You cannot copy content of this page

NCERT solution class 12 chapter 4 Determinants exercise 4.4 mathematics part 1

EXERCISE 4.4


Page No 126:

Question 1:

Write Minors and Cofactors of the elements of following determinants:

(i)  (ii) 

Answer:

(i) The given determinant is.

Minor of element aij is Mij.

∴M11 = minor of element a11 = 3

M12 = minor of element a12 = 0

M21 = minor of element a21 = −4

M22 = minor of element a22 = 2

Cofactor of aij is Aij = (−1)i + j 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) The given determinant is.

Minor of element aij is Mij.

∴M11 = minor of element a11 d

M12 = minor of element a12 b

M21 = minor of element a21 c

M22 = minor of element a22 a

Cofactor of aij is Aij = (−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) 

Answer:

(i) The given determinant is.

By the definition of minors and cofactors, we have:

M11 = minor of a11

M12 = minor of a12

M13 = minor of a13 

M21 = minor of a21 

M22 = minor of a22 

M23 = minor of a23 

M31 = minor of a31

M32 = minor of a32 

M33 = minor of a33 

A11 = cofactor of a11= (−1)1+1 M11 = 1

A12 = cofactor of a12 = (−1)1+2 M12 = 0

A13 = cofactor of a13 = (−1)1+3 M13 = 0

A21 = cofactor of a21 = (−1)2+1 M21 = 0

A22 = cofactor of a22 = (−1)2+2 M22 = 1

A23 = cofactor of a23 = (−1)2+3 M23 = 0

A31 = cofactor of a31 = (−1)3+1 M31 = 0

A32 = cofactor of a32 = (−1)3+2 M32 = 0

A33 = cofactor of a33 = (−1)3+3 M33 = 1

(ii) The given determinant is.

By definition of minors and cofactors, we have:

M11 = minor of a11

M12 = minor of a12

M13 = minor of a13 

M21 = minor of a21 

M22 = minor of a22 

M23 = minor of a23 

M31 = minor of a31

M32 = minor of a32 

M33 = minor of a33 

A11 = cofactor of a11= (−1)1+1 M11 = 11

A12 = cofactor of a12 = (−1)1+2 M12 = −6

A13 = cofactor of a13 = (−1)1+3 M13 = 3

A21 = cofactor of a21 = (−1)2+1 M21 = 4

A22 = cofactor of a22 = (−1)2+2 M22 = 2

A23 = cofactor of a23 = (−1)2+3 M23 = −1

A31 = cofactor of a31 = (−1)3+1 M31 = −20

A32 = cofactor of a32 = (−1)3+2 M32 = 13

A33 = cofactor of a33 = (−1)3+3 M33 = 5

Question 3:

Using Cofactors of elements of second row, evaluate.

Answer:

The given determinant is.

We have:

M21 

∴A21 = cofactor of a21 = (−1)2+1 M21 = 7

M22 

∴A22 = cofactor of a22 = (−1)2+2 M22 = 7

M23 

∴A23 = cofactor of a23 = (−1)2+3 M23 = −7

We know that Δ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.

∴Δ = a21A21 + a22A22 + a23A23 = 2(7) + 0(7) + 1(−7) = 14 − 7 = 7

Question 4:

Using Cofactors of elements of third column, evaluate

Answer:

The given determinant is.

We have:

M13 

M23 

M33 

∴A13 = cofactor of a13 = (−1)1+3 M13 = (z − y)

A23 = cofactor of a23 = (−1)2+3 M23 = − (z − x) = (x − z)

A33 = cofactor of a33 = (−1)3+3 M33 = (y − x)

We know that Δ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.

Hence, 

Question 5:

If  and Aij is Cofactors of aij, then value of Δ is given by

Answer:

Answer: D

We know that:

Δ = Sum of the product of the elements of a column (or a row) with their corresponding cofactors

∴Δ = a11A11 + a21A21 + a31A31

Hence, the value of Δ is given by the expression given in alternative D.

The correct answer is D.


 

Leave a Comment

Your email address will not be published. Required fields are marked *

error: Content is protected !!
Free Web Hosting