Chapter 10 Differentiability Ex 10.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Find whether the following functions is differentiable at

 x = 1 and x = 2 or not:

Solution 6

Question 7(i)

Solution 7(i)

Question 7(ii)

Solution 7(ii)

Question 7(iii)

Solution 7(iii)

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Find the values of a and b, if the function f(x) defined by  is differentiable at x = 1.Solution 11

Given: 

As f(x) is differentiable at x = 1, we have

 … (i)

Now, Rf'(1) exist when (b – 2 – a) = 0 … (ii)

From (i), we get, b = 5

Putting this in (ii), we get, a = 3.

Chapter 10 Differentiability Ex 10.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Examine the differentiability of the function f defined by

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Chapter 9 Continuity Ex 9.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10(i)

Solution 10(i)

Question 10(ii)

Solution 10(ii)

Question 10(iii)

Solution 10(iii)

Question 10(iv)

Solution 10(iv)

Question 10(v)

Solution 10(v)

Question 10(vi)

Solution 10(vi)

Question 10(vii)

Solution 10(vii)

Question 10(viii)

 Discuss the continuity of the following functions at the indicated points:

Solution 10(viii)

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Also sketch the graph of this function.Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Determine the values of a, b, c for which the function

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36(i)

Solution 36(i)

Question 36(ii)

Solution 36(ii)

Question 36(iii)

Solution 36(iii)

Question 36(iv)

Solution 36(iv)

Question 36(v)

Solution 36(v)

Question 36(vi)

Solution 36(vi)

Question 36(vii)

Solution 36(vii)

Question 36(viii)

Solution 36(viii)

Question 36(ix)

In each of the following, find the value of the constant k so that the given function is continuous at the indicated point:

Solution 36(ix)

Question 37

Solution 37

Question 38

Solution 38

Question 39(i)

Solution 39(i)

Question 39(ii)

Solution 39(ii)

Question 40

Solution 40

Question 41

Solution 41

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

is continuous at x = 3.Solution 46

Question 42

For what of  is the function  continuous at x = 0? What about continuity at x = ±1?Solution 42

Given: 

At x = 0, we have

∴ LHL ≠ RHL

So, f(x) is discontinuous at x = 0.

Thus, there is no value of  for which f(x) is continuous at x = 0.

At x = 1, we have

∴ LHL = RHL

So, f(x) is continuous at x = 1.

At x = -1, we have

∴ LHL = RHL

So, f(x) is continuous at x = -1.

Chapter 9 Continuity Ex 9.2

Question 1

Solution 1

Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 3(v)

Solution 3(v)

Question 3(vi)

Solution 3(vi)

Question 3(vii)

Solution 3(vii)

Question 3(viii)

Solution 3(viii)

Question 3(ix)

Solution 3(ix)

Question 3(x)

Solution 3(x)

Question 3(xi)

Solution 3(xi)

Question 3(xii)

Solution 3(xii)

Question 3(xiii)

Solution 3(xiii)

Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 4(iii)

Solution 4(iii)

Question 4(iv)

Solution 4(iv)

Question 4(v)

Solution 4(v)

Question 4(vi)

Solution 4(vi)

Question 4(vii)

Solution 4(vii)

Question 4(viii)

Solution 4(viii)

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Given the function

Find the points of discontinuity of the functions f(f(x)).Solution 18

Question 19

Find all point of discontinuity of the function 

Solution 19

Chapter 8 Solution of Simultaneous Linear Equations Ex. 8.1

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 2(v)

Solution 2(v)

Question 2(vi)

Solution 2(vi)

Question 2(vii)

Solution 2(vii)

Question 2(viii)

Solution 2(viii)

Question 2(ix)

Solution 2(ix)

Question 2(x)

Solution 2(x)

Question 2(xi)

Solution 2(xi)

Question 2(xii)

Solution 2(xii)

Question 2(xiii)

Solution 2(xiii)

Question 2(xiv)

Solution 2(xiv)

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 3(v)

Solution 3(v)

Question 3(vi)

Solution 3(vi)

Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 4(iii)

Solution 4(iii)

Question 4(iv)

Solution 4(iv)

Question 4(v)

Solution 4(v)

Question 4(vi)

Solution 4(vi)

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8(i)

Solution 8(i)

Question 8(ii)

Solution 8(ii)

Question 8(iii)

Solution 8(iii)

Question 8(iv)

Using A^-1, solve the system of linear equations

X – 2y = 10, 2x – y – z = 8 and -2y + z = 7Solution 8(iv)

Question 8(v)

Solution 8(v)

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

The management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for helping and others (say z) for supervising the workers to keep the colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33. If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others, using matrix method, find the number of awardees of each category. Apart from these values, namely, honesty, cooperation and supervision, suggest one more value which the management must include for awards.Solution 13

Question 14

A school wants to award its students for the values of Honesty, Regularity and Hard work with a total cash award of Rs. 6000. Three times the award money for Hard work added to that given for honesty amounts to Rs. 11000. The award money given for Honesty and Hard work together is double the one given for Regularity. Represent the above situation algebraically and find the award for each value, using matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest one more value which the school must include for awards.Solution 14

The school can include an award for creativity and extra-curricular activities.Question 15

Two institutions decided to award their employees for the three values of resourcefulness, competence and determination in the form of prizes at the rate of Rs. x, Rs. y and Rs. z respectively per person. The first institution decided to award respectively 4, 3 and 2 employees with a total prize money of Rs. 37000 and the second institution decided to award respectively 5, 3 and 4 employees with a total prize money of Rs. 47000. If all the three prizes per person together amount to Rs. 12000, then using matrix method find the value of x, y and z. What values are described in these equations?Solution 15

Question 16

Two factories decided to award their employees for three values of (a) adaptable to new techniques, (b) careful and alert in difficult situations and (c) keeping calm in tense situations, at the rate of Rs. x, Rs. y and Rs. z per person respectively. The first factory decided to honour respectively 2, 4 and 3 employees with a total prize money of Rs. 29000. The second factory decided to honour respectively 5, 2 and 3 employees with the prize money of Rs. 30500. If the three prizes per person together cost Rs 9500, then

(i) represent the above situation by matrix equation and form linear equations using matrix multiplication.

(ii) Solve these equations using matrices.

(iii) Which values are reflected in the questions?Solution 16

Keeping calm in a tense situation is more rewarding than carefulness, and carefulness is more rewarding than adaptability.Question 17

Two schools A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. The school A wants to award Rs. x each Rs. y each and Rs. z each for the three respective values to 3, 2 and 1 students respectively with a total award money of Rs. 1,600. School B wants to spend Rs 2,300 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as before). If the total amount of award for one prize on each value is Rs 900, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for award.Solution 17

Question 18

Two schools P and Q want to award their selected students on the values of Discipline, Politeness and Punctuality. The school P wants to award Rs. x each, Rs. y each and Rs. z each for the three respectively values to its 3, 2 and 1 students with a total award money of Rs. 1,000. School Q wants to spend Rs. 1,500 to award its 4, 1 and 3 students on the respective values (by giving the same award money for three values as before). If the total amount of awards for one prize on each value is Rs. 600, using matrices, find the award money for each value. Apart from the above three values, suggest one more value for awards.Solution 18

Question 19

Two schools P and Q want to award their selected students on the values of Tolerance, Kindness and Leadership. The school P wants to award Rs. x each, Rs. y each and Rs. z each for the three respectively values to its 3, 2 and 1 students with a total award money of Rs. 2,200. School Q wants to spend Rs. 3,100 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as school P). If the total amount of award for one prize on each values is Rs. 1,200, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for award.Solution 19

Question 20

A total amount of Rs. 7000 is deposited in three different saving bank accounts with annual interest rates 5%, 8% and 8.5% respectively. The total annual interest from these three accounts is Rs. 550. Equal amounts have been deposited in the 5% and 8% savings accounts. Find the amount deposited in each of the three accounts, with the help of matrices.Solution 20

Let the amount deposited be x, y and z respectively.

As per the data in the question, we get

Question 8(vi)

If  find A^-1 and hence solve the system of equations 2x + y – 3z = 13, 3x + 2y + z = 4, x + 2y – z = 8.Solution 8(vi)

Therefore, A is invertible.

Let C_ij be the co-factors of the elements a_ij.

Now, the given system of equations is expressible as

Here we have |A^T| = |A| = -16 ≠ 0

Therefore, the given system of equations is consistent with a unique solution given by

Question 8(vii)

Use the product  to solve the system of equations x + 3z = -9, -x + 2y – 2z = 4, 2x – 3y + 4z = -3.Solution 8(vii)

Let 

Now, 

Now, the given system of equations is expressible as

Here we have |B^T| = |B| = -1 ≠ 0

Therefore, the given system of equations is consistent with a unique solution given by

Hence, x = 36, y = 5 and z = -15.Question 21

A shopkeeper has 3 varieties of pens ‘A’, ‘B’ and ‘C’. Meenu purchased 1 pen of each variety for a total of Rs. 21. Jeen purchased 4 pens of ‘A’ variety, 3 pens of ‘B’ variety and 2 pens of ‘C’ variety for Rs. 60. Using matrix method find the cost of each pen.Solution 21

From the given information, we can form a matrix as follows

Applying R₂→ R₂ – 4R₁, R₃→ R₃ – 6R₁

Applying R₃→ R₃ + (-4R₁)

From the above matrix form, we get

A + B + C = 21 … (i)

-B – 2C = -24 … (ii)

5C = 40

⇒ C = 8 … (iii)

Putting the value of C in (ii), we get

B = 8

Substituting B and C in (i), we get

C = 5

Hence, cost of variety ‘A’ pen is Rs. 8, cost of variety B pen is Rs. 8 and cost of variety ‘C’ pen is Rs. 5.

Chapter 8 Solution of Simultaneous Linear Equations Ex. 8.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Chapter 7 Adjoint and Inverse of a Matrix Ex 7.1

1. Find the adjoint of each of the following matrices:

Verify that (adj A) A = |A| I = A (adj A) for the above matrices.

Solution:

(i) Let

A = 

Cofactors of A are

C₁₁ = 4

C₁₂ = – 2

C₂₁ = – 5

C₂₂ = – 3

(ii) Let

A = 

Therefore cofactors of A are

C₁₁ = d

C₁₂ = – c

C₂₁ = – b

C₂₂ = a

(iii) Let

A = 

Therefore cofactors of A are

C₁₁ = cos α

C₁₂ = – sin α

C₂₁ = – sin α

C₂₂ = cos α

(iv) Let

A = 

Therefore cofactors of A are

C₁₁ = 1

C₁₂ = tan α/2

C₂₁ = – tan α/2

C₂₂ = 1

2. Compute the adjoint of each of the following matrices.

Solution:

(i) Let

A = 

Therefore cofactors of A are

C₁₁ = – 3

C₂₁ = 2

C₃₁ = 2

C₁₂ = 2

C₂₂ = – 3

C₂₃ = 2

C₁₃ = 2

C₂₃ = 2

C₃₃ = – 3

(ii) Let

A = 

Cofactors of A

C₁₁ = 2

C₂₁ = 3

C₃₁ = – 13

C₁₂ = – 3

C₂₂ = 6

C₃₂ = 9

C₁₃ = 5

C₂₃ = – 3

C₃₃ = – 1

(iii) Let

A = 

Therefore cofactors of A

C₁₁ = – 22

C₂₁ = 11

C₃₁ = – 11

C₁₂ = 4

C₂₂ = – 2

C₃₂ = 2

C₁₃ = 16

C₂₃ = – 8

C₃₃ = 8

(iv) Let

A = 

Therefore cofactors of A

C₁₁ = 3

C₂₁ = – 1

C₃₁ = 1

C₁₂ = – 15

C₂₂ = 7

C₃₂ = – 5

C₁₃ = 4

C₂₃ = – 2

C₃₃ = 2

Solution:

Given

A = 

Therefore cofactors of A

C₁₁ = 30

C₂₁ = 12

C₃₁ = – 3

C₁₂ = – 20

C₂₂ = – 8

C₃₂ = 2

C₁₃ = – 50

C₂₃ = – 20

C₃₃ = 5

Solution:

Given

A = 

Cofactors of A

C₁₁ = – 4

C₂₁ = – 3

C₃₁ = – 3

C₁₂ = 1

C₂₂ = 0

C₃₂ = 1

C₁₃ = 4

C₂₃ = 4

C₃₃ = 3

Solution:

Given

A = 

Cofactors of A are

C₁₁ = – 3

C₂₁ = 6

C₃₁ = 6

C₁₂ = – 6

C₂₂ = 3

C₃₂ = – 6

C₁₃ = – 6

C₂₃ = – 6

C₃₃ = 3

Solution:

Given

A = 

Cofactors of A are

C₁₁ = 9

C₂₁ = 19

C₃₁ = – 4

C₁₂ = 4

C₂₂ = 14

C₃₂ = 1

C₁₃ = 8

C₂₃ = 3

C₃₃ = 2

7. Find the inverse of each of the following matrices:

Solution:

(i) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.

Now, |A| = cos θ (cos θ) + sin θ (sin θ)

= 1

Hence, A ^{– 1} exists.

Cofactors of A are

C₁₁ = cos θ

C₁₂ = sin θ

C₂₁ = – sin θ

C₂₂ = cos θ

(ii) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.

Now, |A| = – 1 ≠ 0

Hence, A ^{– 1} exists.

Cofactors of A are

C₁₁ = 0

C₁₂ = – 1

C₂₁ = – 1

C₂₂ = 0

(iii) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.

(iv) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.

Now, |A| = 2 + 15 = 17 ≠ 0

Hence, A ^{– 1} exists.

Cofactors of A are

C₁₁ = 1

C₁₂ = 3

C₂₁ = – 5

C₂₂ = 2

8. Find the inverse of each of the following matrices.

Solution:

(i) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.

|A| = 

= 1(6 – 1) – 2(4 – 3) + 3(2 – 9)

= 5 – 2 – 21

= – 18≠ 0

Hence, A ^{– 1} exists

Cofactors of A are

C₁₁ = 5

C₂₁ = – 1

C₃₁ = – 7

C₁₂ = – 1

C₂₂ = – 7

C₃₂ = 5

C₁₃ = – 7

C₂₃ = 5

C₃₃ = – 1

(ii) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.

|A| = 

= 1 (1 + 3) – 2 (– 1 + 2) + 5 (3 + 2)

= 4 – 2 + 25

= 27≠ 0

Hence, A ^{– 1} exists

Cofactors of A are

C₁₁ = 4

C₂₁ = 17

C₃₁ = 3

C₁₂ = – 1

C₂₂ = – 11

C₃₂ = 6

C₁₃ = 5

C₂₃ = 1

C₃₃ = – 3

(iii) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.

|A| = 

= 2(4 – 1) + 1(– 2 + 1) + 1(1 – 2)

= 6 – 2

= – 4≠ 0

Hence, A ^{– 1} exists

Cofactors of A are

C₁₁ = 3

C₂₁ = 1

C₃₁ = – 1

C₁₂ = + 1

C₂₂ = 3

C₃₂ = 1

C₁₃ = – 1

C₂₃ = 1

C₃₃ = 3

(iv) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.

|A| = 

= 2(3 – 0) – 0 – 1(5)

= 6 – 5

= 1≠ 0

Hence, A ^{– 1} exists

Cofactors of A are

C₁₁ = 3

C₂₁ = – 1

C₃₁ = 1

C₁₂ = – 15

C₂₂ = 6

C₃₂ = – 5

C₁₃ = 5

C₂₃ = – 2

C₃₃ = 2

(v) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.

|A| = 

= 0 – 1 (16 – 12) – 1 (– 12 + 9)

= – 4 + 3

= – 1≠ 0

Hence, A ^{– 1} exists

Cofactors of A are

C₁₁ = 0

C₂₁ = – 1

C₃₁ = 1

C₁₂ = – 4

C₂₂ = 3

C₃₂ = – 4

C₁₃ = – 3

C₂₃ = 3

C₃₃ = – 4

(vi) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.

|A| = 

= 0 – 0 – 1(– 12 + 8)

= 4≠ 0

Hence, A ^{– 1} exists

Cofactors of A are

C₁₁ = – 8

C₂₁ = 4

C₃₁ = 4

C₁₂ = 11

C₂₂ = – 2

C₃₂ = – 3

C₁₃ = – 4

C₂₃ = 0

C₃₃ = 0

(vii) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.

|A| = 
 – 0 + 0

= – (cos² α – sin² α)

= – 1≠ 0

Hence, A ^{– 1} exists

Cofactors of A are

C₁₁ = – 1

C₂₁ = 0

C₃₁ = 0

C₁₂ = 0

C₂₂ = – cos α

C₃₂ = – sin α

C₁₃ = 0

C₂₃ = – sin α

C₃₃ = cos α

9. Find the inverse of each of the following matrices and verify that A^-1A = I₃.

Solution:

(i) We have

|A| = 

= 1(16 – 9) – 3(4 – 3) + 3(3 – 4)

= 7 – 3 – 3

= 1≠ 0

Hence, A ^{– 1} exists

Cofactors of A are

C₁₁ = 7

C₂₁ = – 3

C₃₁ = – 3

C₁₂ = – 1

C₂₂ = 1

C₃₂ = 0

C₁₃ = – 1

C₂₃ = 0

C₃₃ = 1

(ii) We have

|A| = 

= 2(8 – 7) – 3(6 – 3) + 1(21 – 12)

= 2 – 9 + 9

= 2≠ 0

Hence, A ^{– 1} exists

Cofactors of A are

C₁₁ = 1

C₂₁ = 1

C₃₁ = – 1

C₁₂ = – 3

C₂₂ = 1

C₃₂ = 1

C₁₃ = 9

C₂₃ = – 5

C₃₃ = – 1

10. For the following pair of matrices verify that (AB)^-1 = B^-1A^-1.

Solution:

(i) Given

Hence, (AB)^-1 = B^-1A^-1

(ii) Given

Hence, (AB)^-1 = B^-1A^-1

Solution:

Given

Solution:

Given

Solution:

Given

Solution:

Solution:

Given

A = 
 and B ^{– 1} = 

Here, (AB) ^{– 1 =} B ^{– 1} A ^{– 1}

|A| = – 5 + 4 = – 1

Cofactors of A are

C₁₁ = – 1

C₂₁ = 8

C₃₁ = – 12

C₁₂ = 0

C₂₂ = 1

C₃₂ = – 2

C₁₃ = 1

C₂₃ = – 10

C₃₃ = 15

(i) F(α)^-1 = F (-α)

(ii) G(β)^-1 = G (-β)

(iii) F(α)G(β)^-1 = G (-β) F (-α)

Solution:

(i) Given

F (α) = 

|F (α)| = cos² α + sin² α = 1≠ 0

Cofactors of A are

C₁₁ = cos α

C₂₁ = sin α

C₃₁ = 0

C₁₂ = – sin α

C₂₂ = cos α

C₃₂ = 0

C₁₃ = 0

C₂₃ = 0

C₃₃ = 1

(ii) We have

|G (β)| = cos² β + sin² β = 1

Cofactors of A are

C₁₁ = cos β

C₂₁ = 0

C₃₁ = -sin β

C₁₂ = 0

C₂₂ = 1

C₃₂ = 0

C₁₃ = sin β

C₂₃ = 0

C₃₃ = cos β

(iii) Now we have to show thatF(α)G(β) ^{– 1} = G (– β) F (– α)

We have already know thatG(β) ^{– 1} = G (– β)F(α) ^{– 1} = F (– α)

And LHS = F(α)G(β) ^{– 1}

= G(β) ^{– 1} F(α) ^{– 1}

= G (– β) F (– α)

Hence = RHS

Solution:

Consider,

Solution:

Given

Solution:

Given

Chapter 7 Adjoint and Inverse of a Matrix Ex 7.2

---

Find the inverse of the following matrices by using elementary row transformations:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Chapter 6 Determinants Ex 6.1

1. Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case:

Solution:

(i) Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column.The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, C_ij = (–1)^i+j × M_ij

Given,

From the given matrix we have,

M₁₁ = –1

M₂₁ = 20

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × –1

= –1

C₂₁ = (–1)²⁺¹ × M₂₁

= 20 × –1

= –20

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁

= 5× (–1) + 0 × (–20)

= –5

(ii) Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, C_ij = (–1)^i+j × M_ij

Given

From the above matrix we have

M₁₁ = 3

M₂₁ = 4

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × 3

= 3

C₂₁ = (–1)²⁺¹ × 4

= –1 × 4

= –4

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁

= –1× 3 + 2 × (–4)

= –11

(iii) Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, C_ij = (–1)^i+j × M_ij

Given,

M₃₁ = –3 × 2 – (–1) × 2

M₃₁ = –4

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × –12

= –12

C₂₁ = (–1)²⁺¹ × M₂₁

= –1 × –16

= 16

C₃₁ = (–1)³⁺¹ × M₃₁

= 1 × –4

= –4

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁

= 1× (–12) + 4 × 16 + 3× (–4)

= –12 + 64 –12

= 40

(iv) Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, C_ij = (–1)^i+j × M_ij

Given,

M₃₁ = a × c a – b × bc

M₃₁ = a²c – b²c

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × (ab² – ac²)

= ab² – ac²

C₂₁ = (–1)²⁺¹ × M₂₁

= –1 × (a²b – c²b)

= c²b – a²b

C₃₁ = (–1)³⁺¹ × M₃₁

= 1 × (a²c – b²c)

= a²c – b²c

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁

= 1× (ab² – ac²) + 1 × (c²b – a²b) + 1× (a²c – b²c)

= ab² – ac² + c²b – a²b + a²c – b²c

(v) Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, C_ij = (–1)^i+j × M_ij

Given,

M₃₁ = 2×0 – 5×6

M₃₁ = –30

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × 5

= 5

C₂₁ = (–1)²⁺¹ × M₂₁

= –1 × –40

= 40

C₃₁ = (–1)³⁺¹ × M₃₁

= 1 × –30

= –30

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁

= 0× 5 + 1 × 40 + 3× (–30)

= 0 + 40 – 90

= 50

(vi) Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, C_ij = (–1)^i+j × M_ij

Given,

M₃₁ = h × f – b × g

M₃₁ = hf – bg

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × (bc– f²)

= bc– f²

C₂₁ = (–1)²⁺¹ × M₂₁

= –1 × (hc – fg)

= fg – hc

C₃₁ = (–1)³⁺¹ × M₃₁

= 1 × (hf – bg)

= hf – bg

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁

= a× (bc– f²) + h× (fg – hc) + g× (hf – bg)

= abc– af² + hgf – h²c +ghf – bg²

(vii) Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, C_ij = (–1)^i+j × M_ij

Given,

M₃₁ = –1(1 × 0 – 5 × (–2)) – 0(0 × 0 – (–1) × (–2)) + 1(0 × 5 – (–1) × 1)

M₃₁ = –9

M₄₁ = –1(1×1 – (–1) × (–2)) – 0(0 × 1 – 1 × (–2)) + 1(0 × (–1) – 1 × 1)

M₄₁ = 0

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × (–9)

= –9

C₂₁ = (–1)²⁺¹ × M₂₁

= –1 × 9

= –9

C₃₁ = (–1)³⁺¹ × M₃₁

= 1 × –9

= –9

C₄₁ = (–1)⁴⁺¹ × M₄₁

= –1 × 0

= 0

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁ + a₄₁× C₄₁

= 2 × (–9) + (–3) × –9 + 1 × (–9) + 2 × 0

= – 18 + 27 –9

= 0

2. Evaluate the following determinants:

Solution:

(i) Given

⇒ |A| = x (5x + 1) – (–7) x

|A| = 5x² + 8x

(ii) Given

⇒ |A| = cos θ × cos θ – (–sin θ) x sin θ

|A| = cos²θ + sin²θ

We know that cos²θ + sin²θ = 1

|A| = 1

(iii) Given

⇒ |A| = cos15° × cos75° + sin15° x sin75°

We know that cos (A – B) = cos A cos B + Sin A sin B

By substituting this we get, |A| = cos (75 – 15)°

|A| = cos60°

|A| = 0.5

(iv) Given

⇒ |A| = (a + ib) (a – ib) – (c + id) (–c + id)

= (a + ib) (a – ib) + (c + id) (c – id)

= a² – i² b² + c² – i² d²

We know that i² = -1

= a² – (–1) b² + c² – (–1) d²

= a² + b² + c² + d²

3. Evaluate:

Solution:

Since |AB|= |A||B|

= 2(17 × 12 – 5 × 20) – 3(13 × 12 – 5 × 15) + 7(13 × 20 – 15 × 17)

= 2 (204 – 100) – 3 (156 – 75) + 7 (260 – 255)

= 2×104 – 3×81 + 7×5

= 208 – 243 +35

= 0

Now |A|² = |A|×|A|

|A|²= 0

4. Show that

Solution:

Given

Let the given determinant as A

Using sin (A+B) = sin A × cos B + cos A × sin B

⇒ |A| = sin 10° × cos 80° + cos 10° x sin 80°

|A| = sin (10 + 80)°

|A| = sin90°

|A| = 1

Hence Proved

Solution:

Given,

= 2(1 × 1 – 4 × (–2)) – 3(7 × 1 – (–2) × (–3)) – 5(7 × 4 – 1 × (–3))

= 2(1 + 8) – 3(7 – 6) – 5(28 + 3)

= 2 × 9 – 3 × 1 – 5 × 31

= 18 – 3 – 155

= –140

Now by expanding along the second column

= 2(1 × 1 – 4 × (–2)) – 7(3 × 1 – 4 × (–5)) – 3(3 × (–2) – 1 × (–5))

= 2 (1 + 8) – 7 (3 + 20) – 3 (–6 + 5)

= 2 × 9 – 7 × 23 – 3 × (–1)

= 18 – 161 +3

= –140

Solution:

Given

⇒ |A| = 0 (0 – sinβ (–sinβ)) –sinα (–sinα × 0 – sinβ cosα) – cosα ((–sinα) (–sinβ) – 0 × cosα)

|A| = 0 + sinα sinβ cosα – cosα sinα sinβ

|A| = 0

Chapter 6 Determinants Ex 6.2

1. Evaluate the following determinant:

Solution:

(i) Given

(ii) Given

= 1(109)(12)–(119)(11)

= 1308 – 1309

= – 1

So, Δ = – 1

(iii) Given,

= a (bc – f²) – h (hc – fg) + g (hf – bg)

= abc – af² – ch² + fgh + fgh – bg²

= abc + 2fgh – af² – bg² – ch²

So, Δ = abc + 2fgh – af² – bg² – ch²

(iv) Given

= 21(24–4) = 40

So, Δ = 40

(v) Given

= 1(–7)(–36)–(–20)(–13)

= 252 – 260

= – 8

So, Δ = – 8

(vi) Given,

(vii) Given

(viii) Given,

2. Without expanding, show that the value of each of the following determinants is zero:

Solution:

(i) Given,

(ii) Given,

(iii) Given,

(iv) Given,

(v) Given,

(vi) Given,

(vii) Given,

(viii) Given,

(ix) Given,

As, C₁ = C₂, hence determinant is zero

(x) Given,

(xi) Given,

(xii) Given,

(xiii) Given,

(xiv) Given,

(xv) Given,

(xvi) Given,

(xvii) Given,

Hence proved.

Evaluate the following (3 – 9):

Solution:

Given,

= (a + b + c) (b – a) (c – a) (b – c)

So, Δ = (a + b + c) (b – a) (c – a) (b – c)

Solution:

Given,

Solution:

Given,

Solution:

Given,

Solution:

Given,

Solution:

Given,

Solution:

Given,

= a a(a+x+y)+az + 0 + 0

= a² (a + x + y + z)

So, Δ = a² (a + x + y + z)

Solution:

Prove the following identities (11 – 45):

Solution:

Given,

Solution:

Consider,

= – (a + b + c) (b–c)(a+b–2c)–(c–a)(c+a–2b)

= 3abc – a³ – b³ – c³

Therefore, L.H.S = R.H.S,

Hence the proof.

Solution:

Given,

Solution:

Consider,

,

Solution:

Consider,

L.H.S =

Now by applying, R₁→R₁ + R₂ + R₃, we get,

Solution:

Consider,

Solution:

Consider,

Solution:

Consider,

Hence, the proof.

Solution:

Given,

= – xyz(x – y) (z – y) [z² + y² + zy – x² – y² – xy]

= – xyz(x – y) (z – y) (z–x)(z+x0+y(z–x)

= – xyz(x – y) (z – y) (z – x) (x + y + z)

= R.H.S

Hence, the proof.

Solution:

Consider,

= (a² + b² + c²) (b – a) (c – a) (b+a)(–b)–(–c)(c+a)

= (a² + b² + c²) (a – b) (c – a) (b – c) (a + b + c)

= R.H.S

Hence, the proof.

Solution:

Consider,

= (2a+4)(1)–(1)(2a+6)

= – 2

= R.H.S

Hence, the proof.

Solution:

Consider,

= – (a² + b² + c²) (a – b) (c – a) (–(b+a))(–b)–(c)(c+a)

= (a – b) (b – c) (c – a) (a + b + c) (a² + b² + c²)

= R.H.S

Hence, the proof.

Solution:

Consider,

= R.H.S

Hence, the proof.

Solution:

Consider,

Solution:

Consider,

Solution:

Expanding the determinant along R₁, we have

Δ = 1(1)(7)–(3)(2) – 0 + 0

∴ Δ = 7 – 6 = 1

Thus,

Hence the proof.

Chapter 6 Determinants Ex 6.3

1. Find the area of the triangle with vertices at the points:

(i) (3, 8), (-4, 2) and (5, -1)

(ii) (2, 7), (1, 1) and (10, 8)

(iii) (-1, -8), (-2, -3) and (3, 2)

(iv) (0, 0), (6, 0) and (4, 3)

Solution:

(i) Given (3, 8), (-4, 2) and (5, -1) are the vertices of the triangle.

We know that, if vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), then the area of the triangle is given by:

(ii) Given (2, 7), (1, 1) and (10, 8) are the vertices of the triangle.

We know that if vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), then the area of the triangle is given by:

(iii) Given (-1, -8), (-2, -3) and (3, 2) are the vertices of the triangle.

We know that if vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), then the area of the triangle is given by:

As we know area cannot be negative. Therefore, 15 square unit is the area

Thus area of triangle is 15 square units

(iv) Given (-1, -8), (-2, -3) and (3, 2) are the vertices of the triangle.

We know that if vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), then the area of the triangle is given by:

2. Using the determinants show that the following points are collinear:

(i) (5, 5), (-5, 1) and (10, 7)

(ii) (1, -1), (2, 1) and (10, 8)

(iii) (3, -2), (8, 8) and (5, 2)

(iv) (2, 3), (-1, -2) and (5, 8)

Solution:

(i) Given (5, 5), (-5, 1) and (10, 7)

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), then the area of the triangle is given by

(ii) Given (1, -1), (2, 1) and (10, 8)

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), then the area of the triangle is given by,

(iii) Given (3, -2), (8, 8) and (5, 2)

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), then the area of the triangle is given by,

Now, by substituting given value in above formula

Since, Area of triangle is zero

Hence, points are collinear.

(iv) Given (2, 3), (-1, -2) and (5, 8)

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), then the area of the triangle is given by,

3. If the points (a, 0), (0, b) and (1, 1) are collinear, prove that a + b = ab

Solution:

Given (a, 0), (0, b) and (1, 1) are collinear

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), then the area of the triangle is given by,

⇒

⇒ a + b = ab

Hence Proved

4. Using the determinants prove that the points (a, b), (a’, b’) and (a – a’, b – b) are collinear if a b’ = a’ b.

Solution:

Given (a, b), (a’, b’) and (a – a’, b – b) are collinear

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), then the area of the triangle is given by,

⇒ a b’ = a’ b

Hence, the proof.

5. Find the value of λ so that the points (1, -5), (-4, 5) and (λ, 7) are collinear.

Solution:

Given (1, -5), (-4, 5) and (λ, 7) are collinear

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), then the area of the triangle is given by,

⇒ – 50 – 10λ = 0

⇒ λ = – 5

6. Find the value of x if the area of ∆ is 35 square cms with vertices (x, 4), (2, -6) and (5, 4).

Solution:

Given (x, 4), (2, -6) and (5, 4) are the vertices of a triangle.

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), then the area of the triangle is given by,

⇒ x(–10)–4(–3)+1(8–30) = ± 70

⇒ –10x+12+38 = ± 70

⇒ ±70 = – 10x + 50

Taking positive sign, we get

⇒ + 70 = – 10x + 50

⇒ 10x = – 20

⇒ x = – 2

Taking –negative sign, we get

⇒ – 70 = – 10x + 50

⇒ 10x = 120

⇒ x = 12

Thus x = – 2, 12

Chapter 6 Determinants Ex 6.4

Solve the following system of linear equations by Cramer’s rule:

1. x – 2y = 4

-3x + 5y = -7

Solution:

Given x – 2y = 4

-3x + 5y = -7

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^st row

⇒ D = 5(1) – (– 3) (– 2)

⇒ D = 5 – 6

⇒ D = – 1

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 5(4) – (– 7) (– 2)

⇒ D₁ = 20 – 14

⇒ D₁ = 6

And

Solving determinant, expanding along 1^st row

⇒ D₂ = 1(– 7) – (– 3) (4)

⇒ D₂ = – 7 + 12

⇒ D₂ = 5

Thus by Cramer’s Rule, we have

2. 2x – y = 1

7x – 2y = -7

Solution:

Given 2x – y = 1 and

7x – 2y = -7

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^st row

⇒ D₁ = 1(– 2) – (– 7) (– 1)

⇒ D₁ = – 2 – 7

⇒ D₁ = – 9

And

Solving determinant, expanding along 1^st row

⇒ D₂ = 2(– 7) – (7) (1)

⇒ D₂ = – 14 – 7

⇒ D₂ = – 21

Thus by Cramer’s Rule, we have

3. 2x – y = 17

3x + 5y = 6

Solution:

Given 2x – y = 17 and

3x + 5y = 6

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^st row

⇒ D₁ = 17(5) – (6) (– 1)

⇒ D₁ = 85 + 6

⇒ D₁ = 91

Solving determinant, expanding along 1^st row

⇒ D₂ = 2(6) – (17) (3)

⇒ D₂ = 12 – 51

⇒ D₂ = – 39

Thus by Cramer’s Rule, we have

4. 3x + y = 19

3x – y = 23

Solution:

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^st row

⇒ D = 3(– 1) – (3) (1)

⇒ D = – 3 – 3

⇒ D = – 6

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 19(– 1) – (23) (1)

⇒ D₁ = – 19 – 23

⇒ D₁ = – 42

Solving determinant, expanding along 1^st row

⇒ D₂ = 3(23) – (19) (3)

⇒ D₂ = 69 – 57

⇒ D₂ = 12

Thus by Cramer’s Rule, we have

5. 2x – y = -2

3x + 4y = 3

Solution:

Given 2x – y = -2 and

3x + 4y = 3

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^st row

⇒ D₂ = 3(2) – (– 2) (3)

⇒ D₂ = 6 + 6

⇒ D₂ = 12

Thus by Cramer’s Rule, we have

6. 3x + ay = 4

2x + ay = 2, a ≠ 0

Solution:

Given 3x + ay = 4 and

2x + ay = 2, a ≠ 0

Let there be a system of n simultaneous linear equations and with n unknown given by

3x + ay = 4

2x + ay = 2, a≠0

So by comparing with the theorem, let’s find D, D₁ and D₂

Solving determinant, expanding along 1^st row

⇒ D = 3(a) – (2) (a)

⇒ D = 3a – 2a

⇒ D = a

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 4(a) – (2) (a)

⇒ D = 4a – 2a

⇒ D = 2a

Solving determinant, expanding along 1^st row

⇒ D₂ = 3(2) – (2) (4)

⇒ D = 6 – 8

⇒ D = – 2

Thus by Cramer’s Rule, we have

7. 2x + 3y = 10

x + 6y = 4

Solution:

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^st row

⇒ D = 2 (6) – (3) (1)

⇒ D = 12 – 3

⇒ D = 9

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 10 (6) – (3) (4)

⇒ D = 60 – 12

⇒ D = 48

Solving determinant, expanding along 1^st row

⇒ D₂ = 2 (4) – (10) (1)

⇒ D₂ = 8 – 10

⇒ D₂ = – 2

Thus by Cramer’s Rule, we have

8. 5x + 7y = -2

4x + 6y = -3

Solution:

Let there be a system of n simultaneous linear equations and with n unknown given by

Now, here we have

5x + 7y = – 2

4x + 6y = – 3

So by comparing with the theorem, let’s find D, D₁ and D₂

Solving determinant, expanding along 1^st row

⇒ D = 5(6) – (7) (4)

⇒ D = 30 – 28

⇒ D = 2

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = – 2(6) – (7) (– 3)

⇒ D₁ = – 12 + 21

⇒ D₁ = 9

Solving determinant, expanding along 1^st row

⇒ D₂ = – 3(5) – (– 2) (4)

⇒ D₂ = – 15 + 8

⇒ D₂ = – 7

Thus by Cramer’s Rule, we have

9. 9x + 5y = 10

3y – 2x = 8

Solution:

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^st row

⇒ D = 3(9) – (5) (– 2)

⇒ D = 27 + 10

⇒ D = 37

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 10(3) – (8) (5)

⇒ D₁ = 30 – 40

⇒ D₁ = – 10

Solving determinant, expanding along 1^st row

⇒ D₂ = 9(8) – (10) (– 2)

⇒ D₂ = 72 + 20

⇒ D₂ = 92

Thus by Cramer’s Rule, we have

10. x + 2y = 1

3x + y = 4

Solution:

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^st row

⇒ D = 1(1) – (3) (2)

⇒ D = 1 – 6

⇒ D = – 5

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 1(1) – (2) (4)

⇒ D₁ = 1 – 8

⇒ D₁ = – 7

Solving determinant, expanding along 1^st row

⇒ D₂ = 1(4) – (1) (3)

⇒ D₂ = 4 – 3

⇒ D₂ = 1

Thus by Cramer’s Rule, we have

Solve the following system of linear equations by Cramer’s rule:

11. 3x + y + z = 2

2x – 4y + 3z = -1

4x + y – 3z = -11

Solution:

Let there be a system of n simultaneous linear equations and with n unknown given by

Now, here we have

3x + y + z = 2

2x – 4y + 3z = – 1

4x + y – 3z = – 11

So by comparing with the theorem, let’s find D, D₁, D₂ and D₃

Solving determinant, expanding along 1^st row

⇒ D = 3(–4)(–3)–(3)(1) – 1(2)(–3)–12 + 12–4(–4)

⇒ D = 312–3 – –6–12 + 2+16

⇒ D = 27 + 18 + 18

⇒ D = 63

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 2(–4)(–3)–(3)(1) – 1(–1)(–3)–(–11)(3) + 1(–1)–(–4)(–11)

⇒ D₁ = 212–3 – 13+33 + 1–1–44

⇒ D₁ = 29 – 36 – 45

⇒ D₁ = 18 – 36 – 45

⇒ D₁ = – 63

Again

Solving determinant, expanding along 1^st row

⇒ D₂ = 33+33 – 2–6–12 + 1–22+4

⇒ D₂ = 336 – 2(– 18) – 18

⇒ D₂ = 126

⇒

Solving determinant, expanding along 1^st row

⇒ D₃ = 344+1 – 1–22+4 + 22+16

⇒ D₃ = 345 – 1(– 18) + 2(18)

⇒ D₃ = 135 + 18 + 36

⇒ D₃ = 189

Thus by Cramer’s Rule, we have

12. x – 4y – z = 11

2x – 5y + 2z = 39

-3x + 2y + z = 1

Solution:

Given,

x – 4y – z = 11

2x – 5y + 2z = 39

-3x + 2y + z = 1

Let there be a system of n simultaneous linear equations and with n unknown given by

Now, here we have

x – 4y – z = 11

2x – 5y + 2z = 39

– 3x + 2y + z = 1

So by comparing with theorem, now we have to find D, D₁ and D₂

Solving determinant, expanding along 1^st row

⇒ D = 1(–5)(1)–(2)(2) + 4(2)(1)+6 – 14+5(–3)

⇒ D = 1–5–4 + 48 – –11

⇒ D = – 9 + 32 + 11

⇒ D = 34

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 11(–5)(1)–(2)(2) + 4(39)(1)–(2)(1) – 12(39)–(–5)(1)

⇒ D₁ = 11–5–4 + 439–2 – 178+5

⇒ D₁ = 11–9 + 4(37) – 83

⇒ D₁ = – 99 – 148 – 45

⇒ D₁ = – 34

Again

Solving determinant, expanding along 1^st row

⇒ D₂ = 139–2 – 112+6 – 12+117

⇒ D₂ = 137 – 11(8) – 119

⇒ D₂ = – 170

And,

⇒

Solving determinant, expanding along 1^st row

⇒ D₃ = 1–5–(39)(2) – (– 4) 2–(39)(–3) + 114–(–5)(–3)

⇒ D₃ = 1 –5–78 + 4 (2 + 117) + 11 (4 – 15)

⇒ D₃ = – 83 + 4(119) + 11(– 11)

⇒ D₃ = 272

Thus by Cramer’s Rule, we have

13. 6x + y – 3z = 5

x + 3y – 2z = 5

2x + y + 4z = 8

Solution:

Given

6x + y – 3z = 5

x + 3y – 2z = 5

2x + y + 4z = 8

Let there be a system of n simultaneous linear equations and with n unknown given by

Now, here we have

6x + y – 3z = 5

x + 3y – 2z = 5

2x + y + 4z = 8

So by comparing with theorem, now we have to find D , D₁ and D₂

Solving determinant, expanding along 1^st Row

⇒ D = 6(4)(3)–(1)(–2) – 1(4)(1)+4 – 31–3(2)

⇒ D = 612+2 – 8 – 3–5

⇒ D = 84 – 8 + 15

⇒ D = 91

Again, Solve D₁ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₁ = 5(4)(3)–(–2)(1) – 1(5)(4)–(–2)(8) – 3(5)–(3)(8)

⇒ D₁ = 512+2 – 120+16 – 35–24

⇒ D₁ = 514 – 36 – 3(– 19)

⇒ D₁ = 70 – 36 + 57

⇒ D₁ = 91

Again, Solve D₂ formed by replacing 1^st column by B matrices

Here

Solving determinant

⇒ D₂ = 620+16 – 54–2(–2) + (– 3)8–10

⇒ D₂ = 636 – 5(8) + (– 3) (– 2)

⇒ D₂ = 182

And, Solve D₃ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₃ = 624–5 – 18–10 + 51–6

⇒ D₃ = 619 – 1(– 2) + 5(– 5)

⇒ D₃ = 114 + 2 – 25

⇒ D₃ = 91

Thus by Cramer’s Rule, we have

14. x + y = 5

y + z = 3

x + z = 4

Solution:

Given x + y = 5

y + z = 3

x + z = 4

Let there be a system of n simultaneous linear equations and with n unknown given by

Let D_j be the determinant obtained from D after replacing the j^th column by

Now, here we have

x + y = 5

y + z = 3

x + z = 4

So by comparing with theorem, now we have to find D, D₁ and D₂

Solving determinant, expanding along 1^st Row

⇒ D = 11 – 1–1 + 0–1

⇒ D = 1 + 1 + 0

⇒ D = 2

Again, Solve D₁ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₁ = 51 – 1(3)(1)–(4)(1) + 00–(4)(1)

⇒ D₁ = 5 – 13–4 + 0–4

⇒ D₁ = 5 – 1–1 + 0

⇒ D₁ = 5 + 1 + 0

⇒ D₁ = 6

Again, Solve D₂ formed by replacing 1^st column by B matrices

Here

Solving determinant

⇒ D₂ = 13–4 – 5–1 + 00–3

⇒ D₂ = 1–1 + 5 + 0

⇒ D₂ = 4

And, Solve D₃ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₃ = 14–0 – 10–3 + 50–1

⇒ D₃ = 14 – 1(– 3) + 5(– 1)

⇒ D₃ = 4 + 3 – 5

⇒ D₃ = 2

Thus by Cramer’s Rule, we have

15. 2y – 3z = 0

x + 3y = -4

3x + 4y = 3

Solution:

Given

2y – 3z = 0

x + 3y = -4

3x + 4y = 3

Let there be a system of n simultaneous linear equations and with n unknown given by

Now, here we have

2y – 3z = 0

x + 3y = – 4

3x + 4y = 3

So by comparing with theorem, now we have to find D, D₁ and D₂

Solving determinant, expanding along 1^st Row

⇒ D = 00 – 2(0)(1)–0 – 31(4)–3(3)

⇒ D = 0 – 0 – 34–9

⇒ D = 0 – 0 + 15

⇒ D = 15

Again, Solve D₁ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₁ = 00 – 2(0)(–4)–0 – 34(–4)–3(3)

⇒ D₁ = 0 – 0 – 3–16–9

⇒ D₁ = 0 – 0 – 3(– 25)

⇒ D₁ = 0 – 0 + 75

⇒ D₁ = 75

Again, Solve D₂ formed by replacing 2^nd column by B matrices

Here

Solving determinant

⇒ D₂ = 00 – 0(0)(1)–0 – 31(3)–3(–4)

⇒ D₂ = 0 – 0 + (– 3) (3 + 12)

⇒ D₂ = – 45

And, Solve D₃ formed by replacing 3^rd column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₃ = 09–(–4)4 – 2(3)(1)–(–4)(3) + 01(4)–3(3)

⇒ D₃ = 025 – 2(3 + 12) + 0(4 – 9)

⇒ D₃ = 0 – 30 + 0

⇒ D₃ = – 30

Thus by Cramer’s Rule, we have

16. 5x – 7y + z = 11

6x – 8y – z = 15

3x + 2y – 6z = 7

Solution:

Given

5x – 7y + z = 11

6x – 8y – z = 15

3x + 2y – 6z = 7

Let there be a system of n simultaneous linear equations and with n unknown given by

Now, here we have

5x – 7y + z = 11

6x – 8y – z = 15

3x + 2y – 6z = 7

So by comparing with theorem, now we have to find D, D₁ and D₂

Solving determinant, expanding along 1^st Row

⇒ D = 5(–8)(–6)–(–1)(2) – 7(–6)(6)–3(–1) + 12(6)–3(–8)

⇒ D = 548+2 – 7–36+3 + 112+24

⇒ D = 250 – 231 + 36

⇒ D = 55

Again, Solve D₁ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₁ = 11(–8)(–6)–(2)(–1) – (– 7) (15)(–6)–(–1)(7) + 1(15)2–(7)(–8)

⇒ D₁ = 1148+2 + 7–90+7 + 130+56

⇒ D₁ = 1150 + 7–83 + 86

⇒ D₁ = 550 – 581 + 86

⇒ D₁ = 55

Again, Solve D₂ formed by replacing 2^nd column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₂ = 5(15)(–6)–(7)(–1) – 11 (6)(–6)–(–1)(3) + 1(6)7–(15)(3)

⇒ D₂ = 5–90+7 – 11–36+3 + 142–45

⇒ D₂ = 5–83 – 11(– 33) – 3

⇒ D₂ = – 415 + 363 – 3

⇒ D₂ = – 55

And, Solve D₃ formed by replacing 3^rd column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₃ = 5(–8)(7)–(15)(2) – (– 7) (6)(7)–(15)(3) + 11(6)2–(–8)(3)

⇒ D₃ = 5–56–30 – (– 7) 42–45 + 1112+24

⇒ D₃ = 5–86 + 7–3 + 1136

⇒ D₃ = – 430 – 21 + 396

⇒ D₃ = – 55

Thus by Cramer’s Rule, we have

Chapter 6 Determinants Ex 6.5

Solve each of the following system of homogeneous linear equations:

1. x + y – 2z = 0

2x + y – 3z =0

5x + 4y – 9z = 0

Solution:

Given x + y – 2z = 0

2x + y – 3z =0

5x + 4y – 9z = 0

Any system of equation can be written in matrix form as AX = B

Now finding the Determinant of these set of equations,

= 1(1 × (– 9) – 4 × (– 3)) – 1(2 × (– 9) – 5 × (– 3)) – 2(4 × 2 – 5 × 1)

= 1(– 9 + 12) – 1(– 18 + 15) – 2(8 – 5)

= 1 × 3 –1 × (– 3) – 2 × 3

= 3 + 3 – 6

= 0

Since D = 0, so the system of equation has infinite solution.

Now let z = k

⇒ x + y = 2k

And 2x + y = 3k

Now using the Cramer’s rule

2. 2x + 3y + 4z = 0

x + y + z = 0

2x + 5y – 2z = 0

Solution:

Given

2x + 3y + 4z = 0

x + y + z = 0

2x + 5y – 2z = 0

Any system of equation can be written in matrix form as AX = B

Now finding the Determinant of these set of equations,

= 2(1 × (– 2) – 1 × 5) – 3(1 × (– 2) – 2 × 1) + 4(1 × 5 – 2 × 1)

= 2(– 2 – 5) – 3(– 2 – 2) + 4(5 – 2)

= 1 × (– 7) – 3 × (– 4) + 4 × 3

= – 7 + 12 + 12

= 17

Since D ≠ 0, so the system of equation has infinite solution.

Therefore the system of equation has only solution as x = y = z = 0.

Chapter 5 Algebra of Matrices Ex 5.1

1. If a matrix has 8 elements, what are the possible orders it can have? What if it has 5 elements?

Solution:

If a matrix is of order m × n elements, it has m n elements. So, if the matrix has 8 elements, we will find the ordered pairs m and n.

m n = 8

Then, ordered pairs m and n will be

m × n be (8 × 1),(1 × 8),(4 × 2),(2 × 4)

Now, if it has 5 elements

Possible orders are (5 × 1), (1 × 5).

Solution:

(i)

Now, Comparing with equation (1) and (2)

a₂₂ = 4 and b₂₁ = – 3

a₂₂ + b₂₁ = 4 + (– 3) = 1

(ii)

Now, Comparing with equation (1) and (2)

a₁₁ = 2, a₂₂ = 4, b₁₁ = 2, b₂₂ = 4

a₁₁ b₁₁ + a₂₂ b₂₂ = 2 × 2 + 4 × 4 = 4 + 16 = 20

3. Let A be a matrix of order 3 × 4. If R₁ denotes the first row of A and C₂ denotes its second column, then determine the orders of matrices R₁ and C₂.

Solution:

Given A be a matrix of order 3 × 4.

So, A = [a_{i j}] _3×4

R₁ = first row of A = [a₁₁, a₁₂, a₁₃, a₁₄]

So, order of matrix R₁ = 1 × 4

C₂ = second column of

Therefore order of C₂ = 3 × 1

4. Construct a 2 ×3 matrix A = [a_{j j}] whose elements a_{j j} are given by:

(i) a_{i j} = i × j

(ii) a_{i j} = 2i – j

(iii) a_{i j} = i + j

(iv) a_{i j} = (i + j)²/2

Solution:

(i) Given a_{i j} = i × j

Let A = [a_{i j}]_{2 × 3}

So, the elements in a 2 × 3 matrix are[a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃]

a₁₁ = 1 × 1 = 1

a₁₂ = 1 × 2 = 2

a₁₃ = 1 × 3 = 3

a₂₁ = 2 × 1 = 2

a₂₂ = 2 × 2 = 4

a₂₃ = 2 × 3 = 6

Substituting these values in matrix A we get,

(ii) Given a_{i j} = 2i – j

Let A = [a_{i j}]_2×3

So, the elements in a 2 × 3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃

a₁₁ = 2 × 1 – 1 = 2 – 1 = 1

a₁₂ = 2 × 1 – 2 = 2 – 2 = 0

a₁₃ = 2 × 1 – 3 = 2 – 3 = – 1

a₂₁ = 2 × 2 – 1 = 4 – 1 = 3

a₂₂ = 2 × 2 – 2 = 4 – 2 = 2

a₂₃ = 2 × 2 – 3 = 4 – 3 = 1

Substituting these values in matrix A we get,

(iii) Given a_{i j} = i + j

Let A = [a _{i j}] _2×3

So, the elements in a 2 × 3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃

a₁₁ = 1 + 1 = 2

a₁₂ = 1 + 2 = 3

a₁₃ = 1 + 3 = 4

a₂₁ = 2 + 1 = 3

a₂₂ = 2 + 2 = 4

a₂₃ = 2 + 3 = 5

Substituting these values in matrix A we get,

(iv) Given a_{i j} = (i + j)²/2

Let A = [a_{i j}]_2×3

So, the elements in a 2 × 3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃

Let A = [a_{i j}]_2×3

So, the elements in a 2 × 3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃

a₁₁ =

a₁₂ =

a₁₃ =

a₂₁ =

a₂₂ =

a₂₃ =

Substituting these values in matrix A we get,

5. Construct a 2 × 2 matrix A = [a_{i j}] whose elements a_{i j} are given by:

(i) (i + j)² /2

(ii) a_{i j} = (i – j)² /2

(iii) a_{i j} = (i – 2j)² /2

(iv) a_{i j} = (2i + j)² /2

(v) a_{i j} = |2i – 3j|/2

(vi) a_{i j} = |-3i + j|/2

(vii) a_{i j} = e^2ix sin x j

Solution:

(i) Given (i + j)² /2

Let A = [a_{i j}]_2×2

So, the elements in a 2 × 2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

Substituting these values in matrix A we get,

(ii) Given a_{i j} = (i – j)² /2

Let A = [a_{i j}]_2×2

So, the elements in a 2 × 2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

Substituting these values in matrix A we get,

(iii) Given a_{i j} = (i – 2j)² /2

Let A = [a_{i j}]_2×2

So, the elements in a 2 × 2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

Substituting these values in matrix A we get,

(iv) Given a_{i j} = (2i + j)² /2

Let A = [a_{i j}]_2×2

So, the elements in a 2 × 2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

Substituting these values in matrix A we get,

(v) Given a_{i j} = |2i – 3j|/2

Let A = [a_{i j}]_2×2

So, the elements in a 2×2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

Substituting these values in matrix A we get,

(vi) Given a_{i j} = |-3i + j|/2

Let A = [a_{i j}]_2×2

So, the elements in a 2 × 2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

Substituting these values in matrix A we get,

(vii) Given a_{i j} = e^2ix sin x j

Let A = [a_{i j}]_2×2

So, the elements in a 2 × 2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂,

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

Substituting these values in matrix A we get,

6. Construct a 3×4 matrix A = [a_{i j}] whose elements a_{i j} are given by:
(i) a_{i j} = i + j

(ii) a_{i j} = i – j

(iii) a_{i j} = 2i

(iv) a_{i j} = j

(v) a_{i j} = ½ |-3i + j|

Solution:

(i) Given a_{i j} = i + j

Let A = [a_{i j}]_2×3

So, the elements in a 3 × 4 matrix are

a₁₁, a₁₂, a₁₃, a_14, a₂₁, a₂₂, a₂₃, a₂₄, a_31, a_32, a_33, a₃₄

A =

a₁₁ = 1 + 1 = 2

a₁₂ = 1 + 2 = 3

a₁₃ = 1 + 3 = 4

a₁₄ = 1 + 4 = 5

a₂₁ = 2 + 1 = 3

a₂₂ = 2 + 2 = 4

a₂₃ = 2 + 3 = 5

a_{24 =} 2 + 4 = 6

a₃₁ = 3 + 1 = 4

a₃₂ = 3 + 2 = 5

a₃₃ = 3 + 3 = 6

a₃₄ = 3 + 4 = 7

Substituting these values in matrix A we get,

A =

(ii) Given a_{i j} = i – j

Let A = [a_{i j}]_2×3

So, the elements in a 3×4 matrix are

a₁₁, a₁₂, a₁₃, a_14, a₂₁, a₂₂, a₂₃, a₂₄, a_31, a_32, a_33, a₃₄

A =

a₁₁ = 1 – 1 = 0

a₁₂ = 1 – 2 = – 1

a₁₃ = 1 – 3 = – 2

a₁₄ = 1 – 4 = – 3

a₂₁ = 2 – 1 = 1

a₂₂ = 2 – 2 = 0

a₂₃ = 2 – 3 = – 1

a_{24 =} 2 – 4 = – 2

a₃₁ = 3 – 1 = 2

a₃₂ = 3 – 2 = 1

a₃₃ = 3 – 3 = 0

a₃₄ = 3 – 4 = – 1

Substituting these values in matrix A we get,

A =

(iii) Given a_{i j} = 2i

Let A = [a_{i j}]_2×3

So, the elements in a 3×4 matrix are

a₁₁, a₁₂, a₁₃, a_14, a₂₁, a₂₂, a₂₃, a₂₄, a_31, a_32, a_33, a₃₄

A =

a₁₁ = 2×1 = 2

a₁₂ = 2×1 = 2

a₁₃ = 2×1 = 2

a₁₄ = 2×1 = 2

a₂₁ = 2×2 = 4

a₂₂ = 2×2 = 4

a₂₃ = 2×2 = 4

a_{24 =} 2×2 = 4

a₃₁ = 2×3 = 6

a₃₂ = 2×3 = 6

a₃₃ = 2×3 = 6

a₃₄ = 2×3 = 6

Substituting these values in matrix A we get,

A =

(iv) Given a_{i j} = j

Let A = [a_{i j}]_2×3

So, the elements in a 3×4 matrix are

a₁₁, a₁₂, a₁₃, a_14, a₂₁, a₂₂, a₂₃, a₂₄, a_31, a_32, a_33, a₃₄

A =

a₁₁ = 1

a₁₂ = 2

a₁₃ = 3

a₁₄ = 4

a₂₁ = 1

a₂₂ = 2

a₂₃ = 3

a_{24 =} 4

a₃₁ = 1

a₃₂ = 2

a₃₃ = 3

a₃₄ = 4

Substituting these values in matrix A we get,

A =

(vi) Given a_{i j} = ½ |-3i + j|

Let A = [a_{i j}]_2×3

So, the elements in a 3×4 matrix are

a₁₁, a₁₂, a₁₃, a_14, a₂₁, a₂₂, a₂₃, a₂₄, a_31, a_32, a_33, a₃₄

A =

a₁₁ =
a₁₂ =
a₁₃ =
a₁₄ =
a₂₁ =

a₂₂ =

a₂₃ =

a_{24 =}
a₃₁ =
a₃₂ =
a₃₃ =
a₃₄ =

Substituting these values in matrix A we get,

A =

Multiplying by negative sign we get,

7. Construct a 4 × 3 matrix A = [a_{i j}] whose elements a_{i j} are given by:

(i) a_{i j} = 2i + i/j

(ii) a_{i j} = (i – j)/ (i + j)

(iii) a_{i j} = i

Solution:

(i) Given a_{i j} = 2i + i/j

Let A = [a_{i j}]_4×3

So, the elements in a 4 × 3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃, a_31, a_32, a_33, a_41, a_42, a₄₃

A =

a₁₁ =

a₁₂ =

a₁₃ =

a₂₁ =

a₂₂ =

a₂₃ =

a₃₁ =

a₃₂ =

a₃₃ =

a₄₁ =

a₄₂ =

a₄₃ =

Substituting these values in matrix A we get,

A =

(ii) Given a_{i j} = (i – j)/ (i + j)

Let A = [a_{i j}]_4×3

So, the elements in a 4 × 3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃, a_31, a_32, a_33, a_41, a_42, a₄₃

A =

a₁₁ =

a₁₂ =

a₁₃ =

a₂₁ =

a₂₂ =

a₂₃ =

a₃₁ =

a₃₂ =

a₃₃ =

a₄₁ =

a₄₂ =

a₄₃ =

Substituting these values in matrix A we get,

A =

(iii) Given a_{i j} = i

Let A = [a_{i j}]_4×3

So, the elements in a 4 × 3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃, a_31, a_32, a_33, a_41, a_42, a₄₃

A =

a₁₁ = 1

a₁₂ = 1

a₁₃ = 1

a₂₁ = 2

a₂₂ = 2

a₂₃ = 2

a₃₁ = 3

a₃₂ = 3

a₃₃ = 3

a₄₁ = 4

a₄₂ = 4

a₄₃ = 4

Substituting these values in matrix A we get,

A =

8. Find x, y, a and b if

Solution:

Given

Given that two matrices are equal.

We know that if two matrices are equal then the elements of each matrices are also equal.

Therefore by equating them we get,

3x + 4y = 2 …… (1)

x – 2y = 4 …… (2)

a + b = 5 …… (3)

2a – b = – 5 …… (4)

Multiplying equation (2) by 2 and adding to equation (1), we get

3x + 4y + 2x – 4y = 2 + 8

⇒ 5x = 10

⇒ x = 2

Now, substituting the value of x in equation (1)

3 × 2 + 4y = 2

⇒ 6 + 4y = 2

⇒ 4y = 2 – 6

⇒ 4y = – 4

⇒ y = – 1

Now by adding equation (3) and (4)

a + b + 2a – b = 5 + (– 5)

⇒ 3a = 5 – 5 = 0

⇒ a = 0

Now, again by substituting the value of a in equation (3), we get

0 + b = 5

⇒ b = 5

∴ a = 0, b = 5, x = 2 and y = – 1

9. Find x, y, a and b if

Solution:

We know that if two matrices are equal then the elements of each matrices are also equal.

Given that two matrices are equal.

Therefore by equating them we get,

2a + b = 4 …… (1)

And a – 2b = – 3 …… (2)

And 5c – d = 11 …… (3)

4c + 3d = 24 …… (4)

Multiplying equation (1) by 2 and adding to equation (2)

4a + 2b + a – 2b = 8 – 3

⇒ 5a = 5

⇒ a = 1

Now, substituting the value of a in equation (1)

2 × 1 + b = 4

⇒ 2 + b = 4

⇒ b = 4 – 2

⇒ b = 2

Multiplying equation (3) by 3 and adding to equation (4)

15c – 3d + 4c + 3d = 33 + 24

⇒ 19c = 57

⇒ c = 3

Now, substituting the value of c in equation (4)

4 × 3 + 3d = 24

⇒ 12 + 3d = 24

⇒ 3d = 24 – 12

⇒ 3d = 12

⇒ d = 4

∴ a = 1, b = 2, c = 3 and d = 4

10. Find the values of a, b, c and d from the following equations:

Solution:

Given

We know that if two matrices are equal then the elements of each matrices are also equal.

Given that two matrices are equal.

Therefore by equating them we get,

2a + b = 4 …… (1)

And a – 2b = – 3 …… (2)

And 5c – d = 11 …… (3)

4c + 3d = 24 …… (4)

Multiplying equation (1) by 2 and adding to equation (2)

4a + 2b + a – 2b = 8 – 3

⇒ 5a = 5

⇒ a = 1

Now, substituting the value of a in equation (1)

2 × 1 + b = 4

⇒ 2 + b = 4

⇒ b = 4 – 2

⇒ b = 2

Multiplying equation (3) by 3 and adding to equation (4)

15c – 3d + 4c + 3d = 33 + 24

⇒ 19c = 57

⇒ c = 3

Now, substituting the value of c in equation (4)

4 × 3 + 3d = 24

⇒ 12 + 3d = 24

⇒ 3d = 24 – 12

⇒ 3d = 12

⇒ d = 4

∴ a = 1, b = 2, c = 3 and d = 4

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Chapter 5 Algebra of Matrices Ex 5.2

1. Compute the following sums:

Solution:

(i) Given

Corresponding elements of two matrices should be added

Therefore, we get

Therefore,

(ii) Given

Therefore,

Find each of the following:

(i) 2A – 3B

(ii) B – 4C

(iii) 3A – C

(iv) 3A – 2B + 3C

Solution:

(i) Given

First we have to compute 2A

Now by computing 3B we get,

Now by we have to compute 2A – 3B we get

Therefore

(ii) Given

First we have to compute 4C,

Now,

Therefore we get,

(iii) Given

First we have to compute 3A,

Now,

Therefore,

(iv) Given

First we have to compute 3A

Now we have to compute 2B

By computing 3C we get,

Therefore,

(i) A + B and B + C

(ii) 2B + 3A and 3C – 4B

Solution:

(i) Consider A + B,

A + B is not possible because matrix A is an order of 2 x 2 and Matrix B is an order of 2 x 3, so the Sum of the matrix is only possible when their order is same.

Now consider B + C

(ii) Consider 2B + 3A

2B + 3A also does not exist because the order of matrix B and matrix A is different, so we cannot find the sum of these matrix.

Now consider 3C – 4B,

Solution:

Given

Now we have to compute 2A – 3B + 4C

5. If A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4), find

(i) A – 2B

(ii) B + C – 2A

(iii) 2A + 3B – 5C

Solution:

(i) Given A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4)

(ii) Given A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4)

We have to find B + C – 2A

Here,

Now we have to compute B + C – 2A

(iii) Given A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4)

Now we have to find 2A + 3B – 5C

Here,

Now consider 2A + 3B – 5C

6. Given the matrices

Verify that (A + B) + C = A + (B + C)

Solution:

Given

Now we have to verify (A + B) + C = A + (B + C)

First consider LHS, (A + B) + C,

Now consider RHS, that is A + (B + C)

Therefore LHS = RHS

Hence (A + B) + C = A + (B + C)

7. Find the matrices X and Y,

Solution:

Consider,

Now by simplifying we get,

Therefore,

Again consider,

Now by simplifying we get,

Therefore,

Solution:

Given

Now by transposing, we get

Therefore,

Solution:

Given

Now by multiplying equation (1) and (2) we get,

Now by adding equation (2) and (3) we get,

Now by substituting X in equation (2) we get,

Solution:

Consider

Now, again consider

Therefore,

And

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Chapter 5 Algebra of Matrices Ex 5.3

1. Compute the indicated products:

Solution:

(i) Consider

On simplification we get,

(ii) Consider

On simplification we get,

(iii) Consider

On simplification we get,

2. Show that AB ≠ BA in each of the following cases:

Solution:

(i) Consider,

Again consider,

From equation (1) and (2), it is clear that

AB ≠ BA

(ii) Consider,

Now again consider,

From equation (1) and (2), it is clear that

AB ≠ BA

(iii) Consider,

Now again consider,

From equation (1) and (2), it is clear that

AB ≠ BA

3. Compute the products AB and BA whichever exists in each of the following cases:

Solution:

(i) Consider,

BA does not exist

Because the number of columns in B is greater than the rows in A

(ii) Consider,

Again consider,

(iii) Consider,

AB = 0+(−1)+6+6

AB = 11

Again consider,

(iv) Consider,

4. Show that AB ≠ BA in each of the following cases:

Solution:

(i) Consider,

Again consider,

From equation (1) and (2), it is clear that

AB ≠ BA

(ii) Consider,

Again consider,

From equation (1) and (2) it is clear that,

AB ≠ BA

5. Evaluate the following:

Solution:

(i) Given

First we have to add first two matrix,

On simplifying, we get

(ii) Given,

First we have to multiply first two given matrix,

= 82

(iii) Given

First we have subtract the matrix which is inside the bracket,

Solution:

Given

We know that,

Again we know that,

Now, consider,

We have,

Now, from equation (1), (2), (3) and (4), it is clear that A² = B²= C²= I₂

Solution:

Given

Consider,

Now we have to find,

Solution:

Given

Consider,

Hence the proof.

Solution:

Given,

Consider,

Again consider,

Hence the proof.

Solution:

Given,

Consider,

Hence the proof.

Solution:

Given,

Consider,

We know that,

Again we have,

Solution:

Given,

Consider,

Again consider,

From equation (1) and (2) AB = BA = 0_3×3

Solution:

Given

Consider,

Again consider,

From equation (1) and (2) AB = BA = 0_3×3

Solution:

Given

Now consider,

Therefore AB = A

Again consider, BA we get,

Hence BA = B

Hence the proof.

Solution:

Given,

Consider,

Now again consider, B²

Now by subtracting equation (2) from equation (1) we get,

16. For the following matrices verify the associativity of matrix multiplication i.e. (AB) C = A (BC)

Solution:

(i) Given

Consider,

Now consider RHS,

From equation (1) and (2), it is clear that (AB) C = A (BC)

(ii) Given,

Consider the LHS,

Now consider RHS,

From equation (1) and (2), it is clear that (AB) C = A (BC)

17. For the following matrices verify the distributivity of matrix multiplication over matrix addition i.e. A (B + C) = AB + AC.

Solution:

(i) Given

Consider LHS,

Now consider RHS,

From equation (1) and (2), it is clear that A (B + C) = AB + AC

(ii) Given,

Consider the LHS

Now consider RHS,

Solution:

Given,

Consider the LHS,

Now consider RHS

From the above equations LHS = RHS

Therefore, A (B – C) = AB – AC.

19. Compute the elements a₄₃ and a₂₂ of the matrix:

Solution:

Given

From the above matrix, a₄₃ = 8and a₂₂ = 0

Solution:

Given

Consider,

Again consider,

Now, consider the RHS

Therefore, A³ = p I + q A + rA²

Hence the proof.

21. If ω is a complex cube root of unity, show that

Solution:

Given

It is also given that ω is a complex cube root of unity,

Consider the LHS,

We know that 1 + ω + ω² = 0 and ω³ = 1

Now by simplifying we get,

Again by substituting 1 + ω + ω² = 0 and ω³ = 1 in above matrix we get,

Therefore LHS = RHS

Hence the proof.

Solution:

Given,

Consider A²

Therefore A² = A

Solution:

Given

Consider A²,

Hence A² = I₃

Solution:

(i) Given

= 2x+1+2+x+3 = 0

= 3x+6 = 0

= 3x = -6

x = -6/3

x = -2

(ii) Given,

On comparing the above matrix we get,

x = 13

Solution:

Given

⇒ (2x+4)x+4(x+2)–1(2x+4) = 0

⇒ 2x² + 4x + 4x + 8 – 2x – 4 = 0

⇒ 2x² + 6x + 4 = 0

⇒ 2x² + 2x + 4x + 4 = 0

⇒ 2x (x + 1) + 4 (x + 1) = 0

⇒ (x + 1) (2x + 4) = 0

⇒ x = -1 or x = -2

Hence, x = -1 or x = -2

Solution:

Given

By multiplying we get,

Solution:

Given

Now we have to prove A² – A + 2 I = 0

Solution:

Given

Solution:

Given

Hence the proof.

Solution:

Given

Hence the proof.

Solution:

Given

Solution:

Given

Solution:

Given

Solution:

Given

Solution:

Given

Solution:

Given

I is identity matrix, so

Also given,

Now, we have to find A², we get

Now, we will find the matrix for 8A, we get

So,

Substitute corresponding values from eqn (i) and (ii), we get

And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal

Hence,

Therefore, the value of k is 7

Solution:

Given

To show that f (A) = 0

Substitute x = A in f(x), we get

I is identity matrix, so

Now, we will find the matrix for A², we get

Now, we will find the matrix for 2A, we get

Substitute corresponding values from eqn (ii) and (iii) in eqn (i), we get

So,

Hence Proved

Solution:

Given

So

Now, we will find the matrix for A², we get

Now, we will find the matrix for λ A, we get

But given, A² = λ A + μ I

Substitute corresponding values from equation (i) and (ii), we get

And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal

Hence, λ + 0 = 4 ⇒ λ = 4

And also, 2λ + μ = 7

Substituting the obtained value of λ in the above equation, we get

2(4) + μ = 7 ⇒ 8 + μ = 7 ⇒ μ = – 1

Therefore, the value of λ and μ are 4 and – 1 respectively

39. Find the value of x for which the matrix product

Solution:

We know,

is identity matrix of size 3.

So according to the given criteria

Now we will multiply the two matrices on LHS using the formula c_ij = a_i1b_1j + a_i2b_2j + … + a_in b_nj, we get

And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal

So we get

So the value of x is

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Chapter 5 Algebra of Matrices Ex 5.4

(i) (2A)^T = 2 A^T

(ii) (A + B)^T = A^T + B^T

(iii) (A – B)^T = A^T – B^T

(iv) (AB)^T = B^T A^T

Solution:

(i) Given

Consider,

Put the value of A

L.H.S = R.H.S

(ii) Given

Consider,

L.H.S = R.H.S

Hence proved.

(iii) Given

Consider,

L.H.S = R.H.S

(iv) Given

So,

Solution:

Given

L.H.S = R.H.S

So,

(i) A + B)^T = A^T + B^T

(ii) (AB)^T = B^T A^T

(iii) (2A)^T = 2 A^T

Solution:

(i) Given

Consider,

L.H.S = R.H.S

So,

(ii) Given

Consider,

L.H.S = R.H.S

So,

(iii) Given

Consider,

L.H.S = R.H.S

So,

Solution:

Given

Consider,

L.H.S = R.H.S

So,

Solution:

Given

Now we have to find (AB)^T

So,

---

Chapter 5 Algebra of Matrices Ex 5.5

Solution:

Given

Consider,

 … (i)

 … (ii)

From (i) and (ii) we can see that

A skew-symmetric matrix is a square matrix whose transpose equal to its negative, that is,

X = – X^T

So, A – A^T is a skew-symmetric.

Solution:

Given

Consider,

 … (i)

 … (ii)

From (i) and (ii) we can see that

A skew-symmetric matrix is a square matrix whose transpose equals its negative, that is,

X = – X^T

So, A – A^T is a skew-symmetric matrix.

Solution:

Given,

 is a symmetric matrix.

We know that A = [a_ij]_{m × n} is a symmetric matrix if a_ij = a_ji

So,

Hence, x = 4, y = 2, t = -3 and z can have any value.

4. Let. Find matrices X and Y such that X + Y = A, where X is a symmetric and y is a skew-symmetric matrix.

Solution:

Given,
 Then

Now,

Now,

X is a symmetric matrix.

Now,

-Y ^T = Y

Y is a skew symmetric matrix.

Hence, X + Y = A

Chapter 3 Binary Operations Ex 3.1

1. Determine whether the following operation define a binary operation on the given set or not:

(i) ‘*’ on N defined by a * b = a^b for all a, b ∈ N.

(ii) ‘O’ on Z defined by a O b = a^b for all a, b ∈ Z.

(iii)  ‘*’ on N defined by a * b = a + b – 2 for all a, b ∈ N

(iv) ‘×₆‘ on S = {1, 2, 3, 4, 5} defined by a ×₆ b = Remainder when a b is divided by 6.

(v) ‘+₆’ on S = {0, 1, 2, 3, 4, 5} defined by a +₆ b

(vi) ‘⊙’ on N defined by a ⊙ b= a^b + b^a for all a, b ∈ N

(vii) ‘*’ on Q defined by a * b = (a – 1)/ (b + 1) for all a, b ∈ Q

Solution:

(i) Given ‘*’ on N defined by a * b = a^b for all a, b ∈ N.

Let a, b ∈ N. Then,

a^b ∈ N      [∵ a^b≠0 and a, b is positive integer]

⇒ a * b ∈ N

Therefore,

a * b ∈ N, ∀ a, b ∈ N

Thus, * is a binary operation on N.

(ii) Given ‘O’ on Z defined by a O b = a^b for all a, b ∈ Z.

Both a = 3 and b = -1 belong to Z.

⇒ a * b = 3^-1

= 1/3 ∉ Z

Thus, * is not a binary operation on Z.

(iii)  Given ‘*’ on N defined by a * b = a + b – 2 for all a, b ∈ N

If a = 1 and b = 1,

a * b = a + b – 2

= 1 + 1 – 2

= 0 ∉ N

Thus, there exist a = 1 and b = 1 such that a * b ∉ N

So, * is not a binary operation on N.

(iv) Given ‘×₆‘ on S = {1, 2, 3, 4, 5} defined by a ×₆ b = Remainder when a b is divided by 6.

Consider the composition table,

X6	1	2	3	4	5
1	1	2	3	4	5
2	2	4	0	2	4
3	3	0	3	0	3
4	4	2	0	4	2
5	5	4	3	2	1

Here all the elements of the table are not in S.

⇒ For a = 2 and b = 3,

a ×₆ b = 2 ×₆ 3 = remainder when 6 divided by 6 = 0 ≠ S

Thus, ×₆ is not a binary operation on S.

(v) Given ‘+₆’ on S = {0, 1, 2, 3, 4, 5} defined by a +₆ b

Consider the composition table,

+6	0	1	2	3	4	5
0	0	1	2	3	4	5
1	1	2	3	4	5	0
2	2	3	4	5	0	1
3	3	4	5	0	1	2
4	4	5	0	1	2	3
5	5	0	1	2	3	4

Here all the elements of the table are not in S.

⇒ For a = 2 and b = 3,

a ×₆ b = 2 ×₆ 3 = remainder when 6 divided by 6 = 0 ≠ Thus, ×₆ is not a binary operation on S.

(vi) Given ‘⊙’ on N defined by a ⊙ b= a^b + b^a for all a, b ∈ N

Let a, b ∈ N. Then,

a^b, b^a ∈ N

⇒ a^b + b^a ∈ N      ∵AdditionisbinaryoperationonN

⇒ a ⊙ b ∈ N

Thus, ⊙ is a binary operation on N.

(vii) Given ‘*’ on Q defined by a * b = (a – 1)/ (b + 1) for all a, b ∈ Q

If a = 2 and b = -1 in Q,

a * b = (a – 1)/ (b + 1)

= (2 – 1)/ (- 1 + 1)

= 1/0 whichisnotdefined

For a = 2 and b = -1

a * b does not belongs to Q

So, * is not a binary operation in Q.

2. Determine whether or not the definition of * given below gives a binary operation. In the event that * is not a binary operation give justification of this.
(i) On Z⁺, defined * by a * b = a – b

(ii) On Z⁺, define * by a*b = ab

(iii) On R, define * by a*b = ab²

(iv) On Z⁺ define * by a * b = |a − b|

(v) On Z⁺ define * by a * b = a

(vi) On R, define * by a * b = a + 4b²

Here, Z⁺ denotes the set of all non-negative integers.

Solution:

(i) Given On Z⁺, defined * by a * b = a – b

If a = 1 and b = 2 in Z⁺, then

a * b = a – b

= 1 – 2

= -1 ∉ Z⁺ [because Z⁺ is the set of non-negative integers]

For a = 1 and b = 2,

a * b ∉ Z⁺

Thus, * is not a binary operation on Z⁺.

(ii) Given Z⁺, define * by a*b = a b

Let a, b ∈ Z⁺

⇒ a, b ∈ Z⁺

⇒ a * b ∈ Z⁺

Thus, * is a binary operation on R.

(iii) Given on R, define by a*b = ab²

Let a, b ∈ R

⇒ a, b² ∈ R

⇒ ab² ∈ R

⇒ a * b ∈ R

Thus, * is a binary operation on R.

(iv) Given on Z⁺ define * by a * b = |a − b|

Let a, b ∈ Z⁺

⇒ | a – b | ∈ Z⁺

⇒ a * b ∈ Z⁺

Therefore,

a * b ∈ Z⁺, ∀ a, b ∈ Z⁺

Thus, * is a binary operation on Z⁺.

(v) Given on Z⁺ define * by a * b = a

Let a, b ∈ Z⁺

⇒ a ∈ Z⁺

⇒ a * b ∈ Z⁺

Therefore, a * b ∈ Z⁺ ∀ a, b ∈ Z⁺

Thus, * is a binary operation on Z⁺.

(vi) Given On R, define * by a * b = a + 4b²

Let a, b ∈ R

⇒ a, 4b² ∈ R

⇒ a + 4b² ∈ R

⇒ a * b ∈ R

Therefore, a *b ∈ R, ∀ a, b ∈ R

Thus, * is a binary operation on R.

3. Let * be a binary operation on the set I of integers, defined by a * b = 2a + b − 3. Find the value of 3 * 4.

Solution:

Given a * b = 2a + b – 3

3 * 4 = 2 (3) + 4 – 3

= 6 + 4 – 3

= 7

4. Is * defined on the set {1, 2, 3, 4, 5} by a * b = LCM of a and b a binary operation? Justify your answer.

Solution:

LCM	1	2	3	4	5
1	1	2	3	4	5
2	2	2	6	4	10
3	3	5	3	12	15
4	4	4	12	4	20
5	5	10	15	20	5

In the given composition table, all the elements are not in the set {1, 2, 3, 4, 5}.

If we consider a = 2 and b = 3, a * b = LCM of a and b = 6 ∉ {1, 2, 3, 4, 5}.

Thus, * is not a binary operation on {1, 2, 3, 4, 5}.

5. Let S = {a, b, c}. Find the total number of binary operations on S.

Solution:

Number of binary operations on a set with n elements is (n^{n^{2}})

Here, S = {a, b, c}

Number of elements in S = 3

Number of binary operations on a set with 3 elements is (3^{3^{2}})

Chapter 3 Binary Operations Ex 3.2

1. Let ‘*’ be a binary operation on N defined by a * b = l.c.m. (a, b) for all a, b ∈ N
(i) Find 2 * 4, 3 * 5, 1 * 6.

(ii) Check the commutativity and associativity of ‘*’ on N.

Solution:

(i) Given a * b = 1.c.m. (a, b)

2 * 4 = l.c.m. (2, 4)

= 4

3 * 5 = l.c.m. (3, 5)

= 15

1 * 6 = l.c.m. (1, 6)

= 6

(ii) We have to prove commutativity of *

Let a, b ∈ N

a * b = l.c.m (a, b)

= l.c.m (b, a)

= b * a

Therefore

a * b = b * a ∀ a, b ∈ N

Thus * is commutative on N.

Now we have to prove associativity of *

Let a, b, c ∈ N

a * (b * c ) = a * l.c.m. (b, c)

= l.c.m. (a, (b, c))

= l.c.m (a, b, c)

(a * b) * c = l.c.m. (a, b) * c

= l.c.m. ((a, b), c)

= l.c.m. (a, b, c)

Therefore

(a * (b * c) = (a * b) * c, ∀ a, b , c ∈ N

Thus, * is associative on N.

2. Determine which of the following binary operation is associative and which is commutative:

(i) * on N defined by a * b = 1 for all a, b ∈ N

(ii) * on Q defined by a * b = (a + b)/2 for all a, b ∈ Q

Solution:

(i) We have to prove commutativity of *

Let a, b ∈ N

a * b = 1

b * a = 1

Therefore,

a * b = b * a, for all a, b ∈ N

Thus * is commutative on N.

Now we have to prove associativity of *

Let a, b, c ∈ N

Then a * (b * c) = a * (1)

= 1

(a * b) *c = (1) * c

= 1

Therefore a * (b * c) = (a * b) *c for all a, b, c ∈ N

Thus, * is associative on N.

(ii) First we have to prove commutativity of *

Let a, b ∈ N

a * b = (a + b)/2

= (b + a)/2

= b * a

Therefore, a * b = b * a, ∀ a, b ∈ N

Thus * is commutative on N.

Now we have to prove associativity of *

Let a, b, c ∈ N

a * (b * c) = a * (b + c)/2

= a+(b+c)/2

= (2a + b + c)/4

Now, (a * b) * c = (a + b)/2 * c

= (a+b)/2+c /2

= (a + b + 2c)/4

Thus, a * (b * c) ≠ (a * b) * c

If a = 1, b= 2, c = 3

1 * (2 * 3) = 1 * (2 + 3)/2

= 1 * (5/2)

= 1+(5/2)/2

= 7/4

(1 * 2) * 3 = (1 + 2)/2 * 3

= 3/2 * 3

= (3/2)+3/2

= 4/9

Therefore, there exist a = 1, b = 2, c = 3 ∈ N such that a * (b * c) ≠ (a * b) * c

Thus, * is not associative on N.

3. Let A be any set containing more than one element. Let ‘*’ be a binary operation on A defined by a * b = b for all a, b ∈ A Is ‘*’ commutative or associative on A?

Solution:

Let a, b ∈ A

Then, a * b = b

b * a = a

Therefore a * b ≠ b * a

Thus, * is not commutative on A

Now we have to check associativity:

Let a, b, c ∈ A

a * (b * c) = a * c

= c

Therefore

a * (b * c) = (a * b) * c, ∀ a, b, c ∈ A

Thus, * is associative on A

4. Check the commutativity and associativity of each of the following binary operations:

(i) ‘*’ on Z defined by a * b = a + b + a b for all a, b ∈ Z 

(ii) ‘*’ on N defined by a * b = 2^ab for all a, b ∈ N

(iii) ‘*’ on Q defined by a * b = a – b for all a, b ∈ Q

(iv) ‘⊙’ on Q defined by a ⊙ b = a² + b² for all a, b ∈ Q

(v) ‘o’ on Q defined by a o b = (ab/2) for all a, b ∈ Q

(vi) ‘*’ on Q defined by a * b = ab² for all a, b ∈ Q

(vii) ‘*’ on Q defined by a * b = a + a b for all a, b ∈ Q

(viii) ‘*’ on R defined by a * b = a + b -7 for all a, b ∈ R

(ix) ‘*’ on Q defined by a * b = (a – b)² for all a, b ∈ Q

(x) ‘*’ on Q defined by a * b = a b + 1 for all a, b ∈ Q

(xi) ‘*’ on N defined by a * b = a^b for all a, b ∈ N

(xii) ‘*’ on Z defined by a * b = a – b for all a, b ∈ Z

(xiii) ‘*’ on Q defined by a * b = (ab/4) for all a, b ∈ Q

(xiv) ‘*’ on Z defined by a * b = a + b – ab for all a, b ∈ Z

(xv) ‘*’ on Q defined by a * b = gcd (a, b) for all a, b ∈ Q

Solution:

(i) First we have to check commutativity of *

Let a, b ∈ Z

Then a * b = a + b + ab

= b + a + ba

= b * a

Therefore,

a * b = b * a, ∀ a, b ∈ Z

Now we have to prove associativity of *

Let a, b, c ∈ Z, Then,

a * (b * c) = a * (b + c + b c)

= a + (b + c + b c) + a (b + c + b c)

= a + b + c + b c + a b + a c + a b c

(a * b) * c = (a + b + a b) * c

= a + b + a b + c + (a + b + a b) c

= a + b + a b + c + a c + b c + a b c

Therefore,

a * (b * c) = (a * b) * c, ∀ a, b, c ∈ Z

Thus, * is associative on Z.

(ii) First we have to check commutativity of *

Let a, b ∈ N

a * b = 2^ab

= 2^ba

= b * a

Therefore, a * b = b * a, ∀ a, b ∈ N

Thus, * is commutative on N

Now we have to check associativity of *

Let a, b, c ∈ N

Then, a * (b * c) = a * (2^bc)

=(2^{a*2^{bc}})

(a * b) * c = (2^ab) * c

=(2^{ab*2^{c}})

Therefore, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on N

(iii) First we have to check commutativity of *

Let a, b ∈ Q, then

a * b = a – b

b * a = b – a

Therefore, a * b ≠ b * a

Thus, * is not commutative on Q

Now we have to check associativity of *

Let a, b, c ∈ Q, then

a * (b * c) = a * (b – c)

= a – (b – c)

= a – b + c

(a * b) * c = (a – b) * c

= a – b – c

Therefore,

a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q

(iv) First we have to check commutativity of ⊙

Let a, b ∈ Q, then

a ⊙ b = a² + b²

= b² + a²

= b ⊙ a

Therefore, a ⊙ b = b ⊙ a, ∀ a, b ∈ Q

Thus, ⊙ on Q

Now we have to check associativity of ⊙

Let a, b, c ∈ Q, then

a ⊙ (b ⊙ c) = a ⊙ (b² + c²)

= a² + (b² + c²)²

= a² + b⁴ + c⁴ + 2b²c²

(a ⊙ b) ⊙ c = (a² + b²) ⊙ c

= (a² + b²)² + c²

= a⁴ + b⁴ + 2a²b² + c²

Therefore,

(a ⊙ b) ⊙ c ≠ a ⊙ (b ⊙ c)

Thus, ⊙ is not associative on Q.

(v) First we have to check commutativity of o

Let a, b ∈ Q, then

a o b = (ab/2)

= (b a/2)

= b o a

Therefore, a o b = b o a, ∀ a, b ∈ Q

Thus, o is commutative on Q

Now we have to check associativity of o

Let a, b, c ∈ Q, then

a o (b o c) = a o (b c/2)

= a(bc/2)/2

= a(bc/2)/2

= (a b c)/4

(a o b) o c = (ab/2) o c

= (ab/2)c /2

= (a b c)/4

Therefore a o (b o c) = (a o b) o c, ∀ a, b, c ∈ Q

Thus, o is associative on Q.

(vi) First we have to check commutativity of *

Let a, b ∈ Q, then

a * b = ab²

b * a = ba²

Therefore,

a * b ≠ b * a

Thus, * is not commutative on Q

Now we have to check associativity of *

Let a, b, c ∈ Q, then

a * (b * c) = a * (bc²)

= a (bc²)²

= ab² c⁴

(a * b) * c = (ab²) * c

= ab²c²

Therefore a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(vii) First we have to check commutativity of *

Let a, b ∈ Q, then

a * b = a + ab

b * a = b + ba

= b + ab

Therefore, a * b ≠ b * a

Thus, * is not commutative on Q.

Now we have to prove associativity on Q.

Let a, b, c ∈ Q, then

a * (b * c) = a * (b + b c)

= a + a (b + b c)

= a + ab + a b c

(a * b) * c = (a + a b) * c

= (a + a b) + (a + a b) c

= a + a b + a c + a b c

Therefore a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(viii) First we have to check commutativity of *

Let a, b ∈ R, then

a * b = a + b – 7

= b + a – 7

= b * a

Therefore,

a * b = b * a, for all a, b ∈ R

Thus, * is commutative on R

Now we have to prove associativity of * on R.

Let a, b, c ∈ R, then

a * (b * c) = a * (b + c – 7)

= a + b + c -7 -7

= a + b + c – 14

(a * b) * c = (a + b – 7) * c

= a + b – 7 + c – 7

= a + b + c – 14

Therefore,

a * (b * c ) = (a * b) * c, for all a, b, c ∈ R

Thus, * is associative on R.

(ix) First we have to check commutativity of *

Let a, b ∈ Q, then

a * b = (a – b)²

= (b – a)²

= b * a

Therefore,

a * b = b * a, for all a, b ∈ Q

Thus, * is commutative on Q

Now we have to prove associativity of * on Q

Let a, b, c ∈ Q, then

a * (b * c) = a * (b – c)²

= a * (b² + c² – 2 b c)

= (a – b² – c² + 2bc)²

(a * b) * c = (a – b)² * c

= (a² + b² – 2ab) * c

= (a² + b² – 2ab – c)²

Therefore, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(x) First we have to check commutativity of *

Let a, b ∈ Q, then

a * b = ab + 1

= ba + 1

= b * a

Therefore

a * b = b * a, for all a, b ∈ Q

Thus, * is commutative on Q

Now we have to prove associativity of * on Q

Let a, b, c ∈ Q, then

a * (b * c) = a * (bc + 1)

= a (b c + 1) + 1

= a b c + a + 1

(a * b) * c = (ab + 1) * c

= (ab + 1) c + 1

= a b c + c + 1

Therefore, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(xi) First we have to check commutativity of *

Let a, b ∈ N, then

a * b = a^b

b * a = b^a

Therefore, a * b ≠ b * a

Thus, * is not commutative on N.

Now we have to check associativity of *

a * (b * c) = a * (b^c)

=

(a * b) * c = (a^b) * c

= (a^b)^c

= a^bc

Therefore, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on N

(xii) First we have to check commutativity of *

Let a, b ∈ Z, then

a * b = a – b

b * a = b – a

Therefore,

a * b ≠ b * a

Thus, * is not commutative on Z.

Now we have to check associativity of *

Let a, b, c ∈ Z, then

a * (b * c) = a * (b – c)

= a – (b – c)

= a – (b + c)

(a * b) * c = (a – b) – c

= a – b – c

Therefore, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Z

(xiii) First we have to check commutativity of *

Let a, b ∈ Q, then

a * b = (ab/4)

= (ba/4)

= b * a

Therefore, a * b = b * a, for all a, b ∈ Q

Thus, * is commutative on Q

Now we have to check associativity of *

Let a, b, c ∈ Q, then

a * (b * c) = a * (b c/4)

= a(bc/4)/4

= (a b c/16)

(a * b) * c = (ab/4) * c

= (ab/4)c/4

= a b c/16

Therefore,

a * (b * c) = (a * b) * c for all a, b, c ∈ Q

Thus, * is associative on Q.

(xiv) First we have to check commutativity of *

Let a, b ∈ Z, then

a * b = a + b – ab

= b + a – ba

= b * a

Therefore, a * b = b * a, for all a, b ∈ Z

Thus, * is commutative on Z.

Now we have to check associativity of *

Let a, b, c ∈ Z

a * (b * c) = a * (b + c – b c)

= a + b + c- b c – ab – ac + a b c

(a * b) * c = (a + b – a b) c

= a + b – ab + c – (a + b – ab) c

= a + b + c – ab – ac – bc + a b c

Therefore,

a * (b * c) = (a * b) * c, for all a, b, c ∈ Z

Thus, * is associative on Z.

(xv) First we have to check commutativity of *

Let a, b ∈ N, then

a * b = gcd (a, b)

= gcd (b, a)

= b * a

Therefore, a * b = b * a, for all a, b ∈ N

Thus, * is commutative on N.

Now we have to check associativity of *

Let a, b, c ∈ N

a * (b * c) = a * gcd(a,b)

= gcd (a, b, c)

(a * b) * c = gcd(a,b) * c

= gcd (a, b, c)

Therefore,

a * (b * c) = (a * b) * c, for all a, b, c ∈ N

Thus, * is associative on N.

5. If the binary operation o is defined by a0b = a + b – ab on the set Q – {-1} of all rational numbers other than 1, show that o is commutative on Q – 1.

Solution:

Let a, b ∈ Q – {-1}.

Then aob = a + b – ab

= b+ a – ba

= boa

Therefore,

aob = boa for all a, b ∈ Q – {-1}

Thus, o is commutative on Q – {-1}

6. Show that the binary operation * on Z defined by a * b = 3a + 7b is not commutative?

Solution:

Let a, b ∈ Z

a * b = 3a + 7b

b * a = 3b + 7a

Thus, a * b ≠ b * a

Let a = 1 and b = 2

1 * 2 = 3 × 1 + 7 × 2

= 3 + 14

= 17

2 * 1 = 3 × 2 + 7 × 1

= 6 + 7

= 13

Therefore, there exist a = 1, b = 2 ∈ Z such that a * b ≠ b * a

Thus, * is not commutative on Z.

7. On the set Z of integers a binary operation * is defined by a 8 b = ab + 1 for all a, b ∈ Z. Prove that * is not associative on Z.

Solution:

Let a, b, c ∈ Z

a * (b * c) = a * (bc + 1)

= a (bc + 1) + 1

= a b c + a + 1

(a * b) * c = (ab+ 1) * c

= (ab + 1) c + 1

= a b c + c + 1

Thus, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Z.

Chapter 3 Binary Operations Ex 3.3

1. Find the identity element in the set I⁺ of all positive integers defined by a * b = a + b for all a, b ∈ I⁺.

Solution:

Let e be the identity element in I⁺ with respect to * such that

a * e = a = e * a, ∀ a ∈ I⁺

a * e = a and e * a = a, ∀ a ∈ I⁺

a + e = a and e + a = a, ∀ a ∈ I⁺

e = 0, ∀ a ∈ I⁺

Thus, 0 is the identity element in I⁺ with respect to *.

2. Find the identity element in the set of all rational numbers except – 1 with respect to * defined by a * b = a + b + ab

Solution:

Let e be the identity element in I⁺ with respect to * such that

a * e = a = e * a, ∀ a ∈ Q – {-1}

a * e = a and e * a = a, ∀ a ∈ Q – {-1}

a + e + ae = a and e + a + ea = a, ∀ a ∈ Q – {-1}

e + ae = 0 and e + ea = 0, ∀ a ∈ Q – {-1}

e (1 + a) = 0 and e (1 + a) = 0, ∀ a ∈ Q – {-1}

e = 0, ∀ a ∈ Q – {-1} becauseanotequalto−1

Thus, 0 is the identity element in Q – {-1} with respect to *.

---

Chapter 3 Binary Operations Ex 3.4

1. Let * be a binary operation on Z defined by a * b = a + b – 4 for all a, b ∈ Z.

(i) Show that * is both commutative and associative.

(ii) Find the identity element in Z

(iii) Find the invertible element in Z.

Solution:

(i) First we have to prove commutativity of *

Let a, b ∈ Z. then,

a * b = a + b – 4

= b + a – 4

= b * a

Therefore,

a * b = b * a, ∀ a, b ∈ Z

Thus, * is commutative on Z.

Now we have to prove associativity of Z.

Let a, b, c ∈ Z. then,

a * (b * c) = a * (b + c – 4)

= a + b + c -4 – 4

= a + b + c – 8

(a * b) * c = (a + b – 4) * c

= a + b – 4 + c – 4

= a + b + c – 8

Therefore,

a * (b * c) = (a * b) * c, for all a, b, c ∈ Z

Thus, * is associative on Z.

(ii) Let e be the identity element in Z with respect to * such that

a * e = a = e * a ∀ a ∈ Z

a * e = a and e * a = a, ∀ a ∈ Z

a + e – 4 = a and e + a – 4 = a, ∀ a ∈ Z

e = 4, ∀ a ∈ Z

Thus, 4 is the identity element in Z with respect to *.

(iii) Let a ∈ Z and b ∈ Z be the inverse of a. Then,

a * b = e = b * a

a * b = e and b * a = e

a + b – 4 = 4 and b + a – 4 = 4

b = 8 – a ∈ Z

Thus, 8 – a is the inverse of a ∈ Z

2. Let * be a binary operation on Q₀ (set of non-zero rational numbers) defined by a * b = (3ab/5) for all a, b ∈ Q₀. Show that * is commutative as well as associative. Also, find its identity element, if it exists.

Solution:

First we have to prove commutativity of *

Let a, b ∈ Q₀

a * b = (3ab/5)

= (3ba/5)

= b * a

Therefore, a * b = b * a, for all a, b ∈ Q₀

Now we have to prove associativity of *

Let a, b, c ∈ Q₀

a * (b * c) = a * (3bc/5)

= a(3bc/5) /5

= 3 abc/25

(a * b) * c = (3 ab/5) * c

= (3ab/5)c/ 5

= 3 abc /25

Therefore a * (b * c) = (a * b) * c, for all a, b, c ∈ Q₀

Thus * is associative on Q₀

Now we have to find the identity element

Let e be the identity element in Z with respect to * such that

a * e = a = e * a ∀ a ∈ Q₀

a * e = a and e * a = a, ∀ a ∈ Q₀

3ae/5 = a and 3ea/5 = a, ∀ a ∈ Q₀

e = 5/3 ∀ a ∈ Q₀ becauseaisnotequalto0

Thus, 5/3 is the identity element in Q₀ with respect to *.

3. Let * be a binary operation on Q – {-1} defined by a * b = a + b + ab for all a, b ∈ Q – {-1}. Then,

(i) Show that * is both commutative and associative on Q – {-1}

(ii) Find the identity element in Q – {-1}

(iii) Show that every element of Q – {-1} is invertible. Also, find inverse of an arbitrary element.

Solution:

(i) First we have to check commutativity of *

Let a, b ∈ Q – {-1}

Then a * b = a + b + ab

= b + a + ba

= b * a

Therefore,

a * b = b * a, ∀ a, b ∈ Q – {-1}

Now we have to prove associativity of *

Let a, b, c ∈ Q – {-1}, Then,

a * (b * c) = a * (b + c + b c)

= a + (b + c + b c) + a (b + c + b c)

= a + b + c + b c + a b + a c + a b c

(a * b) * c = (a + b + a b) * c

= a + b + a b + c + (a + b + a b) c

= a + b + a b + c + a c + b c + a b c

Therefore,

a * (b * c) = (a * b) * c, ∀ a, b, c ∈ Q – {-1}

Thus, * is associative on Q – {-1}.

(ii) Let e be the identity element in I⁺ with respect to * such that

a * e = a = e * a, ∀ a ∈ Q – {-1}

a * e = a and e * a = a, ∀ a ∈ Q – {-1}

a + e + ae = a and e + a + ea = a, ∀ a ∈ Q – {-1}

e + ae = 0 and e + ea = 0, ∀ a ∈ Q – {-1}

e (1 + a) = 0 and e (1 + a) = 0, ∀ a ∈ Q – {-1}

e = 0, ∀ a ∈ Q – {-1} becauseanotequalto−1

Thus, 0 is the identity element in Q – {-1} with respect to *.

(iii) Let a ∈ Q – {-1} and b ∈ Q – {-1} be the inverse of a. Then,

a * b = e = b * a

a * b = e and b * a = e

a + b + ab = 0 and b + a + ba = 0

b (1 + a) = – a Q – {-1}

b = -a/1 + a Q – {-1} becauseanotequalto−1

Thus, -a/1 + a is the inverse of a ∈ Q – {-1}

4. Let A = R₀ × R, where R₀ denote the set of all non-zero real numbers. A binary operation ‘O’ is defined on A as follows: (a, b) O (c, d) = (ac, bc + d) for all (a, b), (c, d) ∈ R₀ × R.

(i) Show that ‘O’ is commutative and associative on A

(ii) Find the identity element in A

(iii) Find the invertible element in A.

Solution:

(i) Let X = (a, b) and Y = (c, d) ∈ A, ∀ a, c ∈ R₀ and b, d ∈ R

Then, X O Y = (ac, bc + d)

And Y O X = (ca, da + b)

Therefore,

X O Y = Y O X, ∀ X, Y ∈ A

Thus, O is not commutative on A.

Now we have to check associativity of O

Let X = (a, b), Y = (c, d) and Z = (e, f), ∀ a, c, e ∈ R₀ and b, d, f ∈ R

X O (Y O Z) = (a, b) O (ce, de + f)

= (ace, bce + de + f)

(X O Y) O Z = (ac, bc + d) O (e, f)

= (ace, (bc + d) e + f)

= (ace, bce + de + f)

Therefore, X O (Y O Z) = (X O Y) O Z, ∀ X, Y, Z ∈ A

(ii) Let E = (x, y) be the identity element in A with respect to O, ∀ x ∈ R₀ and y ∈ R

Such that,

X O E = X = E O X, ∀ X ∈ A

X O E = X and EOX = X

(ax, bx +y) = (a, b) and (xa, ya + b) = (a, b)

Considering (ax, bx + y) = (a, b)

ax = a

x = 1

And bx + y = b

y = 0 sincex=1

Considering (xa, ya + b) = (a, b)

xa = a

x = 1

And ya + b = b

y = 0 sincex=1

Therefore (1, 0) is the identity element in A with respect to O.

(iii) Let F = (m, n) be the inverse in A ∀ m ∈ R₀ and n ∈ R

X O F = E and F O X = E

(am, bm + n) = (1, 0) and (ma, na + b) = (1, 0)

Considering (am, bm + n) = (1, 0)

am = 1

m = 1/a

And bm + n = 0

n = -b/a sincem=1/a

Considering (ma, na + b) = (1, 0)

ma = 1

m = 1/a

And na + b = 0

n = -b/a

Therefore the inverse of (a, b) ∈ A with respect to O is (1/a, -b/a)

Chapter 3 Binary Operations Ex 3.5

1. Construct the composition table for ×₄ on set S = {0, 1, 2, 3}.

Solution:

Given that ×₄ on set S = {0, 1, 2, 3}

Here,

1 ×₄ 1 = remainder obtained by dividing 1 × 1 by 4

= 1

0 ×₄ 1 = remainder obtained by dividing 0 × 1 by 4

= 0

2 ×₄ 3 = remainder obtained by dividing 2 × 3 by 4

= 2

3 ×₄ 3 = remainder obtained by dividing 3 × 3 by 4

= 1

So, the composition table is as follows:

×4	0	1	2	3
0	0	0	0	0
1	0	1	2	3
2	0	2	0	2
3	0	3	2	1

2. Construct the composition table for +₅ on set S = {0, 1, 2, 3, 4}

Solution:

1 +₅ 1 = remainder obtained by dividing 1 + 1 by 5

= 2

3 +₅ 1 = remainder obtained by dividing 3 + 1 by 5

= 2

4 +₅ 1 = remainder obtained by dividing 4 + 1 by 5

= 3

So, the composition table is as follows:

+5	0	1	2	3	4
0	0	1	2	3	4
1	1	2	3	4	0
2	2	3	4	0	1
3	3	4	0	1	2
4	4	0	1	2	3

3. Construct the composition table for ×₆ on set S = {0, 1, 2, 3, 4, 5}.

Solution:

Here,

1 ×₆ 1 = remainder obtained by dividing 1 × 1 by 6

= 1

3 ×₆ 4 = remainder obtained by dividing 3 × 4 by 6

= 0

4 ×₆ 5 = remainder obtained by dividing 4 × 5 by 6

= 2

So, the composition table is as follows:

×6	0	1	2	3	4	5
0	0	0	0	0	0	0
1	0	1	2	3	4	5
2	0	2	4	0	2	4
3	0	3	0	3	0	3
4	0	4	2	0	4	2
5	0	5	4	3	2	1

4. Construct the composition table for ×₅ on set Z₅ = {0, 1, 2, 3, 4}

Solution:

Here,

1 ×₅ 1 = remainder obtained by dividing 1 × 1 by 5

= 1

3 ×₅ 4 = remainder obtained by dividing 3 × 4 by 5

= 2

4 ×₅ 4 = remainder obtained by dividing 4 × 4 by 5

= 1

So, the composition table is as follows:

×5	0	1	2	3	4
0	0	0	0	0	0
1	0	1	2	3	4
2	0	2	4	1	3
3	0	3	1	4	2
4	0	4	3	2	1

5. For the binary operation ×₁₀ set S = {1, 3, 7, 9}, find the inverse of 3.

Solution:

Here,

1 ×₁₀ 1 = remainder obtained by dividing 1 × 1 by 10

= 1

3 ×₁₀ 7 = remainder obtained by dividing 3 × 7 by 10

= 1

7 ×₁₀ 9 = remainder obtained by dividing 7 × 9 by 10

= 3

So, the composition table is as follows:

×10	1	3	7	9
1	1	3	7	9
3	3	9	1	7
7	7	1	9	3
9	9	7	3	1

From the table we can observe that elements of first row as same as the top-most row.

So, 1 ∈ S is the identity element with respect to ×₁₀

Now we have to find inverse of 3

3 ×₁₀ 7 = 1

So the inverse of 3 is 7.