1  y  =1 y 





 –1 1 0   z 
 z 
This gives the result y = 2x, z = – 2x and we can take for a normalized
eigenvector
1 
1 
u = 3  2 
 –2 
=–2
 1 2 –1   x 
 x 
 2 1 1   y  =–2 y 

 



 –1 1 0   z 
 z 
This gives the following equations
3x + 2y – z = 0
– 42 –
2x + 3y + z = 0
– x + y + 2z = 0
Adding the first two equations gives x = – y and then the last equation gives
1
1
z = 2(x – y) = 2 (2x) = x
So the general form of the eigenvector is
 x 
 –x 


 x 
with x arbitrary and so we may take a normalized eigenvector as
1
1 
–1
u=
3  1




=3
 1 2 –1   x 
 x 
 2 1 1   y  =3 y 

 



 –1 1 0   z 
 z 
This gives the result x = y, z = 0, and a normalized eigenvector may be taken
to be
1
u=
2
 1 
 1 


 0 
 1 0 0 
ix) The eigenvalues of  2 1 –2  are  =1, 1  2j with corresponding
1 
 3 2
eigenvectors given by:
=1
 1 2 –1   x 
 x 
 2 1 1  y  =1 y 





 –1 1 0   z 
 z 
– 43 –
3
This gives the result x = z, y = – 2 x. The general form of the eigenvector is
therefore
 3x 


u= –2x


 x 
and a normalized form is
2 
1 
–3 
u=
17  2 
 = 1 + 2j
 1 0 0  x 
 x 
 2 1 –2  y  = (1 + 2j)  y 





1  z 
 3 2
 z 
This gives x = 0 and y = jz, and we can take, say, z = 1 to give the eigenvector
(not normalized)
 0 
u =  j 
 1 
 = 1 – 2j
 1 0 0  x 
 x 
 2 1 –2  y  = (1 – 2j)  y 





1  z 
 3 2
 z 
This gives x = 0 and z = jy, and we can take, say, y = 1 to give the eigenvector
(again, we have not bothered to normalize this)
 0 
u =  1 
 j 
– 44 –