DETERMINANT-Notes & Problems - JEE PDF Download

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J E E - M a t h e m a t i c s
1 . INTRODUCTION :
If the equations a
1
x + b
1
 = 0, a
2
x + b
2 
= 0 are satisfied by the same value of x, then a
1
b
2 
– a
2
b
1 
= 0. The
expression a
1
b
2  
– a
2
b
1 
is called a determinant of the second order, and is denoted by :
1 1
2 2
a b
a b
A determinant of second order consists of two rows and two columns.
Next consider the system of equations  a
1
x + b
1
y + c
1 
= 0, a
2
x + b
2
y + c
2 
= 0, a
3
x + b
3
y + c
3 
= 0
If these equations are satisfied by the same values of x and y, then on eliminating x and y we get.
a
1
(b
2
c
3
 – b
3
c
2
) + b
1
(c
2
a
3
 – c
3
a
2
) + c
1
(a
2
b
3
 – a
3
b
2
) = 0
The expression on the left is called a determinant of the third order, and is denoted by
1 1 1
2 2 2
3 3 3
a b c
a b c
a b c
A determinant of third order consists of three rows and three columns.
Illustration 1 : Eliminate ?, m, n from the equations a ? + cm + bn = 0, c ? + bm + an = 0,  b ? + am + cn = 0
and express the result in the simplest form.
Solution : The given set of equations can also be written as (if n ? 0) :
m
a c b 0
n n
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?
;   
m
c b a 0
n n
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?
;   
m
b a c 0
n n
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?
Then, let 
m
x ; y
n n
? ?
?
? System of equations :
ax + cy + b = 0 ...(i)
cx + by + a = 0 ...(ii)
bx + ay + c = 0 ...(iii)
We have to eliminate x & y from these simultaneous linear equations.
Since these equations are satisfied by the same values of x and y, then eliminating x
and y we get,
a c b
c b a 0
b a c
?
2 . VALUE OF A DETERMINANT :
1 1 1
2 2 2
3 3 3
a b c
D a b c
a b c
?
2 2 2 2 2 2
1 1 1
3 3 3 3 3 3
b c a c a b
a b c
b c a c a b
? ? ?
= a
1
(b
2
c
3
 – b
3
c
2
) – b
1
(a
2
c
3
 – a
3
c
2
) + c
1
(a
2
b
3
 – a
3
b
2
)
Note : Sarrus diagram to get the value of determinant of order three :
1 1 1
2 2 2
3 3 3
a b c
D a b c
a b c
? = 
a
1
b
1
c
1
a
2
a
3
b
2
b
3
c
2
c
3
a
1
b
1
a
2
a
3
b
2
b
3
 –ve –ve –ve
 +ve +ve +ve
 = (a
1
b
2
c
3
 + a
2
b
3
c
1
 +a
3
b
1
c
2
) – (a
3
b
2
c
1
 + a
2
b
1
c
3 
+ a
1
b
3
c
2
)
Note that the product of the terms in first bracket (i.e. a
1
a
2
a
3
b
1
b
2
b
3
c
1
c
2
c
3
) is same as the product of the terms
in second bracket.
DETERMINANT
JEEMAIN.GURU
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J E E - M a t h e m a t i c s
1 . INTRODUCTION :
If the equations a
1
x + b
1
 = 0, a
2
x + b
2 
= 0 are satisfied by the same value of x, then a
1
b
2 
– a
2
b
1 
= 0. The
expression a
1
b
2  
– a
2
b
1 
is called a determinant of the second order, and is denoted by :
1 1
2 2
a b
a b
A determinant of second order consists of two rows and two columns.
Next consider the system of equations  a
1
x + b
1
y + c
1 
= 0, a
2
x + b
2
y + c
2 
= 0, a
3
x + b
3
y + c
3 
= 0
If these equations are satisfied by the same values of x and y, then on eliminating x and y we get.
a
1
(b
2
c
3
 – b
3
c
2
) + b
1
(c
2
a
3
 – c
3
a
2
) + c
1
(a
2
b
3
 – a
3
b
2
) = 0
The expression on the left is called a determinant of the third order, and is denoted by
1 1 1
2 2 2
3 3 3
a b c
a b c
a b c
A determinant of third order consists of three rows and three columns.
Illustration 1 : Eliminate ?, m, n from the equations a ? + cm + bn = 0, c ? + bm + an = 0,  b ? + am + cn = 0
and express the result in the simplest form.
Solution : The given set of equations can also be written as (if n ? 0) :
m
a c b 0
n n
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?
;   
m
c b a 0
n n
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?
;   
m
b a c 0
n n
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?
Then, let 
m
x ; y
n n
? ?
?
? System of equations :
ax + cy + b = 0 ...(i)
cx + by + a = 0 ...(ii)
bx + ay + c = 0 ...(iii)
We have to eliminate x & y from these simultaneous linear equations.
Since these equations are satisfied by the same values of x and y, then eliminating x
and y we get,
a c b
c b a 0
b a c
?
2 . VALUE OF A DETERMINANT :
1 1 1
2 2 2
3 3 3
a b c
D a b c
a b c
?
2 2 2 2 2 2
1 1 1
3 3 3 3 3 3
b c a c a b
a b c
b c a c a b
? ? ?
= a
1
(b
2
c
3
 – b
3
c
2
) – b
1
(a
2
c
3
 – a
3
c
2
) + c
1
(a
2
b
3
 – a
3
b
2
)
Note : Sarrus diagram to get the value of determinant of order three :
1 1 1
2 2 2
3 3 3
a b c
D a b c
a b c
? = 
a
1
b
1
c
1
a
2
a
3
b
2
b
3
c
2
c
3
a
1
b
1
a
2
a
3
b
2
b
3
 –ve –ve –ve
 +ve +ve +ve
 = (a
1
b
2
c
3
 + a
2
b
3
c
1
 +a
3
b
1
c
2
) – (a
3
b
2
c
1
 + a
2
b
1
c
3 
+ a
1
b
3
c
2
)
Note that the product of the terms in first bracket (i.e. a
1
a
2
a
3
b
1
b
2
b
3
c
1
c
2
c
3
) is same as the product of the terms
in second bracket.
DETERMINANT
JEEMAIN.GURU
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J E E - M a t h e m a t i c s
Illustration 2 : The value of 
1 2 3
4 3 6
2 7 9
?
?
  is -
(A) 213      (B) – 231 (C) 231 (D) 39
Solution :
1 2 3
4 3 6
2 7 9
?
?
 = 1 
3 6
7 9 ?
 –2
4 6 4 3
3
2 9 2 7
? ?
?
?
= (27 + 42) – 2 (–36 –12) + 3 (28 – 6) = 231
Alternative :  By sarrus diagram
1 2 3
4 3 6
2 7 9
?
?
=
1 2 3
4 3 6
2 7 9
1 2
4 3
2 7
?
?
?
?
= (27 + 24 + 84) – (18 – 42 – 72)= 135 – (18 – 114) = 231 Ans. (C)
3 . MINORS & COFACTORS :
The minor of a given element of determinant is the determinant obtained by deleting the row & the column in
which the given element stands.
For example, the minor of a
1
 in 
1 1 1
2 2 2
3 3 3
a b c
a b c
a b c
 is 
2 2
3 3
b c
b c
 & the minor of b
2
 is 
1 1
3 3
a c
a c
.
Hence a determinant of order three will have “9 minors”.
If M
ij
 represents the minor of the element belonging to i
th 
row and j
th 
column then the cofactor of that element is
given by : C
ij
 = (–1)
i + j
.
 
M
ij
Illustration 3 : Find the minors and cofactors of elements '–3', '5', '–1' & '7' in the determinant 
2 3 1
4 0 5
1 6 7
?
?
Solution : Minor of –3 = 
4 5
1 7 ?
 = 33 ;  Cofactor of – 3 = –33
Minor of 5 = 
2 3
9
1 6
?
?
?
 ; Cofactor of 5 = –9
Minor of –1 = 
3 1
15
0 5
?
? ?
; Cofactor of –1 = –15
Minor of 7 = 
2 3
12
4 0
?
?
; Cofactor of 7 = 12
JEEMAIN.GURU
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J E E - M a t h e m a t i c s
1 . INTRODUCTION :
If the equations a
1
x + b
1
 = 0, a
2
x + b
2 
= 0 are satisfied by the same value of x, then a
1
b
2 
– a
2
b
1 
= 0. The
expression a
1
b
2  
– a
2
b
1 
is called a determinant of the second order, and is denoted by :
1 1
2 2
a b
a b
A determinant of second order consists of two rows and two columns.
Next consider the system of equations  a
1
x + b
1
y + c
1 
= 0, a
2
x + b
2
y + c
2 
= 0, a
3
x + b
3
y + c
3 
= 0
If these equations are satisfied by the same values of x and y, then on eliminating x and y we get.
a
1
(b
2
c
3
 – b
3
c
2
) + b
1
(c
2
a
3
 – c
3
a
2
) + c
1
(a
2
b
3
 – a
3
b
2
) = 0
The expression on the left is called a determinant of the third order, and is denoted by
1 1 1
2 2 2
3 3 3
a b c
a b c
a b c
A determinant of third order consists of three rows and three columns.
Illustration 1 : Eliminate ?, m, n from the equations a ? + cm + bn = 0, c ? + bm + an = 0,  b ? + am + cn = 0
and express the result in the simplest form.
Solution : The given set of equations can also be written as (if n ? 0) :
m
a c b 0
n n
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?
;   
m
c b a 0
n n
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?
;   
m
b a c 0
n n
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?
Then, let 
m
x ; y
n n
? ?
?
? System of equations :
ax + cy + b = 0 ...(i)
cx + by + a = 0 ...(ii)
bx + ay + c = 0 ...(iii)
We have to eliminate x & y from these simultaneous linear equations.
Since these equations are satisfied by the same values of x and y, then eliminating x
and y we get,
a c b
c b a 0
b a c
?
2 . VALUE OF A DETERMINANT :
1 1 1
2 2 2
3 3 3
a b c
D a b c
a b c
?
2 2 2 2 2 2
1 1 1
3 3 3 3 3 3
b c a c a b
a b c
b c a c a b
? ? ?
= a
1
(b
2
c
3
 – b
3
c
2
) – b
1
(a
2
c
3
 – a
3
c
2
) + c
1
(a
2
b
3
 – a
3
b
2
)
Note : Sarrus diagram to get the value of determinant of order three :
1 1 1
2 2 2
3 3 3
a b c
D a b c
a b c
? = 
a
1
b
1
c
1
a
2
a
3
b
2
b
3
c
2
c
3
a
1
b
1
a
2
a
3
b
2
b
3
 –ve –ve –ve
 +ve +ve +ve
 = (a
1
b
2
c
3
 + a
2
b
3
c
1
 +a
3
b
1
c
2
) – (a
3
b
2
c
1
 + a
2
b
1
c
3 
+ a
1
b
3
c
2
)
Note that the product of the terms in first bracket (i.e. a
1
a
2
a
3
b
1
b
2
b
3
c
1
c
2
c
3
) is same as the product of the terms
in second bracket.
DETERMINANT
JEEMAIN.GURU
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J E E - M a t h e m a t i c s
Illustration 2 : The value of 
1 2 3
4 3 6
2 7 9
?
?
  is -
(A) 213      (B) – 231 (C) 231 (D) 39
Solution :
1 2 3
4 3 6
2 7 9
?
?
 = 1 
3 6
7 9 ?
 –2
4 6 4 3
3
2 9 2 7
? ?
?
?
= (27 + 42) – 2 (–36 –12) + 3 (28 – 6) = 231
Alternative :  By sarrus diagram
1 2 3
4 3 6
2 7 9
?
?
=
1 2 3
4 3 6
2 7 9
1 2
4 3
2 7
?
?
?
?
= (27 + 24 + 84) – (18 – 42 – 72)= 135 – (18 – 114) = 231 Ans. (C)
3 . MINORS & COFACTORS :
The minor of a given element of determinant is the determinant obtained by deleting the row & the column in
which the given element stands.
For example, the minor of a
1
 in 
1 1 1
2 2 2
3 3 3
a b c
a b c
a b c
 is 
2 2
3 3
b c
b c
 & the minor of b
2
 is 
1 1
3 3
a c
a c
.
Hence a determinant of order three will have “9 minors”.
If M
ij
 represents the minor of the element belonging to i
th 
row and j
th 
column then the cofactor of that element is
given by : C
ij
 = (–1)
i + j
.
 
M
ij
Illustration 3 : Find the minors and cofactors of elements '–3', '5', '–1' & '7' in the determinant 
2 3 1
4 0 5
1 6 7
?
?
Solution : Minor of –3 = 
4 5
1 7 ?
 = 33 ;  Cofactor of – 3 = –33
Minor of 5 = 
2 3
9
1 6
?
?
?
 ; Cofactor of 5 = –9
Minor of –1 = 
3 1
15
0 5
?
? ?
; Cofactor of –1 = –15
Minor of 7 = 
2 3
12
4 0
?
?
; Cofactor of 7 = 12
JEEMAIN.GURU
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J E E - M a t h e m a t i c s
4 . EXPANSION OF A DETERMINANT IN TERMS OF THE ELEMENTS OF ANY ROW OR COLUMN:
Let D = 
1 1 1
2 2 2
3 3 3
a b c
a b c
a b c
(i) The sum of the product of elements of any row (column) with their corresponding cofactors is always equal
to the value of the determinant.
D can be expressed in any of the six forms :
a
1
A
1
 + b
1
B
1
 + c
1
C
1
, a
1
A
1
 + a
2
A
2
 + a
3
A
3
,
a
2
A
2
 + b
2
B
2
 + c
2
C
2
, b
1
B
1
 + b
2
B
2
 + b
3
B
3
,
a
3
A
3
 + b
3
B
3
 + c
3
C
3
, c
1
C
1
 + c
2
C
2
 + c
3
C
3
,
where A
i
,B
i
 & C
i
 (i = 1,2,3) denote cofactors of a
i
,b
i
 & c
i
 respectively.
(ii) The sum of the product of elements of any row (column) with the cofactors of other row (column) is always
equal to zero.
Hence,
a
2
A
1
 + b
2
B
1
 + c
2
C
1
 = 0,
b
1
A
1
 + b
2
A
2
 + b
3
A
3
 = 0 and so on.
where A
i
,B
i
 & C
i
 (i = 1,2,3) denote cofactors of a
i
,b
i
 & c
i
 respectively.
Do yourself -1 :
(i) Find minors & cofactors of elements '6', '5', '0' & '4' of the determinant 
2 1 3
6 5 7
3 0 4
.
(ii) Calculate the value of the determinant 
5 3 7
2 4 8
9 3 10
?
? ?
?
(iii) The value of the determinant 
a b 0
0 a b
b 0 a
 is equal to -
(A) a
3
 – b
3
(B) a
3
 + b
3
(C) 0 (D) none of these
(iv) Find the value of 'k', if 
1 2 0
2 3 1 4
3 k 2
?
(v) Prove that 
2 2 2
1 z y
z 1 x 1 x y z
y x 1
?
? ? ? ? ?
?
5 . PROPERTIES OF DETERMINANTS :
( a ) The value of a determinant remains unaltered, if the rows & columns are inter-changed,
e.g. if 
1 2 3 1 1 1
1 2 3 2 2 2
1 2 3 3 3 3
a a a a b c
b b b a b c D
c c c a b c
? ?
JEEMAIN.GURU
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J E E - M a t h e m a t i c s
1 . INTRODUCTION :
If the equations a
1
x + b
1
 = 0, a
2
x + b
2 
= 0 are satisfied by the same value of x, then a
1
b
2 
– a
2
b
1 
= 0. The
expression a
1
b
2  
– a
2
b
1 
is called a determinant of the second order, and is denoted by :
1 1
2 2
a b
a b
A determinant of second order consists of two rows and two columns.
Next consider the system of equations  a
1
x + b
1
y + c
1 
= 0, a
2
x + b
2
y + c
2 
= 0, a
3
x + b
3
y + c
3 
= 0
If these equations are satisfied by the same values of x and y, then on eliminating x and y we get.
a
1
(b
2
c
3
 – b
3
c
2
) + b
1
(c
2
a
3
 – c
3
a
2
) + c
1
(a
2
b
3
 – a
3
b
2
) = 0
The expression on the left is called a determinant of the third order, and is denoted by
1 1 1
2 2 2
3 3 3
a b c
a b c
a b c
A determinant of third order consists of three rows and three columns.
Illustration 1 : Eliminate ?, m, n from the equations a ? + cm + bn = 0, c ? + bm + an = 0,  b ? + am + cn = 0
and express the result in the simplest form.
Solution : The given set of equations can also be written as (if n ? 0) :
m
a c b 0
n n
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?
;   
m
c b a 0
n n
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?
;   
m
b a c 0
n n
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?
Then, let 
m
x ; y
n n
? ?
?
? System of equations :
ax + cy + b = 0 ...(i)
cx + by + a = 0 ...(ii)
bx + ay + c = 0 ...(iii)
We have to eliminate x & y from these simultaneous linear equations.
Since these equations are satisfied by the same values of x and y, then eliminating x
and y we get,
a c b
c b a 0
b a c
?
2 . VALUE OF A DETERMINANT :
1 1 1
2 2 2
3 3 3
a b c
D a b c
a b c
?
2 2 2 2 2 2
1 1 1
3 3 3 3 3 3
b c a c a b
a b c
b c a c a b
? ? ?
= a
1
(b
2
c
3
 – b
3
c
2
) – b
1
(a
2
c
3
 – a
3
c
2
) + c
1
(a
2
b
3
 – a
3
b
2
)
Note : Sarrus diagram to get the value of determinant of order three :
1 1 1
2 2 2
3 3 3
a b c
D a b c
a b c
? = 
a
1
b
1
c
1
a
2
a
3
b
2
b
3
c
2
c
3
a
1
b
1
a
2
a
3
b
2
b
3
 –ve –ve –ve
 +ve +ve +ve
 = (a
1
b
2
c
3
 + a
2
b
3
c
1
 +a
3
b
1
c
2
) – (a
3
b
2
c
1
 + a
2
b
1
c
3 
+ a
1
b
3
c
2
)
Note that the product of the terms in first bracket (i.e. a
1
a
2
a
3
b
1
b
2
b
3
c
1
c
2
c
3
) is same as the product of the terms
in second bracket.
DETERMINANT
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Illustration 2 : The value of 
1 2 3
4 3 6
2 7 9
?
?
  is -
(A) 213      (B) – 231 (C) 231 (D) 39
Solution :
1 2 3
4 3 6
2 7 9
?
?
 = 1 
3 6
7 9 ?
 –2
4 6 4 3
3
2 9 2 7
? ?
?
?
= (27 + 42) – 2 (–36 –12) + 3 (28 – 6) = 231
Alternative :  By sarrus diagram
1 2 3
4 3 6
2 7 9
?
?
=
1 2 3
4 3 6
2 7 9
1 2
4 3
2 7
?
?
?
?
= (27 + 24 + 84) – (18 – 42 – 72)= 135 – (18 – 114) = 231 Ans. (C)
3 . MINORS & COFACTORS :
The minor of a given element of determinant is the determinant obtained by deleting the row & the column in
which the given element stands.
For example, the minor of a
1
 in 
1 1 1
2 2 2
3 3 3
a b c
a b c
a b c
 is 
2 2
3 3
b c
b c
 & the minor of b
2
 is 
1 1
3 3
a c
a c
.
Hence a determinant of order three will have “9 minors”.
If M
ij
 represents the minor of the element belonging to i
th 
row and j
th 
column then the cofactor of that element is
given by : C
ij
 = (–1)
i + j
.
 
M
ij
Illustration 3 : Find the minors and cofactors of elements '–3', '5', '–1' & '7' in the determinant 
2 3 1
4 0 5
1 6 7
?
?
Solution : Minor of –3 = 
4 5
1 7 ?
 = 33 ;  Cofactor of – 3 = –33
Minor of 5 = 
2 3
9
1 6
?
?
?
 ; Cofactor of 5 = –9
Minor of –1 = 
3 1
15
0 5
?
? ?
; Cofactor of –1 = –15
Minor of 7 = 
2 3
12
4 0
?
?
; Cofactor of 7 = 12
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4 . EXPANSION OF A DETERMINANT IN TERMS OF THE ELEMENTS OF ANY ROW OR COLUMN:
Let D = 
1 1 1
2 2 2
3 3 3
a b c
a b c
a b c
(i) The sum of the product of elements of any row (column) with their corresponding cofactors is always equal
to the value of the determinant.
D can be expressed in any of the six forms :
a
1
A
1
 + b
1
B
1
 + c
1
C
1
, a
1
A
1
 + a
2
A
2
 + a
3
A
3
,
a
2
A
2
 + b
2
B
2
 + c
2
C
2
, b
1
B
1
 + b
2
B
2
 + b
3
B
3
,
a
3
A
3
 + b
3
B
3
 + c
3
C
3
, c
1
C
1
 + c
2
C
2
 + c
3
C
3
,
where A
i
,B
i
 & C
i
 (i = 1,2,3) denote cofactors of a
i
,b
i
 & c
i
 respectively.
(ii) The sum of the product of elements of any row (column) with the cofactors of other row (column) is always
equal to zero.
Hence,
a
2
A
1
 + b
2
B
1
 + c
2
C
1
 = 0,
b
1
A
1
 + b
2
A
2
 + b
3
A
3
 = 0 and so on.
where A
i
,B
i
 & C
i
 (i = 1,2,3) denote cofactors of a
i
,b
i
 & c
i
 respectively.
Do yourself -1 :
(i) Find minors & cofactors of elements '6', '5', '0' & '4' of the determinant 
2 1 3
6 5 7
3 0 4
.
(ii) Calculate the value of the determinant 
5 3 7
2 4 8
9 3 10
?
? ?
?
(iii) The value of the determinant 
a b 0
0 a b
b 0 a
 is equal to -
(A) a
3
 – b
3
(B) a
3
 + b
3
(C) 0 (D) none of these
(iv) Find the value of 'k', if 
1 2 0
2 3 1 4
3 k 2
?
(v) Prove that 
2 2 2
1 z y
z 1 x 1 x y z
y x 1
?
? ? ? ? ?
?
5 . PROPERTIES OF DETERMINANTS :
( a ) The value of a determinant remains unaltered, if the rows & columns are inter-changed,
e.g. if 
1 2 3 1 1 1
1 2 3 2 2 2
1 2 3 3 3 3
a a a a b c
b b b a b c D
c c c a b c
? ?
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(b ) If any two rows (or columns) of a determinant be interchanged, the value of determinant is changed in sign
only.  e.g.
1 1 1 2 2 2
2 2 2 1 1 1 1
3 3 3 3 3 3
a b c a b c
a b c a b c Let D & D
a b c a b c
? ?
 Then D
1
 = – D.
(c) If all the elements of a row (or column) are zero, then the value of the determinant is zero.
(d ) If all the elements of any row (or column) are multiplied by the same number, then the determinant is
multiplied by that number.
e.g.     If D = 
1 1 1
2 2 2
3 3 3
a b c
a b c
a b c
  and      D
1
 = 
1 1 1
2 2 2
3 3 3
Ka Kb Kc
a b c
a b c
 Then D
1
 = KD
(e) If all the elements of a row (or column) are proportional (or identical) to the element of any other row,
then the determinant vanishes, i.e. its value is zero.
e.g. If D = 
1 1 1
1 1 1
3 3 3
a b c
a b c
a b c
 ? D = 0 ; If 
1 1 1
1 1 1 1 1
3 3 3
a b c
D ka kb kc D 0
a b c
? ? ?
Illustration 4 : Prove that 
a b c y b q
x y z x a p
p q r z c r
?
Solution : D = 
a b c
x y z
p q r
= 
a x p
b y q
c z r
(By interchanging rows & columns)
= 
x a p
y b q
z c r
? (C
1
 ? C
2
)
= 
y b q
x a p
z c r
(R
1
 ? R
2
)
Illustration 5 : Find the value of the determinant 
2
2
2
a ab ac
ab b bc
ac bc c
Solution : D = 
2
2
2
a ab ac
ab b bc
ac bc c
= a
2
2
a b c
ab b bc
ac bc c
=
a b c
abc a b c
a b c
= 0
Since all rows are same, hence value of the determinant is zero.
JEEMAIN.GURU
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J E E - M a t h e m a t i c s
1 . INTRODUCTION :
If the equations a
1
x + b
1
 = 0, a
2
x + b
2 
= 0 are satisfied by the same value of x, then a
1
b
2 
– a
2
b
1 
= 0. The
expression a
1
b
2  
– a
2
b
1 
is called a determinant of the second order, and is denoted by :
1 1
2 2
a b
a b
A determinant of second order consists of two rows and two columns.
Next consider the system of equations  a
1
x + b
1
y + c
1 
= 0, a
2
x + b
2
y + c
2 
= 0, a
3
x + b
3
y + c
3 
= 0
If these equations are satisfied by the same values of x and y, then on eliminating x and y we get.
a
1
(b
2
c
3
 – b
3
c
2
) + b
1
(c
2
a
3
 – c
3
a
2
) + c
1
(a
2
b
3
 – a
3
b
2
) = 0
The expression on the left is called a determinant of the third order, and is denoted by
1 1 1
2 2 2
3 3 3
a b c
a b c
a b c
A determinant of third order consists of three rows and three columns.
Illustration 1 : Eliminate ?, m, n from the equations a ? + cm + bn = 0, c ? + bm + an = 0,  b ? + am + cn = 0
and express the result in the simplest form.
Solution : The given set of equations can also be written as (if n ? 0) :
m
a c b 0
n n
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?
;   
m
c b a 0
n n
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?
;   
m
b a c 0
n n
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?
Then, let 
m
x ; y
n n
? ?
?
? System of equations :
ax + cy + b = 0 ...(i)
cx + by + a = 0 ...(ii)
bx + ay + c = 0 ...(iii)
We have to eliminate x & y from these simultaneous linear equations.
Since these equations are satisfied by the same values of x and y, then eliminating x
and y we get,
a c b
c b a 0
b a c
?
2 . VALUE OF A DETERMINANT :
1 1 1
2 2 2
3 3 3
a b c
D a b c
a b c
?
2 2 2 2 2 2
1 1 1
3 3 3 3 3 3
b c a c a b
a b c
b c a c a b
? ? ?
= a
1
(b
2
c
3
 – b
3
c
2
) – b
1
(a
2
c
3
 – a
3
c
2
) + c
1
(a
2
b
3
 – a
3
b
2
)
Note : Sarrus diagram to get the value of determinant of order three :
1 1 1
2 2 2
3 3 3
a b c
D a b c
a b c
? = 
a
1
b
1
c
1
a
2
a
3
b
2
b
3
c
2
c
3
a
1
b
1
a
2
a
3
b
2
b
3
 –ve –ve –ve
 +ve +ve +ve
 = (a
1
b
2
c
3
 + a
2
b
3
c
1
 +a
3
b
1
c
2
) – (a
3
b
2
c
1
 + a
2
b
1
c
3 
+ a
1
b
3
c
2
)
Note that the product of the terms in first bracket (i.e. a
1
a
2
a
3
b
1
b
2
b
3
c
1
c
2
c
3
) is same as the product of the terms
in second bracket.
DETERMINANT
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J E E - M a t h e m a t i c s
Illustration 2 : The value of 
1 2 3
4 3 6
2 7 9
?
?
  is -
(A) 213      (B) – 231 (C) 231 (D) 39
Solution :
1 2 3
4 3 6
2 7 9
?
?
 = 1 
3 6
7 9 ?
 –2
4 6 4 3
3
2 9 2 7
? ?
?
?
= (27 + 42) – 2 (–36 –12) + 3 (28 – 6) = 231
Alternative :  By sarrus diagram
1 2 3
4 3 6
2 7 9
?
?
=
1 2 3
4 3 6
2 7 9
1 2
4 3
2 7
?
?
?
?
= (27 + 24 + 84) – (18 – 42 – 72)= 135 – (18 – 114) = 231 Ans. (C)
3 . MINORS & COFACTORS :
The minor of a given element of determinant is the determinant obtained by deleting the row & the column in
which the given element stands.
For example, the minor of a
1
 in 
1 1 1
2 2 2
3 3 3
a b c
a b c
a b c
 is 
2 2
3 3
b c
b c
 & the minor of b
2
 is 
1 1
3 3
a c
a c
.
Hence a determinant of order three will have “9 minors”.
If M
ij
 represents the minor of the element belonging to i
th 
row and j
th 
column then the cofactor of that element is
given by : C
ij
 = (–1)
i + j
.
 
M
ij
Illustration 3 : Find the minors and cofactors of elements '–3', '5', '–1' & '7' in the determinant 
2 3 1
4 0 5
1 6 7
?
?
Solution : Minor of –3 = 
4 5
1 7 ?
 = 33 ;  Cofactor of – 3 = –33
Minor of 5 = 
2 3
9
1 6
?
?
?
 ; Cofactor of 5 = –9
Minor of –1 = 
3 1
15
0 5
?
? ?
; Cofactor of –1 = –15
Minor of 7 = 
2 3
12
4 0
?
?
; Cofactor of 7 = 12
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4 . EXPANSION OF A DETERMINANT IN TERMS OF THE ELEMENTS OF ANY ROW OR COLUMN:
Let D = 
1 1 1
2 2 2
3 3 3
a b c
a b c
a b c
(i) The sum of the product of elements of any row (column) with their corresponding cofactors is always equal
to the value of the determinant.
D can be expressed in any of the six forms :
a
1
A
1
 + b
1
B
1
 + c
1
C
1
, a
1
A
1
 + a
2
A
2
 + a
3
A
3
,
a
2
A
2
 + b
2
B
2
 + c
2
C
2
, b
1
B
1
 + b
2
B
2
 + b
3
B
3
,
a
3
A
3
 + b
3
B
3
 + c
3
C
3
, c
1
C
1
 + c
2
C
2
 + c
3
C
3
,
where A
i
,B
i
 & C
i
 (i = 1,2,3) denote cofactors of a
i
,b
i
 & c
i
 respectively.
(ii) The sum of the product of elements of any row (column) with the cofactors of other row (column) is always
equal to zero.
Hence,
a
2
A
1
 + b
2
B
1
 + c
2
C
1
 = 0,
b
1
A
1
 + b
2
A
2
 + b
3
A
3
 = 0 and so on.
where A
i
,B
i
 & C
i
 (i = 1,2,3) denote cofactors of a
i
,b
i
 & c
i
 respectively.
Do yourself -1 :
(i) Find minors & cofactors of elements '6', '5', '0' & '4' of the determinant 
2 1 3
6 5 7
3 0 4
.
(ii) Calculate the value of the determinant 
5 3 7
2 4 8
9 3 10
?
? ?
?
(iii) The value of the determinant 
a b 0
0 a b
b 0 a
 is equal to -
(A) a
3
 – b
3
(B) a
3
 + b
3
(C) 0 (D) none of these
(iv) Find the value of 'k', if 
1 2 0
2 3 1 4
3 k 2
?
(v) Prove that 
2 2 2
1 z y
z 1 x 1 x y z
y x 1
?
? ? ? ? ?
?
5 . PROPERTIES OF DETERMINANTS :
( a ) The value of a determinant remains unaltered, if the rows & columns are inter-changed,
e.g. if 
1 2 3 1 1 1
1 2 3 2 2 2
1 2 3 3 3 3
a a a a b c
b b b a b c D
c c c a b c
? ?
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J E E - M a t h e m a t i c s
(b ) If any two rows (or columns) of a determinant be interchanged, the value of determinant is changed in sign
only.  e.g.
1 1 1 2 2 2
2 2 2 1 1 1 1
3 3 3 3 3 3
a b c a b c
a b c a b c Let D & D
a b c a b c
? ?
 Then D
1
 = – D.
(c) If all the elements of a row (or column) are zero, then the value of the determinant is zero.
(d ) If all the elements of any row (or column) are multiplied by the same number, then the determinant is
multiplied by that number.
e.g.     If D = 
1 1 1
2 2 2
3 3 3
a b c
a b c
a b c
  and      D
1
 = 
1 1 1
2 2 2
3 3 3
Ka Kb Kc
a b c
a b c
 Then D
1
 = KD
(e) If all the elements of a row (or column) are proportional (or identical) to the element of any other row,
then the determinant vanishes, i.e. its value is zero.
e.g. If D = 
1 1 1
1 1 1
3 3 3
a b c
a b c
a b c
 ? D = 0 ; If 
1 1 1
1 1 1 1 1
3 3 3
a b c
D ka kb kc D 0
a b c
? ? ?
Illustration 4 : Prove that 
a b c y b q
x y z x a p
p q r z c r
?
Solution : D = 
a b c
x y z
p q r
= 
a x p
b y q
c z r
(By interchanging rows & columns)
= 
x a p
y b q
z c r
? (C
1
 ? C
2
)
= 
y b q
x a p
z c r
(R
1
 ? R
2
)
Illustration 5 : Find the value of the determinant 
2
2
2
a ab ac
ab b bc
ac bc c
Solution : D = 
2
2
2
a ab ac
ab b bc
ac bc c
= a
2
2
a b c
ab b bc
ac bc c
=
a b c
abc a b c
a b c
= 0
Since all rows are same, hence value of the determinant is zero.
JEEMAIN.GURU
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J E E - M a t h e m a t i c s
Do yourself -2 :
(i) Without expanding the determinant prove that  
a p r n c
b q m q m b 0
c r n p a
? ?
?
?
(ii) If 
D
? ?
?
? ?
, then 
2 2
2 2
? ?
? ?
 is equal to -
(A) D (B) 2D (C) 4D (D) 16D
(iii) If 
p q r
D x y z
m n
?
?
, then KD is equal to -
(A) 
Kp q r
x Ky z
m Kn ?
(B) 
p q r
x y z
K Km Kn ?
(C) 
p Kx
q Ky m
r Kz n
?
(D) 
Kp Kx K
Kq Ky Km
Kr Kz Kn
?
(f) If each element of any row (or column) is expressed as a sum of two (or more) terms, then the determinant
can  be expressed as the sum of two (or more) determinants.
e.g.
1 1 1 1 1 1
2 2 2 2 2 2 2 2 2
3 3 3 3 3 3 3 3 3
a x b y c z a b c x y z
a b c a b c a b c
a b c a b c a b c
? ? ?
? ?
Note that : If    
r
1 1 1
ƒ(r) g(r) h(r)
D a b c
a b c
?
where r ? N and a,b,c, a
1
, b
1
,c
1
 are constants, then
n n n
r 1 r 1 r 1
n
r
r 1
1 1 1
ƒ(r) g(r) h(r)
D a b c
a b c
? ? ?
?
?
? ? ?
?
( g ) Row - column operation : The value of a determinant remains unaltered under a column (C
i
) operation
of the form C
i
 ?  C
i
 + ?C
j
 + ?C
k
 (j, k ? i) or row (R
i
) operation of the form R
i
 ? R
i
 + ?R
j
 + ?R
k
 (j, k ? i).
In other words, the value of a determinant is not altered by adding the elements of any row (or column) to
the same multiples of the  corresponding elements of any other row (or column)
e.g.   Let D = 
1 1 1
2 2 2
3 3 3
a b c
a b c
a b c
D = 
1 2 1 2 1 2
2 2 2
3 1 3 1 3 1
a a b b c c
a b c
a a b b c c
? ? ? ? ? ?
? ? ? ? ? ?
 (R
1
 ??R
1
 + ?R
2
;
 
R
3
 ??R
3
 + ?R
2
)
Note : (i) By using the operation R
i
 ? xR
i
 + yR
j
 + zR
k
 (j, k ? i), the value of the determinant becomes x
times the original one.
(ii) While applying this property ATLEAST ONE ROW (OR COLUMN) must remain unchanged.
JEEMAIN.GURU