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

• Last Updated : 25 Jan, 2021

### 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 = −4

M22 = Minor of elements a22 = 2

Finding cofactor of aij

Let us assume cofactor of aij is Aij Mij

A11 = (−1)1+1 M11 = (−1)2 (3) = 3

A12 = (−1)1+2 M12 = (−1)3 (0) = 0

A21 = (−1)2+1 M21 = (−1)3 (−4) = 4

A22 = (−1)2+2 M22 = (−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 (−1)i+j Mij

A11 = (−1)1+1 M11 = (−1)2 (d) = d

A12 = (−1)1+2 M12 = (−1)3 (b) = −b

A21 = (−1)2+1 M21 = (−1)3 (c) = −c

A22 = (−1)2+2 M22 = (−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 − 0 = 1 and A11 = 1

M12 = Minor of elements a12 = = 0 − 0 = 0 and A12 = 0

M13 = Minor of elements a13 = = 0 − 0 = 0 and A13 = 0

M21 = Minor of elements a21 = = 0 − 0 = 0 and A21 = 0

M22 = Minor of elements a22 = = 1 − 0 = 1 and A22 = 1

M23 = Minor of elements a23 = = 0 − 0 = 0 and A23 = 0

M31 = Minor of elements a31 = = 0 − 0 = 0 and A31 = 0

M32 = Minor of elements a32 = = 0 − 0 = 0 and A32 = 0

M33 = Minor of elements a33 = = 1 − 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 − (−1) = 11 and A11 = 11

M12 = Minor of elements a12 = = 6 − 0 = 6 and A12 = −6

M13 = Minor of elements a13 = = 3 − 0 = 3 and A13 = 3

M21 = Minor of elements a21 = = 0 − 4 = −4 and A21 = 4

M22 = Minor of elements a22 = = 2 − 0 = 2 and A22 = 2

M23 = Minor of elements a23 = = 1 − 0 = 1 and A23 = −1

M31 = Minor of elements a31 = = 0 − 20 = −20 and A31 = −20

M32 = Minor of elements a32 = = −1 − 12 = −13 and A32 = 13

M33 = Minor of elements a33 = = 5 − 0 = 5 and A33 = 5

### Question 3. Using Cofactors of elements of second row, evaluate △? Solution:

Finding the Cofactors of elements of second row:

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

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

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

Now, △ = a21 A21 + a22 A22 + a23 A23 = 14 + 0 − 7 = 7

### Question 4. Using Cofactors of elements of third column, evaluate △? Solution:

Finding the Cofactors of elements of third column:

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

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

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

Now, △ = a13 A13 + a23 A23 + a33 A33

= yz (z − y) + zx (x − z) + xy (y − x)

= (yz2 − y2z) + (xy2 − xz2) + (xz2 − x2y)

= (y − z)[−yz + x(y + z) − x2]

= (y − z)[−yz + x (z − x) + x (z − x)]

= (x − y)(y − x)(z − x)

### Question 5. If and Aij is cofactor of aij then value of △ 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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