Linear algebra using matrix analysis

1. Chapter 3
Determinants
Linear Algebra

2. Ch03_2
3.1 Introduction to Determinants
Definition
The determinant of a 2  2 matrix A is denoted |A| and is given
by
Observe that the determinant of a 2  2 matrix is given by the
different of the products of the two diagonals of the matrix.
The notation det(A) is also used for the determinant of A.
21
12
22
11
22
21
12
11
a
a
a
a
a
a
a
a


Example 1








1
3
4
2
A
14
12
2
))
3
(
4
(
)
1
2
(
1
3
4
2
)
det( 









A

3. Ch03_3
Definition
Let A be a square matrix.
The minor of the element aij is denoted Mij and is the determinant
of the matrix that remains after deleting row i and column j of A.
The cofactor of aij is denoted Cij and is given by
Cij = (–1)i+j Mij
Note that Cij = Mij or Mij .

4. Ch03_4
10
)
4
3
(
)
2
1
(
2
4
3
1
1
2
0
2
1
4
3
0
1
:
of
Minor 32
32 









M
a
3
))
2
(
2
(
)
1
1
(
1
2
2
1
1
2
0
2
1
4
3
0
1
:
of
Minor 11
11 












M
a
Example 2
Solution
Determine the minors and cofactors of the elements a11 and a32
of the following matrix A.











1
2
0
2
1
4
3
0
1
A
3
)
3
(
)
1
(
)
1
(
:
of
Cofactor 2
11
1
1
11
11 



 
M
C
a
10
)
10
(
)
1
(
)
1
(
:
of
Cofactor 5
32
2
3
32
32 




 
M
C
a

5. Ch03_5
Definition
The determinant of a square matrix is the sum of the products
of the elements of the first row and their cofactors.
These equations are called cofactor expansions of |A|.
n
nC
a
C
a
C
a
C
a
A
n
n
A
C
a
C
a
C
a
C
a
A
A
C
a
C
a
C
a
A
A
1
1
13
13
12
12
11
11
14
14
13
13
12
12
11
11
13
13
12
12
11
11
,
is
If
4,
4
is
If
3,
3
is
If


















6. Ch03_6
Example 3
Evaluate the determinant of the following matrix A.







 

1
2
4
1
0
3
1
2
1
A
Solution
2
4
0
3
)
1
)(
1
(
1
4
1
3
)
1
(
2
1
2
1
0
)
1
(
1 4
3
2
13
13
12
12
11
11









 C
a
C
a
C
a
A
6
6
2
2
)]
4
0
(
)
2
3
[(
)]
4
1
(
)
1
3
[(
2
)]
2
1
(
)
1
0
[(



















7. Ch03_7
Theorem 3.1
The determinant of a square matrix is the sum of the products of
the elements of any row or column and their cofactors.
ith row expansion:
jth column expansion: nj
nj
j
j
j
j
in
in
i
i
i
i
C
a
C
a
C
a
A
C
a
C
a
C
a
A










2
2
1
1
2
2
1
1
Example 4
Find the determinant of the following matrix using the second row.







 

1
2
4
1
0
3
1
2
1
A
Solution
6
6
0
12
)]
4
2
(
)
2
1
[(
1
)]
4
1
(
)
1
1
[(
0
)]
2
1
(
)
1
2
[(
3
2
4
2
1
1
1
4
1
1
0
1
2
1
2
3
23
23
22
22
21
21





























 C
a
C
a
C
a
A

8. Ch03_8
Example 5
Evaluate the determinant of the following 4  4 matrix.













3
0
1
0
5
3
2
7
2
0
1
0
4
0
1
2
Solution
6
)
2
3
(
6
3
1
2
1
)
2
(
3
3
1
0
2
1
0
4
1
2
3









)
(
0
)
(
3
)
(
0
)
(
0 43
33
23
13
43
43
33
33
23
23
13
13
C
C
C
C
C
a
C
a
C
a
C
a
A









9. Ch03_9
Example 6
Solve the following equation for the variable x.
7
2
1
1 



x
x
x
Solution
Expand the determinant to get the equation
7
)
1
)(
1
(
)
2
( 



 x
x
x
Proceed to simplify this equation and solve for x.
3
or
2
0
)
3
)(
2
(
0
6
7
1
2
2
2












x
x
x
x
x
x
x
x
There are two solutions to this equation, x = – 2 or 3.

10. Ch03_10
Computing Determinants of 2  2
and 3  3 Matrices







22
21
12
11
a
a
a
a
A 21
12
22
11
A a
a
a
a 











33
32
31
23
22
21
13
12
11
a
a
a
a
a
a
a
a
a
A
32
31
22
21
12
11
33
32
31
23
22
21
13
12
11
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a









left)
right to
from
products
(diagonal
right)
left to
from
products
(diagonal
A
33
21
12
32
23
11
31
22
13
32
21
13
31
23
12
33
22
11
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a








11. Ch03_11
Homework
Exercise 3.1 pages 161-162:
1, 3, 5, 7, 9, 11, 13

12. Ch03_12
Let A be an n  n matrix and c be a nonzero scalar.
(a) If then |B| = c|A|.
(b) If then |B| = –|A|.
(c) If then |B| = |A|.
3.2 Properties of Determinants
Theorem 3.2
Proof (a)
|A| = ak1Ck1 + ak2Ck2 + … + aknCkn
|B| = cak1Ck1 + cak2Ck2 + … + caknCkn
|B| = c|A|.
B
A
cRk

B
A
Rj
Ri

B
A
cRj
Ri


13. Ch03_13
Example 1
.
3
9
2
3
6
1
2
4
3




Solution
21
3
1
2
3
)
3
(
3
3
2
3
0
1
2
0
3
3
9
2
3
6
1
2
4
3
C3
2
C2















Evaluate the determinant

14. Ch03_14
Example 2
If |A| = 12 is known.
Evaluate the determinants of the following matrices.
,
10
4
2
5
2
0
3
4
1











A





































16
4
0
5
2
0
3
4
1
)
c
(
5
2
0
10
4
2
3
4
1
(b)
10
12
2
5
6
0
3
12
1
)
a
( 3
2
1 B
B
B
Solution
(a) Thus |B1| = 3|A| = 36.
(b) Thus |B2| = – |A| = –12.
(c) Thus |B3| = |A| = 12.
1
2
3
B
A
C

2
3
2
B
A
R
R 

3
1
2
3
B
A
R
R 


15. Ch03_15
Theorem 3.3
Let A be a square matrix. A is singular if
(a) all the elements of a row (column) are zero.
(b) two rows (columns) are equal.
(c) two rows (columns) are proportional. (i.e., Ri=cRj)
Proof
(a) Let all elements of the kth row of A be zero.
0
0
0
0 2
1
2
2
1
1 







 kn
k
k
kn
kn
k
k
k
k C
C
C
C
a
C
a
C
a
A 

(c) If Ri=cRj, then , row i of B is [0 0 … 0].
 |A|=|B|=0
B
A
cRj
Ri

Definition
A square matrix A is said to be singular if |A|=0.
A is nonsingular if |A|0.

16. Ch03_16
Example 3
Show that the following matrices are singular.









 














8
4
2
4
2
1
3
1
2
(b)
9
0
4
1
0
3
7
0
2
)
(a B
A
Solution
(a) All the elements in column 2 of A are zero. Thus |A| = 0.
(b) Row 2 and row 3 are proportional. Thus |B| = 0.

17. Ch03_17
Theorem 3.4
Let A and B be n  n matrices and c be a nonzero scalar.
(a) |cA| = cn|A|.
(b) |AB| = |A||B|.
(c) |At| = |A|.
(d) (assuming A–1 exists)
A
A
1
1


Proof
(a)
A
c
cA
cA
A n
cRn
cR
cR



...,
,
2
,
1
(d)
A
A
I
A
A
A
A
1
1 1
1
1






 



18. Ch03_18
Example 4
If A is a 2  2 matrix with |A| = 4, use Theorem 3.4 to compute
the following determinants.
(a) |3A| (b) |A2| (c) |5AtA–1|, assuming A–1 exists
Solution
(a) |3A| = (32)|A| = 9  4 = 36.
(b) |A2| = |AA| =|A| |A|= 4  4 = 16.
(c) |5AtA–1| = (52)|AtA–1| = 25|At||A–1| .
25
1
25 

A
A
Example 5
Prove that |A–1AtA| = |A|
Solution
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A t
t
t
t




 


 1
)
( 1
1
1
1

19. Ch03_19
Example 6
Prove that if A and B are square matrices of the same size, with A
being singular, then AB is also singular. Is the converse true?
Solution
()
|A| = 0  |AB| = |A||B| = 0
Thus the matrix AB is singular.
()
|AB| = 0  |A||B| = 0  |A| = 0 or |B| = 0
Thus AB being singular implies that either A or B is singular.
The inverse is not true.

20. Ch03_20
Homework
Exercise 3.2 pp. 170-171: 1, 3, 5, 7, 9, 13
Exercise 11
Prove the following identity without evaluating the determinants.
w
v
u
r
q
p
f
d
b
w
v
u
r
q
p
e
c
a
w
v
u
r
q
p
f
e
d
c
b
a





v
u
q
p
f
e
w
u
r
p
d
c
w
v
r
q
b
a
w
v
u
r
q
p
f
e
d
c
b
a
)
(
)
(
)
( 









21. Ch03_21
3.3 Numerical Evaluation of a
Determinant
Definition
A square matrix is called an upper triangular matrix if all the
elements below the main diagonal are zero.
It is called a lower triangular matrix if all the elements above the
main diagonal are zero.
triangular
upper
1
0
0
0
9
0
0
0
5
3
2
0
7
0
4
1
,
9
0
0
5
1
0
2
8
3























triangular
lower
1
8
5
4
0
2
0
7
0
0
4
1
0
0
0
8
,
8
9
3
0
1
2
0
0
7
























22. Ch03_22
.
find
,
5
0
0
4
3
0
9
1
2
Let A
A














Theorem 3.5
The determinant of a triangular matrix is the product of its
diagonal elements.
Example 1
Proof
nn
nn
n
n
nn
n
n
nn
n
n
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a























22
11
4
44
3
34
33
22
11
3
33
2
23
22
11
2
22
1
12
11
0
0
0
0
0
0
0
0
0




.
30
)
5
(
3
2
ol. 





A
S
Numerical Evaluation of a Determinant

23. Ch03_23
Numerical Evaluation of a Determinant
Example 2
Evaluation the determinant.










10
9
4
4
5
2
1
4
2
8
1
0
5
1
0
1
4
2
R1
)
2
(
3
R
R1
R2
10
9
4
4
5
2
1
4
2







13
0
0
5
1
0
1
4
2
2
R3 


R
26
13
)
1
(
2 





Solution (elementary row operations）

24. Ch03_24
Example 3
Evaluation the determinant.
1
2
0
1
3
0
0
1
0
1
1
2
1
2
0
1


Solution
2
4
0
0
2
2
0
0
2
3
1
0
1
2
0
1
R1
R4
R1
)
1
(
R3
R1
)
2
(
R2
1
2
0
1
3
0
0
1
0
1
1
2
1
2
0
1












6
0
0
0
2
2
0
0
2
3
1
0
1
2
0
1
2R3
R4 





12
6
)
2
(
)
1
(
1 







25. Ch03_25
Example 4
Evaluation the determinant.
11
2
2
5
2
1
4
2
1




Solution
3
2
0
1
0
0
4
2
1
1
R
)
2
(
3
R
R1
R2
11
2
2
5
2
1
4
2
1










1
0
0
3
2
0
4
2
1
)
1
(
R3
R2





2
)
1
(
2
1
)
1
( 







26. Ch03_26
Example 5
Evaluation the determinant.
1
5
6
6
4
3
2
2
3
2
1
1
2
0
1
1




Solution
11
5
0
0
0
3
0
0
5
2
0
0
2
0
1
1
R1
)
6
(
R4
R1
)
2
(
R3
R1
R2
1
5
6
6
4
3
2
2
3
2
1
1
2
0
1
1












diagonal element is zero and
all elements below this
diagonal element are zero.
0


27. Ch03_27
3.3 Determinants, Matrix Inverse,
and Systems of Linear Equations
Definition
Let A be an n  n matrix and Cij be the cofactor of aij.
The matrix whose (i, j)th element is Cij is called the matrix of
cofactors of A.
The transpose of this matrix is called the adjoint of A and is
denoted adj(A).
cofactors
of
matrix
2
1
2
22
21
1
12
11












nn
n
n
n
n
C
C
C
C
C
C
C
C
C






matrix
adjoint
2
1
2
22
21
1
12
11
t
C
C
C
C
C
C
C
C
C
nn
n
n
n
n



















28. Ch03_28
Example 1
Give the matrix of cofactors and the adjoint matrix of the
following matrix A.












5
3
1
2
4
1
3
0
2
A
Solution The cofactors of A are as follows.
14
5
3
2
4
11 



C 3
5
1
2
1
12 




C 1
3
1
4
1
13 




C
9
5
3
3
0
21 




C 7
5
1
3
2
22 

C 6
3
1
0
2
23 



C
12
2
4
3
0
31 



C 1
2
1
3
2
32 




C 8
4
1
0
2
33 


C
The matrix of cofactors
of A is












8
6
1
1
7
3
12
9
14
)
(
adj A













8
1
12
6
7
9
1
3
14
The adjoint of A is

29. Ch03_29
Theorem 3.6
Let A be a square matrix with |A|  0. A is invertible with
)
(
adj
1
1
A
A
A 

Proof
Consider the matrix product Aadj(A). The (i, j)th element of this
product is
 
jn
in
j
i
j
i
jn
j
j
in
i
i
C
a
C
a
C
a
C
C
C
a
a
a
A
j
A
i
j
i




















2
2
1
1
2
1
2
1
))
adj(
of
(column
)
of
(row
element
th
)
,
(

30. Ch03_30
Therefore






j
i
j
i
A
j
i
if
0
if
element
th
)
.
(  A adj(A) = |A|In
Proof of Theorem 3.6
.
2
2
1
1 A
C
a
C
a
C
a jn
in
j
i
j
i 


 
If i = j,
If i  j, let .
by
replaced
is
B
A
Ri
Rj

.
0
So 2
2
1
1 



 B
C
a
C
a
C
a jn
in
j
i
j
i 
Since |A|  0, n
I
A
A
A 








)
(
adj
1
Similarly, .
n
I
A
A
A









)
(
adj
1
Thus )
(
adj
1
1
A
A
A 

row i = row j in B
Matrices A and B have the same cofactors
Cj1, Cj2, …, Cjn.

31. Ch03_31
Theorem 3.7
A square matrix A is invertible if and only if |A|  0.
Proof
() Assume that A is invertible.
 AA–1 = In.
 |AA–1| = |In|.
 |A||A–1| = 1
 |A|  0.
() Theorem 3.6 tells us that if |A|  0, then A is invertible.
A–1 exists if and only if |A|  0.

32. Ch03_32
Example 2
Use a determinant to find out which of the following matrices are
invertible.



































 

0
8
2
2
1
1
1
2
1
1
0
1
7
12
4
3
4
2
1
2
2
4
2
3
1
1 D
C
B
A
Solution
|A| = 5  0. A is invertible.
|B| = 0. B is singular. The inverse does not exist.
|C| = 0. C is singular. The inverse does not exist.
|D| = 2  0. D is invertible.

33. Ch03_33
Example 3
Use the formula for the inverse of a matrix to compute the inverse
of the matrix












5
3
1
2
4
1
3
0
2
A
Solution
|A| = 25, so the inverse of A exists.We found adj(A) in Example 1












8
6
1
1
7
3
12
9
14
)
(
adj A




































25
8
25
6
25
1
25
1
25
7
25
3
25
12
25
9
25
14
8
6
1
1
7
3
12
9
14
25
1
)
adj(
1
1
A
A
A

34. Ch03_34
Homework
Exercise 3.3 page 178-179: 1, 3, 5, 7.
Exercise
Show that if A = A-1, then |A| = 1.
Show that if At = A-1, then |A| = 1.

35. Ch03_35
Theorem 3.8
Let AX = B be a system of n linear equations in n variables.
(1) If |A|  0, there is a unique solution.
(2) If |A| = 0, there may be many or no solutions.
Proof
(1) If |A|  0
 A–1 exists (Thm 3.7)
 there is then a unique solution given by X = A–1B (Thm 2.9).
(2) If |A| = 0
 since A  C implies that if |A|0 then |C|0 (Thm 3.2).
 the reduced echelon form of A is not In.
 The solution to the system AX = B is not unique.
 many or no solutions.

36. Ch03_36
Example 4
Determine whether or not the following system of equations has
an unique solution.
9
4
7
5
3
4
2
2
3
3
3
2
1
3
2
1
3
2
1










x
x
x
x
x
x
x
x
x
Solution
Since
0
1
4
7
3
1
4
2
3
3


Thus the system does not have an unique solution.

37. Ch03_37
Theorem 3.9 Cramer’s Rule
Let AX = B be a system of n linear equations in n variables such
that |A|  0. The system has a unique solution given by
Where Ai is the matrix obtained by replacing column i of A with B.
A
A
x
A
A
x
A
A
x n
n 

 ,
...
,
, 2
2
1
1
Proof
|A|  0  the solution to AX = B is unique and is given by
B
A
A
B
A
X
)
(
adj
1
1

 

38. Ch03_38
xi, the ith element of X, is given by
 
)
(
1
1
)]
(
adj
of
row
[
1
2
2
1
1
2
1
2
1
ni
n
i
i
n
ni
i
i
i
C
b
C
b
C
b
A
b
b
b
C
C
C
A
B
A
i
A
x




















Thus
A
A
x i
i 
Proof of Cramer’s Rule
the cofactor expansion of |Ai|
in terms of the ith column

39. Ch03_39
Example 5
Solving the following system of equations using Cramer’s rule.
6
3
2
5
5
2
2
3
3
2
1
3
2
1
3
2
1











x
x
x
x
x
x
x
x
x
Solution
The matrix of coefficients A and column matrix of constants B are




















6
5
2
and
3
2
1
1
5
2
1
3
1
B
A
It is found that |A| = –3  0. Thus Cramer’s rule be applied. We
get

































6
2
1
5
5
2
2
3
1
3
6
1
1
5
2
1
2
1
3
2
6
1
5
5
1
3
2
3
2
1 A
A
A

40. Ch03_40
Giving
Cramer’s rule now gives
9
,
6
,
3 3
2
1 



 A
A
A
3
3
9
,
2
3
6
,
1
3
3 3
3
2
2
1
1 














A
A
x
A
A
x
A
A
x
The unique solution is .
3
,
2
,
1 3
2
1 


 x
x
x

41. Ch03_41
Example 6
Determine values of  for which the following system of
equations has nontrivial solutions.Find the solutions for each
value of .
0
)
1
(
2
0
)
4
(
)
2
(
2
1
2
1







x
x
x
x



Solution
homogeneous system
 x1 = 0, x2 = 0 is the trivial solution.
 nontrivial solutions exist  many solutions

  
  = – 3 or  = 2.
0
1
2
4
2







0
)
3
)(
2
( 

 

0
6
2


 

0
)
4
(
2
)
1
)(
2
( 



 



42. Ch03_42
 = – 3 results in the system
This system has many solutions, x1 = r, x2 = r.
0
2
2
0
2
1
2
1





x
x
x
x
 = 2 results in the system
This system has many solutions, x1 = – 3r/2, x2 = r.
0
3
2
0
6
4
2
1
2
1




x
x
x
x

43. Ch03_43
Homework
Exercise 3.3 pages 179-180:
9, 11, 13, 15.