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NCERT Solutions Class 12 Maths Chapter 4 - Determinants

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 Page 2


A
12
 = (– 1)
1 + 2
 M
12
 = (– 1)
3
 0 = 0
M
13
 = Minor of a
13
 = 
01
00
 = 0 – 0 = 0,
A
13
 = (– 1)
1 + 3
 M
13
 = (– 1)
4
 0 = 0
M
21
 = Minor of a
21
 = 
00
01
 = 0 – 0 = 0,
A
21
 = (– 1)
2 + 1
 M
21
 = (– 1)
3
 0 = 0
M
22
 = Minor of a
22
 = 
10
01
 = 1 – 0 = 1,
A
22
 = (– 1)
2 + 2
 M
22
 = (– 1)
4
 1 = 1
M
23
 = Minor of a
23
 = 
10
00
 = 0 – 0 = 0,
A
23
 = (– 1)
2 + 3
 M
23
 = (– 1)
5
 0 = 0
M
31
 = Minor of a
31
 = 
00
10
 = 0 – 0 = 0,
A
31
 = (– 1)
3 + 1
 M
31
 = (– 1)
4
 0 = 0
M
32
 = Minor of a
32
 = 
10
00
 = 0 – 0 = 0,
A
32
 = (– 1)
3 + 2
 M
32
 = (– 1)
5
 0 = 0
M
33
 = Minor of a
33
 = 
10
01
 = 1 – 0 = 1,
A
33
 = (– 1)
3 + 3
 M
33
 = (– 1)
6
 1 = 1.
(ii) Let ? =
10 4
35 1
01 2
-
M
11
 = Minor of a
11
 = 
51
12
-
 = 10 – (– 1)  = 10 + 1 = 11,
A
11
 = (– 1)
1 + 1
 M
11
 = (– 1)
2
 11 = 11
M
12
 = Minor of a
12
 = 
31
02
-
 = 6 – 0 = 6,
A
12
 = (– 1)
1 + 2
 M
12
 = (– 1)
3
 6 = – 6
M
13
 = Minor of a
13
 = 
35
01
= 3 – 0 = 3,
A
13
 = (– 1)
1 + 3
 M
13
 = (– 1)
4
 3 = 3
24
Page 3


A
12
 = (– 1)
1 + 2
 M
12
 = (– 1)
3
 0 = 0
M
13
 = Minor of a
13
 = 
01
00
 = 0 – 0 = 0,
A
13
 = (– 1)
1 + 3
 M
13
 = (– 1)
4
 0 = 0
M
21
 = Minor of a
21
 = 
00
01
 = 0 – 0 = 0,
A
21
 = (– 1)
2 + 1
 M
21
 = (– 1)
3
 0 = 0
M
22
 = Minor of a
22
 = 
10
01
 = 1 – 0 = 1,
A
22
 = (– 1)
2 + 2
 M
22
 = (– 1)
4
 1 = 1
M
23
 = Minor of a
23
 = 
10
00
 = 0 – 0 = 0,
A
23
 = (– 1)
2 + 3
 M
23
 = (– 1)
5
 0 = 0
M
31
 = Minor of a
31
 = 
00
10
 = 0 – 0 = 0,
A
31
 = (– 1)
3 + 1
 M
31
 = (– 1)
4
 0 = 0
M
32
 = Minor of a
32
 = 
10
00
 = 0 – 0 = 0,
A
32
 = (– 1)
3 + 2
 M
32
 = (– 1)
5
 0 = 0
M
33
 = Minor of a
33
 = 
10
01
 = 1 – 0 = 1,
A
33
 = (– 1)
3 + 3
 M
33
 = (– 1)
6
 1 = 1.
(ii) Let ? =
10 4
35 1
01 2
-
M
11
 = Minor of a
11
 = 
51
12
-
 = 10 – (– 1)  = 10 + 1 = 11,
A
11
 = (– 1)
1 + 1
 M
11
 = (– 1)
2
 11 = 11
M
12
 = Minor of a
12
 = 
31
02
-
 = 6 – 0 = 6,
A
12
 = (– 1)
1 + 2
 M
12
 = (– 1)
3
 6 = – 6
M
13
 = Minor of a
13
 = 
35
01
= 3 – 0 = 3,
A
13
 = (– 1)
1 + 3
 M
13
 = (– 1)
4
 3 = 3
24
M
21
 = Minor of a
21
 = 
04
12
 = 0 – 4 = – 4,
A
21
 = (– 1)
2 + 1
 M
21
 = (– 1)
3
 (– 4) = 4
M
22
 = Minor of a
22
 = 
14
02
 = 2 – 0 = 2,
A
22
 = (– 1)
2 + 2
 M
22
 = (– 1)
4
 2 = 2
M
23
 = Minor of a
23
 = 
10
01
 = 1 – 0 = 1,
A
23
 = (– 1)
2 + 3
 M
23
 = (– 1)
5
 1 = – 1
M
31
 = Minor of a
31
 = 
04
51 -
 = 0 – 20 = – 20,
A
31
 = (– 1)
3 + 1
 M
31
 = (– 1)
4
 (– 20) = – 20
M
32
 = Minor of a
32
 = 
14
31 -
 = – 1 – 12 = – 13,
A
32
 = (– 1)
3 + 2
 M
32
 = (– 1)
5
 (– 13) = 13
M
33
 = Minor of a
33
 = 
10
35
 = 5 – 0 = 5,
A
33
 = (– 1)
3 + 3
 M
33
 = (– 1)
6
 5 = 5.
Note. Two Most Important Results
1. Sum of the products of the elements of any row or column
of a determinant ? with their corresponding factors is = ?.
i.e., ? ? ? ? ? = a
11
A
11
 + a
12
A
12
 + a
13
A
13
 etc.
2. Sum of the products of the elements of any row or column
of a determinant ? with the cofactors of any other row or
column of ? is zero.
For example, a
11
A
21
 + a
12
A
22
 + a
13
A
23
 = 0.
3. Using Cofactors of elements of second row, evaluate
? ? ? ? ? =
53 8
20 1
12 3
.
Sol. ? = 
53 8
201
12 3
Elements of second row of ? are a
21
 = 2, a
22
 = 0, a
23
 = 1
A
21
 = Cofactor of a
21
 = (– 1)
2 + 1
 
38
23
(
.
.
.
A
ij
 = (– 1)
i + j
 M
ij
]
? ?
(determinant obtained by omitting second row and first column of ?)
= (– 1)
3
 (9 – 16) = – (– 7) = 7
25
Page 4


A
12
 = (– 1)
1 + 2
 M
12
 = (– 1)
3
 0 = 0
M
13
 = Minor of a
13
 = 
01
00
 = 0 – 0 = 0,
A
13
 = (– 1)
1 + 3
 M
13
 = (– 1)
4
 0 = 0
M
21
 = Minor of a
21
 = 
00
01
 = 0 – 0 = 0,
A
21
 = (– 1)
2 + 1
 M
21
 = (– 1)
3
 0 = 0
M
22
 = Minor of a
22
 = 
10
01
 = 1 – 0 = 1,
A
22
 = (– 1)
2 + 2
 M
22
 = (– 1)
4
 1 = 1
M
23
 = Minor of a
23
 = 
10
00
 = 0 – 0 = 0,
A
23
 = (– 1)
2 + 3
 M
23
 = (– 1)
5
 0 = 0
M
31
 = Minor of a
31
 = 
00
10
 = 0 – 0 = 0,
A
31
 = (– 1)
3 + 1
 M
31
 = (– 1)
4
 0 = 0
M
32
 = Minor of a
32
 = 
10
00
 = 0 – 0 = 0,
A
32
 = (– 1)
3 + 2
 M
32
 = (– 1)
5
 0 = 0
M
33
 = Minor of a
33
 = 
10
01
 = 1 – 0 = 1,
A
33
 = (– 1)
3 + 3
 M
33
 = (– 1)
6
 1 = 1.
(ii) Let ? =
10 4
35 1
01 2
-
M
11
 = Minor of a
11
 = 
51
12
-
 = 10 – (– 1)  = 10 + 1 = 11,
A
11
 = (– 1)
1 + 1
 M
11
 = (– 1)
2
 11 = 11
M
12
 = Minor of a
12
 = 
31
02
-
 = 6 – 0 = 6,
A
12
 = (– 1)
1 + 2
 M
12
 = (– 1)
3
 6 = – 6
M
13
 = Minor of a
13
 = 
35
01
= 3 – 0 = 3,
A
13
 = (– 1)
1 + 3
 M
13
 = (– 1)
4
 3 = 3
24
M
21
 = Minor of a
21
 = 
04
12
 = 0 – 4 = – 4,
A
21
 = (– 1)
2 + 1
 M
21
 = (– 1)
3
 (– 4) = 4
M
22
 = Minor of a
22
 = 
14
02
 = 2 – 0 = 2,
A
22
 = (– 1)
2 + 2
 M
22
 = (– 1)
4
 2 = 2
M
23
 = Minor of a
23
 = 
10
01
 = 1 – 0 = 1,
A
23
 = (– 1)
2 + 3
 M
23
 = (– 1)
5
 1 = – 1
M
31
 = Minor of a
31
 = 
04
51 -
 = 0 – 20 = – 20,
A
31
 = (– 1)
3 + 1
 M
31
 = (– 1)
4
 (– 20) = – 20
M
32
 = Minor of a
32
 = 
14
31 -
 = – 1 – 12 = – 13,
A
32
 = (– 1)
3 + 2
 M
32
 = (– 1)
5
 (– 13) = 13
M
33
 = Minor of a
33
 = 
10
35
 = 5 – 0 = 5,
A
33
 = (– 1)
3 + 3
 M
33
 = (– 1)
6
 5 = 5.
Note. Two Most Important Results
1. Sum of the products of the elements of any row or column
of a determinant ? with their corresponding factors is = ?.
i.e., ? ? ? ? ? = a
11
A
11
 + a
12
A
12
 + a
13
A
13
 etc.
2. Sum of the products of the elements of any row or column
of a determinant ? with the cofactors of any other row or
column of ? is zero.
For example, a
11
A
21
 + a
12
A
22
 + a
13
A
23
 = 0.
3. Using Cofactors of elements of second row, evaluate
? ? ? ? ? =
53 8
20 1
12 3
.
Sol. ? = 
53 8
201
12 3
Elements of second row of ? are a
21
 = 2, a
22
 = 0, a
23
 = 1
A
21
 = Cofactor of a
21
 = (– 1)
2 + 1
 
38
23
(
.
.
.
A
ij
 = (– 1)
i + j
 M
ij
]
? ?
(determinant obtained by omitting second row and first column of ?)
= (– 1)
3
 (9 – 16) = – (– 7) = 7
25
A
22
 = Cofactor of a
22
 = (– 1)
2 + 2
 
58
13
  = (– 1)
4
 (15 – 8) = 7
 A
23
 = Cofactor a
23
 = (– 1)
2 + 3
 
53
12
   = (– 1)
5
 (10 – 3) = – 7
Now by Result I of Note after the solution of Q. No. 2,
? ? ? ? ? = a
21
 A
21
 + a
22
A
22
 + a
23
A
23
 = 2(7) + 0(7) + 1(– 7) = 14 – 7 = 7.
Remark. The above method of finding the value of ? is equivalent
to expanding ? along second row.
4. Using Cofactors of elements of third column, evaluate
? ? ? ? ? =
1
1
1
xyz
yzx
zxy
.
Sol. ? = 
1
1
1
xyz
yzx
zxy
Here elements of third column of ? are
 a
13
 = yz, a
23
 = zx, a
33
 = xy
A
13
 = Cofactor of a
13
 = (– 1)
1 + 3
 
1
1
y
z
= (– 1)
4
 (z – y) = z – y
?
(determinant obtained by omitting first row and third column of ?)
A
23
 = Cofactor of a
23
 = (– 1)
2 + 3
1
1
x
z
 = (– 1)
5
 (z – x) = – (z – x)
A
33
 = Cofactor of a
33
 = (– 1)
3 + 3
 
1
1
x
y
= (– 1)
6
 (y – x) = y – x
Now by Result I of Note after the solution of Q. NO. 2,
? ? ? ? ? = a
13
A
13
 + a
23
A
23
 + a
33
A
33
= yz(z – y) + zx[– (z – x)] + xy( y – x)
= yz
2
 – y
2
z – z
2
x + zx
2
 + xy
2
 – x
2
y
= (yz
2
 – y
2
z) + (xy
2
 – xz
2
) + (zx
2
 – x
2
y)
= yz(z – y) + x( y
2
 – z
2
) – x
2
( y – z)
= – yz(y – z) + x( y + z)( y – z) – x
2
( y – z)
= ( y – z) [– yz + xy + xz – x
2
]
= ( y – z)[– y(z – x) + x(z – x)]
= ( y – z) (z – x)(– y + x) = (x – y)( y – z)(z – x)
Remark. The above method of finding the value of ? is equivalent
to expanding ? along third column.
26
Page 5


A
12
 = (– 1)
1 + 2
 M
12
 = (– 1)
3
 0 = 0
M
13
 = Minor of a
13
 = 
01
00
 = 0 – 0 = 0,
A
13
 = (– 1)
1 + 3
 M
13
 = (– 1)
4
 0 = 0
M
21
 = Minor of a
21
 = 
00
01
 = 0 – 0 = 0,
A
21
 = (– 1)
2 + 1
 M
21
 = (– 1)
3
 0 = 0
M
22
 = Minor of a
22
 = 
10
01
 = 1 – 0 = 1,
A
22
 = (– 1)
2 + 2
 M
22
 = (– 1)
4
 1 = 1
M
23
 = Minor of a
23
 = 
10
00
 = 0 – 0 = 0,
A
23
 = (– 1)
2 + 3
 M
23
 = (– 1)
5
 0 = 0
M
31
 = Minor of a
31
 = 
00
10
 = 0 – 0 = 0,
A
31
 = (– 1)
3 + 1
 M
31
 = (– 1)
4
 0 = 0
M
32
 = Minor of a
32
 = 
10
00
 = 0 – 0 = 0,
A
32
 = (– 1)
3 + 2
 M
32
 = (– 1)
5
 0 = 0
M
33
 = Minor of a
33
 = 
10
01
 = 1 – 0 = 1,
A
33
 = (– 1)
3 + 3
 M
33
 = (– 1)
6
 1 = 1.
(ii) Let ? =
10 4
35 1
01 2
-
M
11
 = Minor of a
11
 = 
51
12
-
 = 10 – (– 1)  = 10 + 1 = 11,
A
11
 = (– 1)
1 + 1
 M
11
 = (– 1)
2
 11 = 11
M
12
 = Minor of a
12
 = 
31
02
-
 = 6 – 0 = 6,
A
12
 = (– 1)
1 + 2
 M
12
 = (– 1)
3
 6 = – 6
M
13
 = Minor of a
13
 = 
35
01
= 3 – 0 = 3,
A
13
 = (– 1)
1 + 3
 M
13
 = (– 1)
4
 3 = 3
24
M
21
 = Minor of a
21
 = 
04
12
 = 0 – 4 = – 4,
A
21
 = (– 1)
2 + 1
 M
21
 = (– 1)
3
 (– 4) = 4
M
22
 = Minor of a
22
 = 
14
02
 = 2 – 0 = 2,
A
22
 = (– 1)
2 + 2
 M
22
 = (– 1)
4
 2 = 2
M
23
 = Minor of a
23
 = 
10
01
 = 1 – 0 = 1,
A
23
 = (– 1)
2 + 3
 M
23
 = (– 1)
5
 1 = – 1
M
31
 = Minor of a
31
 = 
04
51 -
 = 0 – 20 = – 20,
A
31
 = (– 1)
3 + 1
 M
31
 = (– 1)
4
 (– 20) = – 20
M
32
 = Minor of a
32
 = 
14
31 -
 = – 1 – 12 = – 13,
A
32
 = (– 1)
3 + 2
 M
32
 = (– 1)
5
 (– 13) = 13
M
33
 = Minor of a
33
 = 
10
35
 = 5 – 0 = 5,
A
33
 = (– 1)
3 + 3
 M
33
 = (– 1)
6
 5 = 5.
Note. Two Most Important Results
1. Sum of the products of the elements of any row or column
of a determinant ? with their corresponding factors is = ?.
i.e., ? ? ? ? ? = a
11
A
11
 + a
12
A
12
 + a
13
A
13
 etc.
2. Sum of the products of the elements of any row or column
of a determinant ? with the cofactors of any other row or
column of ? is zero.
For example, a
11
A
21
 + a
12
A
22
 + a
13
A
23
 = 0.
3. Using Cofactors of elements of second row, evaluate
? ? ? ? ? =
53 8
20 1
12 3
.
Sol. ? = 
53 8
201
12 3
Elements of second row of ? are a
21
 = 2, a
22
 = 0, a
23
 = 1
A
21
 = Cofactor of a
21
 = (– 1)
2 + 1
 
38
23
(
.
.
.
A
ij
 = (– 1)
i + j
 M
ij
]
? ?
(determinant obtained by omitting second row and first column of ?)
= (– 1)
3
 (9 – 16) = – (– 7) = 7
25
A
22
 = Cofactor of a
22
 = (– 1)
2 + 2
 
58
13
  = (– 1)
4
 (15 – 8) = 7
 A
23
 = Cofactor a
23
 = (– 1)
2 + 3
 
53
12
   = (– 1)
5
 (10 – 3) = – 7
Now by Result I of Note after the solution of Q. No. 2,
? ? ? ? ? = a
21
 A
21
 + a
22
A
22
 + a
23
A
23
 = 2(7) + 0(7) + 1(– 7) = 14 – 7 = 7.
Remark. The above method of finding the value of ? is equivalent
to expanding ? along second row.
4. Using Cofactors of elements of third column, evaluate
? ? ? ? ? =
1
1
1
xyz
yzx
zxy
.
Sol. ? = 
1
1
1
xyz
yzx
zxy
Here elements of third column of ? are
 a
13
 = yz, a
23
 = zx, a
33
 = xy
A
13
 = Cofactor of a
13
 = (– 1)
1 + 3
 
1
1
y
z
= (– 1)
4
 (z – y) = z – y
?
(determinant obtained by omitting first row and third column of ?)
A
23
 = Cofactor of a
23
 = (– 1)
2 + 3
1
1
x
z
 = (– 1)
5
 (z – x) = – (z – x)
A
33
 = Cofactor of a
33
 = (– 1)
3 + 3
 
1
1
x
y
= (– 1)
6
 (y – x) = y – x
Now by Result I of Note after the solution of Q. NO. 2,
? ? ? ? ? = a
13
A
13
 + a
23
A
23
 + a
33
A
33
= yz(z – y) + zx[– (z – x)] + xy( y – x)
= yz
2
 – y
2
z – z
2
x + zx
2
 + xy
2
 – x
2
y
= (yz
2
 – y
2
z) + (xy
2
 – xz
2
) + (zx
2
 – x
2
y)
= yz(z – y) + x( y
2
 – z
2
) – x
2
( y – z)
= – yz(y – z) + x( y + z)( y – z) – x
2
( y – z)
= ( y – z) [– yz + xy + xz – x
2
]
= ( y – z)[– y(z – x) + x(z – x)]
= ( y – z) (z – x)(– y + x) = (x – y)( y – z)(z – x)
Remark. The above method of finding the value of ? is equivalent
to expanding ? along third column.
26
5. If ? ? ? ? ? =
11 12 13
21 22 23
31 32 33
aa a
aa a
aa a
 and A
ij
 is Cofactor of a
ij
, then value
of ? ? ? ? ? is given by
(A) a
11
A
31
 + a
12
A
32
 + a
13
A
33
(B) a
11
A
11
 + a
12
A
21
 + a
13
A
31
(C) a
21
A
11
 + a
22
A
12
 + a
23
A
13
(D) a
11
A
11
 + a
21
A
21
 + a
31
A
31
.
Sol. Option (D) is correct answer as given in Result I of Note after
solution of Q. No. 2 and used in the solution of Q. No. 3 and 4
above.
Remark. The values of expressions given in options (A) and
(C) are each equal to zero as given in Result II of Note after
solution of Q. No. 2.
27

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FAQs on NCERT Solutions Class 12 Maths Chapter 4 - Determinants

1. What are determinants and why are they important in mathematics?

Ans. Determinants are mathematical objects that are used to solve systems of linear equations, calculate areas and volumes, and determine the invertibility of matrices. They are important because they provide crucial information about the properties and behavior of linear transformations and matrices.

2. How are determinants calculated for matrices of different dimensions?

Ans. The calculation of determinants depends on the dimensions of the matrix. For 2x2 matrices, the determinant is found by subtracting the product of the diagonal elements from each other. For 3x3 matrices, the determinant can be calculated using a formula involving the cross products of the matrix elements. For higher dimensions, determinants can be calculated recursively using cofactor expansion.

3. What is the significance of the determinant value?

Ans. The determinant value provides important information about the matrix. If the determinant is zero, it implies that the matrix is not invertible and the system of equations represented by the matrix does not have a unique solution. A non-zero determinant indicates that the matrix is invertible and the system of equations has a unique solution.

4. How can determinants be used to find the area of a triangle?

Ans. Determinants can be used to find the area of a triangle by considering the coordinates of its vertices. By forming a matrix with the x-coordinates of the vertices in the first column, the y-coordinates in the second column, and a column of ones in the third column, the absolute value of half the determinant of this matrix gives the area of the triangle.

5. Can determinants be negative?

Ans. Yes, determinants can be negative. The sign of the determinant depends on the arrangement of the elements in the matrix. If the arrangement of the elements leads to an odd number of row exchanges during the calculation of the determinant, the determinant will be negative. If there are an even number of row exchanges, the determinant will be positive.

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