This document provides information about determinants of square matrices: - It defines the determinant of a matrix as a scalar value associated with the matrix. Determinants are computed using minors and cofactors. - Properties of determinants are described, such as how determinants change with row/column operations or identical rows/columns. - Examples are provided to demonstrate computing determinants by expanding along rows or columns and using cofactors and minors. - Applications of determinants include finding the area of triangles and solving systems of linear equations.

1. Determinants
Prepared by:
Joey F. Valdriz

2. Matrix
Recall:
Inverse of a

3. SUB-TOPICS
Determinant of a Square Matrix
Minors and Cofactors
Properties of Determinants
Applications of Determinants
Area of a Triangle
Condition of Collinearity of Three Points
Solution of System of Linear Equations
(Cramer’s Rule)

4. DETERMINANT
Every square matrix has associated with it a scalar called
its determinant.
Given a matrix A, we use det(A) or |A| to designate its
determinant.
We can also designate the determinant of matrix A by
replacing the brackets by vertical straight lines. For
example,







30
12
A
30
12
)det( A
Definition 1: The determinant of a 11 matrix [a] is the scalar
a.
Definition 2: The determinant of a 22 matrix is the
scalar ad-bc.
For higher order matrices, we will use a recursive procedure to
compute determinants.






dc
ba

5. Example
4 - 3
Evaluate the determinant :
2 5
 
4 - 3
Solution : = 4×5 - 2× -3 = 20 + 6 = 26
2 5

6. Solution
The determinant of a 3 × 3 matrix A,
where









333231
232221
131211
aaa
aaa
aaa
A
is a real number defined as
 
  
  
det 11 22 33 12 23 31 13 21 32
31 22 13 32 23 11 33 21 12
A a a a a a a a a a
a a a a a a a a a .

7. Solution
If A = is a square matrix of order 3, then
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
 
 
 
  
[Expanding along first row]
11 12 13
22 23 21 23 21 22
21 22 23 11 12 13
32 33 31 33 31 32
31 32 33
a a a
a a a a a a
| A |= a a a = a - a + a
a a a a a a
a a a
     11 22 33 32 23 12 21 33 31 23 13 21 32 31 22= a a a - a a - a a a - a a + a a a - a a
   11 22 33 12 31 23 13 21 32 11 23 32 12 21 33 13 31 22a a a a a a a a a a a a a a a a a a     

8. Example
2 3 - 5
Evaluate the determinant : 7 1 - 2
-3 4 1
 
2 3 - 5
1 - 2 7 - 2 7 1
7 1 - 2 = 2 - 3 + -5
4 1 -3 1 -3 4
-3 4 1
     = 2 1+ 8 - 3 7 - 6 - 5 28 + 3
= 18 - 3 - 155
= -140
[Expanding along first row]
Solution :

9. Properties of Determinants
1. The value of a determinant remains unchanged, if its
rows and columns are interchanged.
1 1 1 1 2 3
2 2 2 1 2 3
3 3 3 1 2 3
a b c a a a
a b c = b b b
a b c c c c
i e A A. . '
2. If any two rows (or columns) of a determinant are interchanged,
then the value of the determinant is changed by minus sign.
 
1 1 1 2 2 2
2 2 2 1 1 1 2 1
3 3 3 3 3 3
a b c a b c
a b c = - a b c R R
a b c a b c
Applying 

10. Properties
3. If all the elements of a row (or column) is multiplied by a
non-zero number k, then the value of the new determinant
is k times the value of the original determinant.
1 1 1 1 1 1
2 2 2 2 2 2
3 3 3 3 3 3
ka kb kc a b c
a b c = k a b c
a b c a b c
which also implies
1 1 1 1 1 1
2 2 2 2 2 2
3 3 3 3 3 3
a b c ma mb mc
1
a b c = a b c
m
a b c a b c

11. Properties
4. If each element of any row (or column) consists of
two or more terms, then the determinant can be
expressed as the sum of two or more determinants.
1 1 1 1 1 1 1 1
2 2 2 2 2 2 2 2
3 3 3 3 3 3 3 3
a +x b c a b c x b c
a +y b c = a b c + y b c
a +z b c a b c z b c
5. The value of a determinant is unchanged, if any row
(or column) is multiplied by a number and then added
to any other row (or column).
 
1 1 1 1 1 1 1 1
2 2 2 2 2 2 2 2 1 1 2 3
3 3 3 3 3 3 3 3
a b c a +mb - nc b c
a b c = a +mb - nc b c C C + mC -nC
a b c a +mb - nc b c
Applying 

12. Properties
6. If any two rows (or columns) of a determinant
are identical, then its value is zero.
2 2 2
3 3 3
0 0 0
a b c = 0
a b c
7. If each element of a row (or column) of a determinant is
zero, then its value is zero.
1 1 1
2 2 2
1 1 1
a b c
a b c = 0
a b c

13. Properties
 
a 0 0
8 Let A = 0 b 0 be a diagonal matrix, then
0 0 c
 
 
 
  
a 0 0
= 0 b 0
0 0 c
A abc

14. Copyright © 2011 Pearson Education, Inc. Slide 7.5-14
The Minor of an Element
• The determinant of each 3 × 3 matrix is called
a minor of the associated element.
• The symbol Mij represents the minor when the
ith row and jth column are eliminated.

15. Copyright © 2011 Pearson Education, Inc. Slide 7.5-15
• To find the determinant of a 3 × 3 or larger square
matrix:
1. Choose any row or column,
2. Multiply the minor of each element in that row or
column by a +1 or –1, depending on whether the
sum of i + j is even or odd,
3. Then, multiply each cofactor by its corresponding
element in the matrix and find the sum of these
products. This sum is the determinant of the
matrix.
The Cofactor of an Element
Let Mij be the minor for element aij in an n × n matrix.
The cofactor of aij, written Aij, is
  .MA ij
ji
ij 

1

16. Copyright © 2011 Pearson Education, Inc. Slide 7.5-16
Example Evaluate det , expanding
by the second column.
Solution First find the minors of each element in the
second column.
Finding the Determinant











201
341
232
8)2)(1()3(2
31
22det
2)2)(1()2(2
21
22det
5)3)(1()2(1
21
31det
32
22
12





















M
M
M

17. Copyright © 2011 Pearson Education, Inc. Slide 7.5-17
Now, find the cofactor.
The determinant is found by multiplying each cofactor by its
corresponding element in the matrix and finding the sum of
these products.
Finding the Determinant
8)8()1()1(
2)2()1()1(
5)5()1()1(
5
32
23
32
4
22
22
22
3
12
21
12






MA
MA
MA
23
)8)(0()2)(4()5(3
201
341
232
det 323222221212














AaAaAa

18. VALUE OF DETERMINANT IN TERMS OF
MINORS AND COFACTORS
11 12 13
21 22 23
31 32 33
a a a
If A = a a a , then
a a a
 
 
 
  
 
3 3
i j
ij ij ij ij
j 1 j 1
A 1 a M a C

 
   
i1 i1 i2 i2 i3 i3= a C +a C +a C , for i=1 or i=2 or i=3

19. ROW (COLUMN) OPERATIONS
Following are the notations to evaluate a determinant:
Similar notations can be used to denote column
operations by replacing R with C.
(i) Ri to denote ith row
(ii) Ri Rj to denote the interchange of ith and jth
rows.
(iii) Ri Ri + lRj to denote the addition of l times the
elements of jth row to the corresponding elements
of ith row.
(iv) lRi to denote the multiplication of all elements of
ith row by l.



20. EVALUATION OF DETERMINANTS
If a determinant becomes zero on putting
is the factor of the determinant. x = , then x - 
2
3
x 5 2
For example, if Δ = x 9 4 , then at x =2
x 16 8
, because C1 and C2 are identical at x = 2
Hence, (x – 2) is a factor of determinant .
  0


21. SIGN SYSTEM FOR EXPANSION OF DETERMINANT
Sign System for order 2 and order 3 are given by
+ – +
+ –
, – + –
– +
+ – +

22.  
42 1 6 6×7 1 6
i 28 7 4 = 4×7 7 4
14 3 2 2×7 3 2
 1
6 1 6
=7 4 7 4 Taking out 7 common from C
2 3 2
EXAMPLE - 1
6 -3 2
2 -1 2
-10 5 2
42 1 6
28 7 4
14 3 2
Find the value of the following determinants
(i) (ii)
Solution :
1 3= 7×0 C and C are identical
= 0
  

23. EXAMPLE –1 (II)
6 -3 2
2 -1 2
-10 5 2
(ii)
 
 
 
   
    
 
3 2 3 2
1 2 1 2
5 2 5 2
 
      
     

1
1 2
3 3 2
( 2) 1 1 2 Taking out 2 common from C
5 5 2
( 2) 0 C and C are identical
0

24. Evaluate the determinant
1 a b+c
1 b c+a
1 c a+b
Solution :
 3 2 3
1 a b+c 1 a a+b+c
1 b c+a = 1 b a+b+c Applying c c +c
1 c a+b 1 c a+b+c

    3
1 a 1
= a+b+c 1 b 1 Taking a+b+c common from C
1 c 1
  
EXAMPLE - 2
  1 3= a+ b + c ×0 C and C are identical
= 0
  

25. 2 2 2
a b c
We have a b c
bc ca ab
 2
1 1 2 2 2 3
(a-b) b-c c
= (a-b)(a+b) (b-c)(b+c) c Applying C C -C and C C -C
-c(a-b) -a(b-c) ab
 
   2
1 2
1 1 c
Taking a-b and b-c common
=(a-b)(b-c) a+b b+c c
from C and C respectively
-c -a ab
 
 
 
EXAMPLE - 3
bc
2 2 2
a b c
a b c
ca ab
Evaluate the determinant:
Solution:

26.  2
1 1 2
0 1 c
=(a-b)(b-c) -(c-a) b+c c Applying c c -c
-(c-a) -a ab

2
0 1 c
=-(a-b)(b-c)(c-a) 1 b+c c
1 -a ab
 2
2 2 3
0 1 c
= -(a-b)(b-c)(c-a) 0 a+b+c c -ab Applying R R -R
1 -a ab

Now expanding along C1 , we get
(a-b) (b-c) (c-a) [- (c2 – ab – ac – bc – c2)]
= (a-b) (b-c) (c-a) (ab + bc + ac)
SOLUTION CONT.

27. Without expanding the determinant,
prove that
3
3x+y 2x x
4x+3y 3x 3x =x
5x+6y 4x 6x
3x+y 2x x 3x 2x x y 2x x
L.H.S= 4x+3y 3x 3x = 4x 3x 3x + 3y 3x 3x
5x+6y 4x 6x 5x 4x 6x 6y 4x 6x
3 2
3 2 1 1 2 1
= x 4 3 3 +x y 3 3 3
5 4 6 6 4 6
EXAMPLE - 4
Solution :
 3 2
1 2
3 2 1
= x 4 3 3 +x y×0 C and C are identical in II determinant
5 4 6

28. SOLUTION CONT.
 3
1 1 2
1 2 1
= x 1 3 3 Applying C C -C
1 4 6

 3
2 2 1 3 3 2
1 2 1
= x 0 1 2 ApplyingR R -R and R R -R
0 1 3
 
 3
1
3
= x ×(3-2) Expanding along C
=x =R.H.S.
3
3 2 1
=x 4 3 3
5 4 6

29. Prove that : = 0 , where w is cube root of unity.
3 5
3 4
5 5
1 ω ω
ω 1 ω
ω ω 1
3 5 3 3 2
3 4 3 3
5 5 3 2 3 2
1 ω ω 1 ω ω .ω
L.H.S= ω 1 ω = ω 1 ω .ω
ω ω 1 ω .ω ω .ω 1
 
2
3
2 2
1 2
1 1 ω
= 1 1 ω ω =1
ω ω 1
=0=R.H.S. C and C are identical
  
EXAMPLE - 5
Solution :

30. EXAMPLE - 6
2
x+a b c
a x+b c =x (x+a+b+c)
a b x+C
Prove that :
 1 1 2 3
x+a b c x+a+b+c b c
L.H.S= a x+b c = x+a+b+c x+b c
a b x+C x+a+b+c b x+c
Applying C C +C +C
Solution :
 
  1
1 b c
= x+a+b+c 1 x+b c
1 b x+c
Taking x+a+b+c commonfrom C  

31. SOLUTION CONT.
 2 2 1 3 3 1
1 b c
=(x+a+b+c) 0 x 0
0 0 x
Applying R R -R and R R -R 
Expanding along C1 , we get
(x + a + b + c) [1(x2)] = x2 (x + a + b + c)
= R.H.S

32.  1 1 2 3
2(a+b+c) 2(a+b+c) 2(a+b+c)
= c+a a+b b+c Applying R R +R +R
a+b b+c c+a

1 1 1
=2(a+b+c) c+a a+b b+c
a+b b+c c+a
EXAMPLE - 7
Solution :
Using properties of determinants, prove that
2 2 2
b+c c+a a+b
c+a a+b b+c =2(a+b+c)(ab+bc+ca-a -b -c ).
a+b b+c c+a
b+c c+a a+b
L.H.S= c+a a+b b+c
a+b b+c c+a

33.  1 1 2 2 2 3
0 0 1
=2(a+b+c) (c-b) (a-c) b+c Applying C C -C and C C -C
(a-c) (b-a) c+a
 
Now expanding along R1 , we get
2
2(a+b+c) (c-b)(b-a)-(a-c)  
2 2 2
=2(a+b+c) bc-b -ac+ab-(a +c -2ac)  
SOLUTION CONT.
2 2 2
=2(a+b+c) ab+bc+ac-a -b -c
=R.H.S
  

34. Using properties of determinants prove that
2
x+4 2x 2x
2x x+4 2x =(5x+4)(4-x)
2x 2x x+4
EXAMPLE - 8
1 2x 2x
=(5x+4)1 x+4 2x
1 2x x+4
Solution :
 1 1 2 3
x+4 2x 2x 5x+4 2x 2x
L.H.S= 2x x+4 2x =5x+4 x+4 2x Applying C C +C +C
2x 2x x+4 5x+4 2x x+4


35. SOLUTION CONT.
 2 2 1 3 3 2
1 2x 2x
=(5x+4) 0 -(x-4) 0 ApplyingR R -R and R R -R
0 x-4 -(x-4)
 
Now expanding along C1 , we get
2
(5x+4) 1(x-4) -0  
2
=(5x+4)(4-x)
=R.H.S

36. EXAMPLE - 9
Using properties of determinants, prove that
x+9 x x
x x+9 x =243 (x+3)
x x x+9
x+9 x x
L.H.S= x x+9 x
x x x+9
 1 1 2 3
3x+9 x x
= 3x+9 x+9 x Applying C C +C +C
3x+9 x x+9

Solution :

37.  1=3(x+3) 81 Expanding along C
=243(x+3)
=R.H.S.

1 x x
=(3x+9)1 x+9 x
1 x x+9
SOLUTION CONT.
  2 2 1 3 3 2
1 x x
=3 x+3 0 9 0 Applying R R -R and R R -R
0 -9 9
   

38. SOLUTION CONT.
 
2
2 2 2
2 2 1 3 3 2
1 a bc
=(a +b +c ) 0 (b-a)(b+a) c(a-b) Applying R R -R and R R -R
0 (c-b)(c+b) a(b-c)
 
 2 2 2 2 2
1=(a +b +c )(a-b)(b-c)(-ab-a +bc+c ) Expanding along C
2 2 2
=(a +b +c )(a-b)(b-c)(c-a)(a+b+c)=R.H.S.
2
2 2 2
1 a bc
=(a +b +c )(a-b)(b-c) 0 -(b+a) c
0 -(b+c) a
    2 2 2
=(a +b +c )(a-b)(b-c) b c-a + c-a c+a  

39. EXAMPLE - 10
Solution :
2 2 2 2 2
2 2 2 2 2
1 1 3
2 2 2 2 2
(b+c) a bc b +c a bc
L.H.S.= (c+a) b ca = c +a b ca Applying C C -2C
(a+b) c ab a +b c ab
  
 
2 2 2 2
2 2 2 2
1 1 2
2 2 2 2
a +b +c a bc
a +b +c b ca Applying C C +C
a +b +c c ab
 
2
2 2 2 2
2
1 a bc
=(a +b +c )1 b ca
1 c ab
2 2
2 2 2 2 2
2 2
(b+c) a bc
(c+a) b ca =(a +b +c )(a-b)(b-c)(c-a)(a+b+c)
(a+b) c ab
Show that

40. Applications of Determinants
(Area of a Triangle)
The area of a triangle whose vertices are
is given by the expression1 1 2 2 3 3(x , y ), (x , y ) and (x , y )
1 1
2 2
3 3
x y 1
1
Δ= x y 1
2
x y 1
1 2 3 2 3 1 3 1 2
1
= [x (y - y )+ x (y - y )+ x (y - y )]
2

41. Example
Find the area of a triangle whose vertices are (-1, 8), (-2, -3)
and (3, 2).
Solution :
1 1
2 2
3 3
x y 1 -1 8 1
1 1
Area of triangle= x y 1 = -2 -3 1
2 2
x y 1 3 2 1
 
1
= -1(-3-2)-8(-2-3)+1(-4+9)
2
 
1
= 5+40+5 =25 sq.units
2

42. If are three points,
then A, B, C are collinear
1 1 2 2 3 3A (x , y ), B (x , y ) and C (x , y )
1 1 1 1
2 2 2 2
3 3 3 3
Area of triangle ABC = 0
x y 1 x y 1
1
x y 1 = 0 x y 1 = 0
2
x y 1 x y 1

 
Condition of Collinearity of Three Points

43. If the points (x, -2) , (5, 2), (8, 8) are collinear, find x , using
determinants.
Example
Solution :
x -2 1
5 2 1 =0
8 8 1

      x 2-8 - -2 5-8 +1 40-16 =0
-6x-6+24=0
6x=18 x=3 
Since the given points are collinear.

44. Solution of System of 2 Linear
Equations (Cramer’s Rule)
Let the system of linear equations be
 2 2 2a x+b y = c ... ii
 1 1 1a x+b y = c ... i
1 2D D
Then x = , y = provided D 0,
D D

1 1 1 1 1 1
1 2
2 2 2 2 2 2
a b c b a c
where D = , D = and D =
a b c b a c

45. Cramer’s Rule
then the system is consistent and has infinitely many
solutions.
  1 22 If D = 0 and D =D = 0,
then the system is inconsistent and has no solution.
 1 If D 0
Note :
,
then the system is consistent and has unique solution.
  1 23 If D=0 and one of D , D 0,

46. Example
2 -3
D= =2+9=11 0
3 1

1
7 -3
D = =7+15=22
5 1
2
2 7
D = =10-21=-11
3 5
Solution :
1 2
D 0
D D22 -11
By Cramer's Rule x= = =2 and y= = =-1
D 11 D 11


Using Cramer's rule , solve the following
system of equations 2x-3y=7, 3x+y=5

47. Solution of System of 3 Linear
Equations (Cramer’s Rule)
Let the system of linear equations be
 2 2 2 2a x+b y+c z = d ... ii
 1 1 1 1a x+b y+c z = d ... i
 3 3 3 3a x+b y+c z = d ... iii
31 2 DD D
Then x = , y = z = provided D 0,
D D D
, 
1 1 1 1 1 1 1 1 1
2 2 2 1 2 2 2 2 2 2 2
3 3 3 3 3 3 3 3 3
a b c d b c a d c
where D = a b c , D = d b c , D = a d c
a b c d b c a d c
1 1 1
3 2 2 2
3 3 3
a b d
and D = a b d
a b d

48. Cramer’s Rule
Note:
(1) If D  0, then the system is consistent and has a unique
solution.
(2) If D=0 and D1 = D2 = D3 = 0, then the system has infinite
solutions or no solution.
(3) If D = 0 and one of D1, D2, D3  0, then the system
is inconsistent and has no solution.
(4) If d1 = d2 = d3 = 0, then the system is called the system of
homogeneous linear equations.
(i) If D  0, then the system has only trivial solution x = y = z = 0.
(ii) If D = 0, then the system has infinite solutions.

49. Example
Using Cramer's rule , solve the following
system of equations
5x - y+ 4z = 5
2x + 3y+ 5z = 2
5x - 2y + 6z = -1
Solution :
5 -1 4
D= 2 3 5
5 -2 6
1
5 -1 4
D = 2 3 5
-1 -2 6
= 5(18+10)+1(12+5)+4(-4 +3)
= 140 +17 –4
= 153
= 5(18+10) + 1(12-25)+4(-4 -15)
= 140 –13 –76 =140 - 89
= 51 0

50. 3
5 -1 5
D = 2 3 2
5 -2 -1
= 5(-3 +4)+1(-2 - 10)+5(-4-15)
= 5 – 12 – 95 = 5 - 107
= - 102
Solution
1 2
3
D 0
D D153 102
By Cramer's Rule x= = =3, y= = =2
D 51 D 51
D -102
and z= = =-2
D 51


2
5 5 4
D = 2 2 5
5 -1 6
= 5(12 +5)+5(12 - 25)+ 4(-2 - 10)
= 85 + 65 – 48 = 150 - 48
= 102

51. Example
Solve the following system of homogeneous linear equations:
x + y – z = 0, x – 2y + z = 0, 3x + 6y + -5z = 0
Solution:
     
1 1 - 1
We have D = 1 - 2 1 = 1 10 - 6 - 1 -5 - 3 - 1 6 + 6
3 6 - 5
= 4 + 8 - 12 = 0
 
 
 
  
The systemhas infinitely many solutions.
Putting z = k, in first two equations, we get
x + y = k, x – 2y = -k

52. 1
k 1
D -k - 2 -2k + k k
By Cramer's rule x = = = =
D -2 - 1 31 1
1 - 2

2
1 k
D 1 - k -k - k 2k
y = = = =
D -2 - 1 31 1
1 - 2
k 2k
x = , y = , z = k , where k R
3 3
 
These values of x, y and z = k satisfy (iii) equation.

53. Seatwork
Find the determinant of each matrix.
42 1 6
28 7 4
14 3 2
6 -3 2
2 -1 2
-10 5 2
2 3 1
0 2 4
2 5 6
 
 
 
  
8 2 1 4
3 5 3 11
0 0 4 0
2 2 7 1
  
  
 
 
 

54. Thank you