# Class 12 NCERT Solutions – Mathematics Part I – Chapter 4 Determinants – Exercise 4.4

### Question 1.

(i)

(ii)

Solution:

(i)

Finding minors of the elements of the determinant:

Let us assume Mij is Minors of elements aij

M11 = Minor of elements a11 = 3

M12 = Minor of elements a12 = 0

M21 = Minor of elements a21 = âˆ’4

M22 = Minor of elements a22 = 2

Finding cofactor of aij

Let us assume cofactor of aij is Aij Mij

A11 = (âˆ’1)1+1 M11 = (âˆ’1)2 (3) = 3

A12 = (âˆ’1)1+2 M12 = (âˆ’1)3 (0) = 0

A21 = (âˆ’1)2+1 M21 = (âˆ’1)3 (âˆ’4) = 4

A22 = (âˆ’1)2+2 M22 = (âˆ’1)4 (2) = 2

(ii)

Finding minors of the elements of the determinant:

Let us assume Mij is Minors of elements aij

M11 = Minor of element a11 = d

M12 = Minor of elements a12 = b

M21 = Minor of elements a21 = c

M22 = Minor of elements a22 = a

Finding cofactor of aij

Let us assume cofactor of aij is Aij, which is (âˆ’1)i+j Mij

A11 = (âˆ’1)1+1 M11 = (âˆ’1)2 (d) = d

A12 = (âˆ’1)1+2 M12 = (âˆ’1)3 (b) = âˆ’b

A21 = (âˆ’1)2+1 M21 = (âˆ’1)3 (c) = âˆ’c

A22 = (âˆ’1)2+2 M22 = (âˆ’1)4 (a) = a

### Question 2.

(i)

(ii)

Solution:

(i)

Let us find the Minors and cofactors of the elements:

Assume, Mij is minor of element aij and Aij is cofactor of aij

M11 = Minor of elements a11 == 1 âˆ’ 0 = 1 and A11 = 1

M12 = Minor of elements a12 == 0 âˆ’ 0 = 0 and A12 = 0

M13 = Minor of elements a13 == 0 âˆ’ 0 = 0 and A13 = 0

M21 = Minor of elements a21 == 0 âˆ’ 0 = 0 and A21 = 0

M22 = Minor of elements a22 == 1 âˆ’ 0 = 1 and A22 = 1

M23 = Minor of elements a23 == 0 âˆ’ 0 = 0 and A23 = 0

M31 = Minor of elements a31 == 0 âˆ’ 0 = 0 and A31 = 0

M32 = Minor of elements a32 == 0 âˆ’ 0 = 0 and A32 = 0

M33 = Minor of elements a33 == 1 âˆ’ 0 = 1 and A33 = 1

(ii)

Let us find the Minors and cofactors of the elements:

Assume, Mij is minor of element aij and Aij is cofactor of aij

M11 = Minor of elements a11 == 10 âˆ’ (âˆ’1) = 11 and A11 = 11

M12 = Minor of elements a12 == 6 âˆ’ 0 = 6 and A12 = âˆ’6

M13 = Minor of elements a13 == 3 âˆ’ 0 = 3 and A13 = 3

M21 = Minor of elements a21 == 0 âˆ’ 4 = âˆ’4 and A21 = 4

M22 = Minor of elements a22 == 2 âˆ’ 0 = 2 and A22 = 2

M23 = Minor of elements a23 == 1 âˆ’ 0 = 1 and A23 = âˆ’1

M31 = Minor of elements a31 == 0 âˆ’ 20 = âˆ’20 and A31 = âˆ’20

M32 = Minor of elements a32 == âˆ’1 âˆ’ 12 = âˆ’13 and A32 = 13

M33 = Minor of elements a33 == 5 âˆ’ 0 = 5 and A33 = 5

### Question 3. Using Cofactors of elements of second row, evaluate â–³?

Solution:

Finding the Cofactors of elements of second row:

A21 = Cofactor of elements a21 = (âˆ’1)2+1  = (âˆ’1)3 (9 âˆ’ 16) = 7

A22 = Cofactor of elements a22 = (âˆ’1)2+2  = (âˆ’1)4 (15 âˆ’ 8) = 7

A23 = Cofactor of elements a23 = (âˆ’1)2+3  = (âˆ’1)5 (10 âˆ’ 3) = 7

Now, â–³ = a21 A21 + a22 A22 + a23 A23 = 14 + 0 âˆ’ 7 = 7

### Question 4. Using Cofactors of elements of third column, evaluate â–³?

Solution:

Finding the Cofactors of elements of third column:

A13 = Cofactor of elements a13 = (âˆ’1)1+3  = (âˆ’1)4 (z âˆ’ y) = âˆ’y

A23 = Cofactor of elements a23 = (âˆ’1)2+3  = (âˆ’1)5 (z âˆ’ x) = x âˆ’ z

A33 = Cofactor of elements a33 = (âˆ’1)3+3  = (âˆ’1)6 (y âˆ’ x) = y âˆ’ x

Now, â–³ = a13 A13 + a23 A23 + a33 A33

= yz (z âˆ’ y) + zx (x âˆ’ z) + xy (y âˆ’ x)

= (yz2 âˆ’ y2z) + (xy2 âˆ’ xz2) + (xz2 âˆ’ x2y)

= (y âˆ’ z)[âˆ’yz + x(y + z) âˆ’ x2]

= (y âˆ’ z)[âˆ’yz + x (z âˆ’ x) + x (z âˆ’ x)]

= (x âˆ’ y)(y âˆ’ x)(z âˆ’ x)

### Question 5. If and Aij is cofactor of aij then value of â–³ is given by:

(A) a11A31 + a12A32 + a13A33

(B) a11A11 + a12A21 + a13A31

(C) a21A11 + a22A12 + a23A13

(D) a11A11 + a21A21 + a31A31

Solution:

Option (D) is correct.

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