The document is a maths project report for class 12th student Tabrez Khan on the topic of determinants. It contains definitions and properties of determinants of order 1, 2 and 3 matrices. It discusses minors, cofactors and applications of determinants like solving systems of linear equations using Cramer's rule. It also contains examples of evaluating determinants and applying properties of determinants to simplify expressions.

1. MATHS PROJECT
CLASS - 12TH
ROLL NO. - 23
STUDENT’S NAME - TABREZ
KHAN
TEACHER’S NAME – SIR RAJ
KUSHWAHA
TOPIC - DETERMINANTS

2.  Determinant of a Square Matrix
 Minors and Cofactors
 Properties of Determinants
 Applications of Determinants
 Area of a Triangle
 Solution of System of Linear Equations (Cramer’s Rule)
 Class Exercise

3. If is a square matrix of order 1,
then |A| = | a11 | = a11
ij
A = a
 
 
If is a square matrix of order 2, then
11 12
21 22
a a
A =
a a
 
 
 
|A| = = a11a22 – a21a12
a a
a a
1
1 1
2
2
1 2
2
Example 1 :
4 - 3
Evaluate the determinant :
2 5
 
4 - 3
Solution : = 4 × 5 - 2 × -3 = 20 + 6 = 26
2 5

4. 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 22
a a a a a a a a a a a a a a a a a a
     
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
[Expanding along first row]
     
= 2 1 + 8 - 3 7 - 6 - 5 28 + 3
= 18 - 3 - 155
= -140
Solution :

5. 4 7 8
If A = -9 0 0 , then
2 3 4
 
 
 
 
 
M11 = Minor of a11 = determinant of the order 2 × 2 square
sub-matrix is obtained by leaving first
row and first column of A
0 0
= = 0
3 4
Similarly, M23 = Minor of a23
4 7
= =12-14=-2
2 3
M32 = Minor of a32 etc.
4 8
= = 0+72 = 72
-9 0

6.  i+j
ij ij ij
C = Cofactor of a in A = -1 M ,
ij ij
where M is minor of a in A
4 7 8
A = -9 0 0
2 3 4
 
 
 
 
 
C11 = Cofactor of a11 = (–1)1 + 1 M11 = (–1)1 +1
0 0
= 0
3 4
C23 = Cofactor of a23 = (–1)2 + 3 M23 =  
4 7
2
2 3
C32 = Cofactor of a32 = (–1)3 + 2M32 = etc.
4 8
- = -72
-9 0

7. 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

8. 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 
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.

9. 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
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).

10.  
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 
6. If any two rows (or columns) of a determinant are
identical, then its value is zero.
1 1 1
2 2 2
1 1 1
a b c
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.
2 2 2
3 3 3
0 0 0
a b c = 0
a b c

11.  
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


12. 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.



13. 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


14. Sign System for order 2 and order 3 are
given by
+ – +
+ –
, – + –
– +
+ – +

15. 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
 
 
 
bc
2 2 2
a b c
a b c
ca ab
Evaluate the determinant:
Solution:
 
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

16.  
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)
Without expanding the determinant,
prove that
3
3x+y 2x x
4x+3y 3x 3x =x
5x+6y 4x 6x
Solution :
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

17.  
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
 
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
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

18. 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
 
 
Solution :
2
x+a b c
a x+b c =x (x+a+b+c)
a b x+C
Prove that :
Solution :
 
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


19.  
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
 
  1
1 b c
= x+a+b+c 1 x+b c
1 b x+c
Taking x+a+b+c commonfrom C
 
 

20. The area of a triangle whose vertices are
is given by the expression
1 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
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

21. If are three points,
then A, B, C are collinear
1 1 2 2 3 3
A (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

 
 
1
= -1(-3-2)-8(-2-3)+1(-4+9)
2
 
1
= 5+40+5 =25 sq.units
2

22. If the points (x, -2) , (5, 2), (8, 8) are collinear,
find x , using determinants.
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.

23. Let the system of linear equations be
 
2 2 2
a x+b y = c ... ii
 
1 1 1
a x+b y = c ... i
1 2
D 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

24. then the system is consistent and has infinitely many
solutions.
  1 2
2 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 2
3 If D=0 and one of D , D 0,

Using Cramer's rule , solve the following
system of equations 2x-3y=7, 3x+y=5
Solution :
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
1 2
D 0
D D
22 -11
By Cramer's Rule x= = =2 and y= = =-1
D 11 D 11



25. Let the system of linear equations be
 
2 2 2 2
a x+b y+c z = d ... ii
 
1 1 1 1
a x+b y+c z = d ... i
 
3 3 3 3
a x+b y+c z = d ... iii
3
1 2 D
D 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

26. 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.

27. 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

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
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

28. 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
1 2
3
D 0
D D
153 102
By Cramer's Rule x = = =3, y = = =2
D 51 D 51
D -102
and z= = =-2
D 51



29. 1
k 1
D -k - 2 -2k + k k
By Cramer's rule x = = = =
D -2 - 1 3
1 1
1 - 2

2
1 k
D 1 - k -k - k 2k
y = = = =
D -2 - 1 3
1 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.