NCERT Solution Exercise 5

Question 1: Write Minors and Cofactors of the elements of following determinants:

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:

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:

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

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

∴ 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:

∴ 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.

∴ Δ = a13A13 + a23A23 + a33A33

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

= yz2 – y2z + x2z – xz2 + xy2 – x2y

= (x2z – y2z) + (yz2 – xz2) + (xy2 – x2y)

= z(x2 – y2) + z2 (y – x) + xy(y – x)

= z(x - y) (x + y) + z2(y - x) + xy (y - x)

= (x - y)[zx + zy – z2 - xy]

= (x - y)[z(x + y) +y  ( z – x)]

= (x - y) (z - x)[-z + y]

=(x - y) (y - z) (z - x)

Hence, Δ =(x - y) (y - z) (z - x)

 

 Question 5: For the matrices A and B, verify that (AB)′ = B’A’ where

Answer

Hence, we have verified that (AB)’ = B’A’.

Hence, we have verified that (AB)’ = B’A’.

 

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