Download Chapter 2 Determinants

Chapter 2: Determinants
1
Chapter 2 Determinants
SECTION A Determinant of a Matrix
By the end of this section you will be able to
 evaluate the determinant of various square matrices
 understand what is meant by the terms cofactor and minor of a matrix
 determine the inverse of a square matrix
T. Seki was the first person to study determinants which arose
naturally out of a system of linear equations and he highlighted this
work in 1683.
Seki taught himself mathematics from a very young age but was
initially introduced to the subject by a household servant. He was
from a family of Samurai warriors.
Lewis Carroll (or to use his real name Charles Dodgson) wrote a
book on determinants called ‘An Elementary Theory of
Determinants’ in 1867. Clearly he is better known for his popular
book ‘Alice’s Adventures in Wonderland’.
Fig 1 T. Seki 1642 to 1708
Lewis Carroll proposed that Oxford University set up a Mathematical
Institute 65 years before it was eventually built. He also wrote to the Dean proposing his salary
be lowered from £300 to £200 per year because his Oxford College was suffering a financial
crisis.
In this chapter we associate a number for every square matrix. Can you remember what is
meant by a square matrix?
A square matrix is a matrix where the number of rows and columns are equal, that is a
matrix of size n by n or n  n . The number associated with each square matrix is called the
determinant of the matrix and tells us whether the matrix is invertible or not. Generally in
this chapter, a matrix will mean a square matrix.
A1 Determinant and Inverse of a 2 by 2 Matrix
We first find the determinant of a 2  2 matrix and then expand to 3  3, , n  n size
matrices.
Example 1
a b
 d b 
Consider the general 2  2 matrix A  
 and the matrix B  
 . Evaluate AB.
a
c d
 c
What do you notice about your result?
Solution
 a b   d b 
AB  


a
 c d   c
 ad  bc  ab  ba 


 cd  cd bc  ad 
0 
 ad  bc
1 0


 ad  bc  

   ad  bc  I
ad  bc  Taking out a common
 0
0 1
factor ad bc
The matrix multiplication AB gives a multiple ad  bc of the identity matrix, I. This
Chapter 2: Determinants
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a b
multiple  ad  bc  is called the determinant of matrix A  
.
c d
The determinant of a matrix A is normally denoted by det  A  or A and is a scalar not a
matrix.
a b
Hence the determinant of the general 2  2 matrix A  
 is defined as
c d

b
a
minus
(2.1)
det  A   ad  bc
c
d
What does this formula (2.1) mean?
It means the determinant of a 2  2 matrix is the result of multiplying the entries of the
leading diagonal and subtracting the product of the other diagonal. Remember the leading
diagonal are the entries of the matrix which slope downwards to the right,
.
Example 2
a b
1  d b 
Again consider the 2  2 matrices A  

.
 and B 
a
det  A   c
c d
Evaluate the matrix multiplication AB provided det  A   0 . What do you notice about your
result?
Solution
a b
1  d b 
AB  



 c d  det  A   c a 

1  a b  d


det  A   c d  c

1
 ad  bc  I
det  A 

1
det  A  I
det  A 
I
b 

a
 Remember A  kB   k  AB  


 where k  1/ det  A  is a scalar 
 By Example 1 because

 we have the same matrices 


 By (2.1) we have



det  A   ad  bc  0 
Cancelling Out det  A  
1 0
Note that the matrix multiplication AB gives the identity matrix I  
.
0 1
Since AB  I , what conclusions can we draw about the matrices A and B?
a b
1  d b 
1
The given matrix A  

 because
 has an inverse matrix A  B 
a
det  A   c
c d
we have AB  I which means B is the inverse of matrix A, that is B  A1 .
a b
Hence the inverse of the general 2  2 matrix A  
 is given by
c d
1  d b 
A 1 
(2.2)

 provided det  A   0
a
det  A   c
What does this formula mean?
Chapter 2: Determinants
3
The inverse of a 2  2 matrix is determined by interchanging entries along the leading
diagonal and placing a negative sign in the other and then multiplying this matrix by
1 det  A  .
What can we say if the determinant is zero, that is det  A   0 ?
If det  A   0 then the matrix A is non-invertible (singular), it has no inverse.
Example 3
Find the inverses of the following matrices:
 2 1 
 2 3
  
(a) A  
(b) B  
(c) C  



 1
2 
 1 5 
  

Solution
(a) Before we can find the inverse we need to evaluate the determinant. Why?
Because if the determinant is 0 then the matrix does not have an inverse. Therefore by
a b 
minus
(2.1)
det 
  ad  bc
c d 
we have
 2 3
det  A   det 
   2  5    1 3  13
 1 5 
The inverse matrix A 1 is given by the above formula (2.2) with det  A   13 :
1

a b
1  d
 By (2.2) 
 

det  A   c

c d
(b) We adopt the same procedure as part (a) to find B1 . By
a b 
minus
(2.1)
det 
  ad  bc
c
d


we have
 2 1 
det  B   det 
  2  2   11  2  1  3
 1

2


By substituting det  B   3 into the inverse formula (2.2) we have
1
 2 3
1  5 3 
A 
  

13  1 2 
 1 5 
1

1
b  

a 


1

 2 1 
a b
1  2 1 
1  d
1
B 

 By (2.2) 
  



 1
3  1
det  B   c
2 
2 

c d 

(c) Similarly applying (2.1) det  C  ad  bc we have
b  

a 

  
det  C   det 
            0
  
What can we conclude about the matrix C?
Since det  C  0 therefore the matrix C is non-invertible (singular). This means it does not
have an inverse.
A2 Applications to Transformations
Example 4
Chapter 2: Determinants
4
Consider a triangle given by the coordinates P  0, 0 , Q  2, 0 and R  0, 3 . Let the
matrix A represent this triangle PQR and determine the image of this triangle under the
2 0
transformation given by BA where B  
.
 3 4
By illustrating this transformation determine the areas of the triangle PQR and the
transformed triangle P’Q’R’. How does this transformation change the size of the area?
Solution
0 2 0
We are given coordinates P  0, 0 , Q  2, 0 and R  0, 3 therefore A  
.
0 0 3
Evaluating the matrix multiplication BA:
P Q R
P' Q' R'
 2 0  0 2 0   0
BA  


 3 4  0 0 3   0
y
Plotting this:
0

6 12 
4
14
R'
12
10
8
Q'
6
R
Fig 1
4
2
P and P'
Q
1
2
3
4
5
x
23
12  4
 3 and area of large triangle P ' Q ' R ' 
 24 .
2
2
The transformation B increases the area by a factor of 8 because 24/ 3  8 .
This factor 8 is called the determinant of the matrix B and it is evaluated by
 2 0
Determinant of matrix 
   2  4  3 0
 3 4
The area of shaded triangle PQR 
Before we discuss the inverse of a 3 by 3 matrix, we examine what is meant by the terms
‘Minors’ and ‘Cofactors’.
A2 Minors and Cofactors
a b c 


Consider the general 3 by 3 matrix A   d e f  . The determinant of the remaining
g h i 


matrix after deleting the row and column of an entry is called the minor of that entry.
For example, in the case of the matrix A we have
e f 
det 
 is the minor of entry a
h i 
Chapter 2: Determinants
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d
det 
g
f
 is the minor of entry b
i
What is the minor of entry c?
d
det 
g
What is the minor of entry e?
a


g

e

h
Deleting the row and column
containing the entry e.
c

e

i 
 a c
Hence det 
 is the minor of entry e.
g i
Example 4
Determine the minor of 1 in
 3 5 7


 1 2 3 
 4 4 9 


Solution
After deleting the rows and columns containing 1 ,
5 7 

Deleting these.


 1


4 9 

5 7
we obtain the matrix 
 . The minor of 1 is the determinant of this matrix:
 4 9 
5 7
det 
   5   9     4  7   73
 4 9  By (2.1)
Next we give the general definition of minor.
Definition (2.3). Consider a square matrix A. Let aij be the entry in the ith row and jth
column of matrix A. The minor M ij of entry aij is the determinant of the remaining matrix
after deleting the entries in the ith row and jth column.
a1n 
 a11


M ij  det 
aij
ith row

a

a
nn 
 n1
jth column
For entries in a 3  3 matrix we will need to find the determinant of a 2  2 matrix as seen in
the above Example 4. For entries in a 4  4 matrix we will need to find the determinant of a
3  3 matrix which we have not yet stated.
Next we define the term cofactor which is a number associated with the minor M ij .
Definition (2.4). Consider a square matrix A. Let aij be the entry in the ith row and jth
column of matrix A. The cofactor Cij of the entry aij is defined as
Cij   1
i j
M ij
Chapter 2: Determinants
6
where M ij is the minor of entry aij .
a1n 
 a11

.
The cofactor is given by Cij   1 det 
aij

a

a
nn 
 n1
This definition might be difficult to follow because of the ij notation but this ij only locates
the entry of a matrix. There is no easier way to locate an entry.
The cofactor is just the minor of an entry with a plus or minus sign depending on the entry.
For example the cofactor of the first entry a11 is equal to the minor, C11  M11 because
i j
 1
11
  1  1 . What is the cofactor of the entry a12 ?
2
In this case a12 means i  1 , j  2 and i  j  1  2  3 therefore C12   1 M 12   M 12 .
3
What is the cofactor of the entry a13 ?
Since i  j  1  3  4 therefore C13   1 M 13  M 13 . If we carry on developing the
4
cofactors we find they are just the minors with a place sign. Generally for a n  n matrix we
have minors with the following place signs:
  



  

  





What do you notice about the place signs?
The first entry of a matrix has a positive sign and then the place signs alternate.
Example 5
 3 5 7
Determine the cofactor of 5 in  1 2 3 
 4 4 9 


Solution
After deleting the row and column containing 5 we have the minor of 5 is


 1 3 
a b
det 
(2.1) det 
   1  9     4  3  21
  ad  bc 
 4 9  By (2.1)
c d


According to the rule, the place sign is negative, so the cofactor of 5 is 21 .
Note that the minor of 5 is 21 but the cofactor is 21 because the position of 5 in the matrix
gives it a negative sign.
We can write the determinant of a
general 3  3 matrix:
a b c 


A  d e f 
g h i 


3  3 matrix in terms of its cofactors, that is if A is a


 Remember place signs are

    


    
    


then
(2.5)
det  A   a  cofactor of a   b  cofactor of b   c  cofactor of c 
Expanding out (2.5) gives
Chapter 2: Determinants
7
  e f 
  d f    d e 
det  A   a det 
   b det 
   c det 

  h i 
  g i    g h 
Why is there a minus sign in front of the b in formula (2.6)?
This is no mistake, the minus sign comes about because the place sign for b is minus.
We can find the determinant of a matrix by expanding along any of the rows or any of the
columns. For example, the formula for expanding along the second row is
det  A   d  cofactor of d   e  cofactor of e   f  cofactor of f 
What is the formula for expanding along the bottom row?
det  A   g  cofactor of g   h  cofactor of h   i  cofactor of i 
We can also expand along any column. The formula for expanding along the first column is
det  A   a  cofactor of a   d  cofactor of d   g  cofactor of g 
(2.6)
 a11

a21
The general formula for the determinant of a n  n matrix A  


 an1
n  3 is defined as
(2.7)
det  A   a11C11  a12C12  a13C13 
a12
a22
an 2
a1n 

a2 n 
where


ann 
n
 a1nC1n   a1k C1k
k 1
where the a ’s are the entries of the given matrix and C ’s are the corresponding cofactors.
Don’t be put off by the sigma  notation. This is just a compact way of writing the above
sum given to us by the great Swiss mathematician Euler, pronounced ‘Oiler’, (1707 to 1783).
n
a
k 1
1k
C1k means summing a1k C1k from k  1 to k  n .
(2.7) is the formula of the determinant for expanding along the first row of the matrix A.
What is the formula for expanding along the ith row?
For expanding along the ith row of the matrix, the formula is
(2.8)
det  A   ai1Ci1  ai 2Ci 2  ai 3Ci 3 
n
 ainCin   aik Cik
k 1
This is sometimes called the Laplacian Expansion named after the French mathematician Pierre
Laplace (1749 to 1827).
Of course you can write a formula of the determinant for expanding along the jth column.
Example 6
Find the determinant of
3
  1 6


A   5 6 7 
 2 0 1 


Solution
Which row or column should we expand along?
Since there is a 0 in the bottom row it is easier to expand along this row:
Chapter 2: Determinants
8
 1 6 3 
 6 3 
 1 3 
 1 6 


det  5 6 7   2 det 
  0 det 
  1det 

 6 7 
 5 7 
 5 6
 2 0 1 


 2  42  18   0   6  30   24
Expanding along
this row
By (2.1)
By (2.1)
Note that the middle term on the Right-Hand Side is zero so we just need to evaluate the
other two terms. We could have also found the determinant of A by expanding along the
second column because this also contains (the same) zero.
In general if a row or column contains zero(s) then expanding along that row or column
makes the arithmetic easier.
We can also obtain the determinant of a 4  4, 5  5, 6  6 etc matrix but it becomes very
laborious to do this just using pen and paper unless we can establish zeros in the matrix. In
these cases it is more convenient to use a graphical calculator or MATLAB.
The MATLAB command for finding the determinant of a matrix A is det(A).
A3 Cofactor Matrix
Let C be the new matrix consisting of the cofactors of the general matrix A. If
a b c 
 A B C


A   d e f  then C   D E F 
g h i 
G H I 




where A is the cofactor of a, B is the cofactor of b, C is the cofactor of c etc. The matrix C is
called the cofactor matrix and it is used in finding the inverse matrix. Note that bold C is
the cofactor matrix and plain C is the cofactor of the entry c.
Example 7
Find the cofactor matrix C of

 1 1 5 
    





A   3 9 7
 Remember the place signs are      
    
 2 1 0 





Solution
Cofactor of the first entry, 1, is
9 7
det 
   9  0   1 7    7
 1 0  By (2.1)
Cofactor of 1 is

Minus
place sign
 3 7
det 
    3  0    2  7    14
 2 0  By (2.1)
Cofactor of 5 is
 3 9
det 
   3 1   2  9    21
 2 1  By (2.1)
a b 
det 
  ad  bc
c d 
Cofactor of 3 is
(2.1)
Chapter 2: Determinants
9

Minus
place sign
 1 5 
det 
    1 0   1 5    5
 1 0  By (2.1)
Cofactor of 9 is
 1 5
det 
  1 0    2  5    10
 2 0  By (2.1)
Cofactor of 7 is

Minus
place sign
 1 1
det 
   11   2   1    1
 2 1  By (2.1)
Cofactor of 2 is
 1 5 
det 
   1 7    9  5    52
 9 7  By (2.1)
Cofactor of 1 (the 1 on the bottom row of the given matrix) is
1 5
 det 
   1 7    3  5    8
Minus
 3 7  By (2.1)
place sign
Cofactor of the last entry 0 is
 1 1
det 
  1 9    3   1    12
 3 9  By (2.1)
Hence by collecting these together and placing them in the corresponding position gives the
cofactor matrix:
 7 14 21


C 5
10 1 
 52
8 12 

As stated above, we use the cofactor matrix to find the inverse of an invertible matrix.
Definition (2.9). Let A be a square matrix then the matrix consisting of the cofactors of each
entry in A is called the cofactor matrix and is normally denoted by C. The transpose of this
cofactor matrix is called the adjoint of A and is denoted by adj  A  . That is
adj  A   CT
Remember we discussed the transpose of a matrix in the last chapter and it means swapping
the rows and columns around.
Example 8
Find the adjoint of the matrix A given in Example 7 above.
Solution
We have already done all the hard work in evaluating the cofactor matrix C above. The
adjoint is the transpose of this matrix C:

 7 5 52 
 7 14 21 





T
adj  A   C   14 10
8
10 1  
 Because C   5
 21 1
 52

12 
8 12  


A4 Inverse of a Matrix
Chapter 2: Determinants
10
The remainder of this section is very demanding because you are required to understand the
proofs provided. It is going to be challenging to follow the remaining proofs but proving
results gives you a better understanding of the concepts involved.
If you are struggling to understand the chain of arguments in the proof then come back and
go over the proof a second time. It is not necessary that you understand every detail on the
first reading. Good luck with your journey on this difficult terrain.
Proposition (2.10). If a square matrix A consists of two identical rows then det  A   0 .
What does this proposition mean?
Means if a matrix has two rows which are the same then the determinant of the matrix is 0.
Proof.
To be proved in the next section.
Proposition (2.11). Let A be a n by n square matrix. If C jk denotes the cofactor of the entry
a jk for k  1, 2, 3,
and n then
ai1C j1  ai 2C j 2  ai 3C j 3 
det  A 
 ainC jn  
0
if i  j
if i  j
Proof.
How do we prove this result?
We consider the two cases i  j and i  j then show the required result in each case.
Case 1: Let i  j then by the formula for determinant
(2.8)
det  A   ai1Ci1  ai 2Ci 2  ai 3Ci 3   ainCin
we have
ai1Ci1  ai 2Ci 2  ai 3Ci 3   ainCin  det  A 
Case 2: Consider the case when i  j :
Let A * be the matrix obtained from matrix A by copying the entries of the ith row into the
jth row of matrix A. That is the matrix A * is the matrix A but with jth row being identical to
the ith row:
a1n 
a1n 
 a11 a12
 a11 a12
ith row








 ai1 ai 2
 ai1 ai 2
ain 
ain 
A
and A*  


ith row =jth row




jth row
 a j1 a j 2
 ai1 ai 2
a jn 
ain 




ann 
ann 
 an1 an 2
 an1 an 2
We have det  A *  0 . Why?
Because we have two identical rows, i and j, in A * therefore by the previous Proposition
(2.10) the determinant is zero, that is det  A *  0 .
Therefore if we expand along the jth row in the Right Hand matrix A * we have
(†)
0  det  A *  ai1C * j1 ai 2C * j 2 ai 3C * j 3   ainC * jn
where C * jk is the cofactor of entry a jk  aik in the matrix A * .
Consider the cofactor C j1 which is the place sign multiplied by the determinant of the
remaining matrix after deleting the row and column containing the entry a j1 in the Left Hand
Chapter 2: Determinants
11
matrix A. Similarly the cofactor C * j1 is the place sign multiplied by the determinant of the
remaining matrix after deleting the row and column containing the entry a j1  ai1 in the Right
Hand matrix A * . We have
a12
a1n 
a12
a1n 












ai 2
ain 
ai 2
ain 
j 1
j 1
C j1   1 det 
 and C * j1   1 det 





 a j1



a
jth row


 i1

an 2
ann 
an 2
ann 


What can you conclude about C j1 and C * j1 ?
C * j1  C j1 because the cofactor is made up of the same entries of matrix A and A * . In both
cases you delete the jth row and first column. Similarly we have C * j 2  C j 2 , C * j 3  C j 3 ,
and C * jn  C jn . Substituting these, C * j1  C j1 , C * j 2  C j 2 ,
gives
0  det  A *  ai1C * j1  ai 2C * j 2 ai 3C * j 3 
 ai1C j1  ai 2C j 2  ai 3C j 3 
and C * jn  C jn , into (†)
 ainC * jn
 ainC jn
Hence in the case where i  j we have
ai1C j1  ai 2C j 2  ai 3C j 3 
which is our required result.
 ainC jn  0
■
Proposition (2.12). Let A be a n  n square matrix. Then
A adj  A   det  A  I
Proof.
Writing out the entries of the general n by n matrix A and the transpose of the corresponding
cofactors of each entry gives
jth column
a1n 
 a11 a12


C j1
Cn1 
a2 n 
 C11 C21
 a21 a22




C12 C22
C j2
Cn 2 

A
 and adj  A   

ai1 ai 2
ain 

ith row


C


C2 n
C jn
Cnn 
1n



ann 
 an1 an 2
Consider the ij entry of the Left Hand matrix multiplication in A adj  A  . How do we
evaluate the ij entry in the matrix multiplication A adj  A  ?
Remember for matrix multiplication it is row times column. So the ij entry of A adj  A  is
the ith row times the jth column. For an ij entry we have
Chapter 2: Determinants
12
 A adj  A   ij  ai1C j1  ai 2C j 2  ai 3C j 3 
 ainC jn
if i  j
det  A 
 By the above


 Proposition (2.11) 
if i  j
0


Repeating this for each ij entry and writing out the matrix A adj  A  means we have det  A 
when i  j , along the leading diagonal, and 0 everywhere else in the matrix A adj  A  :
0
 det  A 

0
det  A 
A adj  A   

0

 0
1 0

0 1

det  A  

0
Taking Out
Common Factor

0
Hence we have our result, A adj  A   det  A  I .


0 


det  A  
0

0
 det  A  I


1
0
I
■
Proposition (2.13). Let A be a square matrix. If det  A   0 then we have
A 1 
1
adj  A 
det  A 
Proof.
This follows from the previous proposition (2.12). Since det  A   0 we can divide the above
formula given in Proposition (2.12) A adj  A   det  A  I by det  A 
 1

A
adj  A    I
 det  A 

Remember from the first chapter we know that A is invertible (non-singular) matrix if
AA 1  I
1
where A 1 is unique. Therefore A 1 
adj  A  provided det  A   0 .
det  A 
■
What does Proposition (2.13) mean?
1
adj  A  . To
det  A 
determine the inverse of a matrix you need to find the cofactors and the determinant of the
given matrix. What is the point of finding the inverse of a matrix?
As discussed in the last chapter, we need the inverse to solve linear system of equations,
which will be discussed later in this chapter.
The inverse of an invertible matrix A is given by the formula A 1 
Chapter 2: Determinants
13
Example 9
Find the inverse of the matrix given in Example 7 which is
 1 1 5 


A   3 9 7
 2 1 0 


Solution
We need to find A 1 which is given by the above Proposition: A 1 
1
adj  A 
det  A 
What is adj  A  equal to?
Remember adj  A  is the cofactor matrix transposed and was found in Example 8 above:
 7 5 52 


adj  A   C   14 10
8
 21 1
12 

We only need to find det  A  . Expanding along the bottom row of the given matrix A
because it contains a 0:

 1 1 5  
 1 5 
1 5



det  A   2 det 
Because det  A   det  3 9 7  
  1det 
0

 9 7
3 7
 2 1 0  



T
 2  7  45    7  15   112
Substituting these into formula (2.13) gives
Expanding along
this row
 7 5 52 
1
1 

A 
adj  A  
14 10
8

det  A 
112 
12 
 21 1
1
Normally we would find the determinant first and then the adjoint of the matrix because if
the determinant is zero then the matrix is non-invertible (singular).
Check that the matrix found in Example 9 is indeed the inverse of A. How?
1 0 0
1
Check the matrix multiplication A  A  I   0 1 0  . If you think this is a tedious task
0 0 1


use MATLAB.
Later in this chapter we use the inverse matrix to solve linear system of equations.
SUMMARY
a b
The determinant and inverse of a 2  2 matrix A  
 is given respectively by
c
d


(2.1)
det  A   ad  bc
(2.2)
A 1 
1  d

det  A   c
b 
 provided det  A   0
a
Chapter 2: Determinants
14
The minor denoted by M ij of an entry aij is the determinant of the remaining matrix after
deleting the entries in the ith row and jth column containing that entry.
The cofactor of an entry is the minor multiplied by a place sign. The place sign for an entry
i j
aij is given by  1 .
The cofactor matrix of a given matrix consists of cofactor of each entry in its corresponding
position.
If we expand along the ith row of a matrix A, then the formula for determinant is
(2.8)
det  A   ai1Ci1  ai 2Ci 2  ai 3Ci 3   ainCin
The adjoint of a matrix A is the cofactor matrix transposed and it is denoted by adj  A  .
The inverse of a square matrix is defined as
1
(2.13)
provided det  A   0 .
A 1 
adj  A 
det  A 