# NCERT Solutions Class 12 maths Chapter-4 (Determinants)Exercise 4.4

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-4 (Determinants)Exercise 4.4 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of  NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

### Exercise 4.4

Write minors and cofactors of the elements of the following determinants:

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)(3) = 3

A12 = (−1)1+2 M12 = (−1)(0) = 0

A21 = (−1)2+1 M21 = (−1)(−4) = 4

A22 = (−1)2+2 M22 = (−1)(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)(d) = d

A12 = (−1)1+2 M12 = (−1)(b) = −b

A21 = (−1)2+1 M21 = (−1)(c) = −c

A22 = (−1)2+2 M22 = (−1)(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)(9 − 16) = 7

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

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

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

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

Now, △ = a13 A13 + a23 A23 + a33 A33

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

= (yz− 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.