Exercise 4.1 Page: 81

Evaluate the following determinants in Exercise 1 and 2.

Question 1. 

Solution
  = 2(-1) – 4(-5) = -2 + 20 = 18

Question 2. (i) 

(ii) 

Solution
(i) 

= (cosθ)(cosθ) – (-sinθ) (sinθ)= cos² θ + sin² θ= 1

(ii) 

= (x² − x + 1)(x + 1) − (x − 1)(x + 1)

= x³ − x² + x + x² − x + 1 − (x² − 1)

= x³ + 1 − x² + 1

= x³ − x² + 2

Question 3. If A =  then show that |2A| = 4|A|

Solution
Given: A = 

then 2A = 2 x 

Hence, proved.

Question 4. If A =  then show that 3|A| = 27|A|

Solution
Given: A =  then 3A =3

It can be observed that in the first column, two entries are zero. Thus, we expand along the first column (C₁) for easier calculation.

 

Hence, proved.

Question 5. Evaluate the determinants:

(i) 

(ii)  

(iii) 

(iv)  

Solution
Evaluate the determinants:

(i) Given: 

It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation.

=

(ii) Given: 

By expanding along the first row, we have:

=

(iii) Given: 

Expanding along first row,

=

= 0 + 6 – 6 = 0

(iv) Given: 

Expanding along first row,

=

= -10 + 15 = 5

Question 6. If A =  find |A|

Solution
Given:     A = 

Expanding along first row,

=

Question7. Find the value of x if:

(i) 

(ii) 

Solution
(i) Given: 

⇒ 2 x 1 – 5 x 4 = 2x * x – 6 x 4

⇒ 2 – 20 = 2x² – 24

⇒ 2x² = 6

⇒  x² = 3

⇒ x = ± √3

(ii) 

⇒ 2 x 5 – 4 x 3 = x * 5 – 2x – 3

⇒10 – 12 = 5x – 6x

⇒ – 2 = -x

⇒ x = 2

Question 8. If  then x is equal to:

(A) 6

(B) ± 6

(C) – 6

(D) 0

Solution
Given: 

⇒x * x – 18 x 2 = 6 x 6 – 18 x 2

⇒x² – 36 = 36 – 36

⇒x² – 36 = 0

⇒x = ± 6

Therefore, option (B) is correct.

Exercise 4.2 Page: 83

Question 1. Find the area of the triangle with vertices at the points given in each of the following:

(i) (1, 0), (6, 0), (4, 3)

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

(iii) (−2, −3), (3, 2), (−1, −8)

Solution :

(i) The area of the triangle with vertices (1, 0), (6, 0), (4, 3) is given by the relation,

=

(ii) The area of the triangle with vertices (2, 7), (1, 1), (10, 8) is given by the relation,

=

(iii) The area of the triangle with vertices (−2, −3), (3, 2), (−1, −8)

=

Question 2. Show that the points A(a,b + c), B(b, c + a), C(c, a+b) are collinear.

Solution :

Therefore, points A, B and C are collinear.

Question 3. Find values of k if area of triangle is 4 sq. units and vertices are:

(i) (k, 0), (4, 0), (0, 2)

(ii) (−2, 0), (0, 4), (0, k)

Solution :

We know that the area of a triangle whose vertices are (x₁, y₁), (x₂, y₂), and

(x₃, y₃) is the absolute value of the determinant (Δ), where

When −k + 4 = − 4, k = 8.

When −k + 4 = 4, k = 0.

Hence, k = 0, 8.

(ii) The area of the triangle with vertices (−2, 0), (0, 4), (0, k) is given by the relation,

∴k − 4 = ± 4

When k − 4 = − 4, k = 0.

When k − 4 = 4, k = 8.

Hence, k = 0, 8.

Question4. (i) Find the equation of the line joining (1, 2) and (3, 6) using determinants.

(ii) Find the equation of the line joining (3, 1) and (9, 3) using determinants.

Solution
(i) Let P(x, y) be any point on the line joining the points (1, 2) and (3, 6).

Then, Area of triangle that could be formed by these points is zero.

Hence, the equation of the line joining the given points is y = 2x.

(ii) Let P (x, y) be any point on the line joining points A (3, 1) and

B (9, 3). Then, the points A, B, and P are collinear. Therefore, the area of triangle ABP will be zero.

Hence, the equation of the line joining the given points is x − 3y = 0.

Question 5. If area of triangle is 35 square units with vertices (2, −6), (5, 4), and (k, 4). Then k is

(A). 12

(B). −2

(C). −12, −2

(D). 12, −2

Solution :

The area of the triangle with vertices (2, −6), (5, 4), and (k, 4) is given by the relation,

It is given that the area of the triangle is ±35.

Therefore, we have:

⇒ 25 – 5k = ± 35

⇒ 5(5 – k) = ± 35

⇒ 5 – k = ± 7

When 5 − k = −7, k = 5 + 7 = 12.

When 5 − k = 7, k = 5 − 7 = −2.

Hence, k = 12, −2.

The correct answer is D.

Therefore, option (D) is correct.

Exercise 4.3 Page: 87

Question 1. Write minors and cofactors of the elements of the following determinants:

(i) 

(ii) 

Solution
(i) Let 

Minor of element a_ij is M_ij.

∴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 a_ij is A_ij = (−1)^{i + j} M_ij.

∴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) Let 

Minor of element a_ij is M_ij.

∴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 a_ij is A_ij = (−1)^{i + j} M_ij.

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

 

Solution

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

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:

Solution :

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:

Solution

Question 5. If  and A_ij is Cofactors of a_ij, then value of Δ is given by

Solution :

We know that:

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

∴Δ = a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁

Hence, the value of Δ is given by the expression given in alternative D.
Option (D) is correct.

Exercise 4.4 Page: 92

Find adjoint of each of the matrices in Exercise 1 and 2.

Question1.

Solution

Question2.

Solution

Question 3.

Verify A (adj A) = (adj A) A = |A| I .

Solution

Question 4.

Verify A (adj A) = (adj A) A = |A| I .

Solution
Let A =

Find the inverse of the matrix (if it exists) given in Exercise 5 to 11.

Question 5.

Solution :

Question6.

Solution :

Question 7.

Solution :

Question 8.

Solution :

Question 9.

Solution :

Question 10.

Solution
Let A =

Question 11.

Solution

Question 12. Let 

Solution

Question 13. If A =, show that A2 – 5A + 7I = 0. Hence find A^-1

Solution

Question 14. For the matrix A = find numbers a and b such that A² + aA + bI = O.

Solution

 

Question 15. For the matrix A =, show that A³ − 6A² + 5A + 11 I = O. Hence, find A^−1.

Solution

Question 16. If A =, verify that A³ − 6A² + 9A − 4I = O and hence find A⁻¹

Solution

Question 17. Let A be a non-singular matrix of order 3 x 3. Then |adjA| is equal to:

(A) |A|

(B) |A|²

(C) |A|³

(D) 3|A|

Solution

Therefore, option (B) is correct.

Question 18. If A is an invertible matrix of order 2, then det (A⁻¹) is equal to:

(A) det A  

(B) 1/det A

(C) 1   

(D) 0

Solution :

Therefore, option (B) is correct.

Exercise 4.5 Page: 97

Examine the consistency of the system of equations in Exercises 1 to 3.

Question 1.

x + 2y = 2

2x + 3y = 3

Solution
Matrix form of given equations is AX = B

∴ A is non-singular.

Therefore, A⁻¹ exists.

Question 2.

2x − y = 5

x + y = 4

Solution
Matrix form of given equations is AX = B

∴ A is non-singular.

Therefore, A⁻¹ exists.

Hence, the given system of equations is consistent.

Hence, the given system of equations is consistent.

Question 3.

x + 3y = 5

2x + 6y = 8

Solution
Matrix form of given equations is AX = B

∴ A is a singular matrix.

Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.

Examine the consistency of the system of equations in Exercises 4 to 6.

Question 4.

x + y + z = 1

2x + 3y + 2z = 2

ax + ay + 2az = 4

Solution
Matrix form of given equations is AX = B

∴ A is non-singular.

Therefore, A⁻¹ exists.

Hence, the given system of equations is consistent.

Question 5.

3x − y − 2z = 2

2y − z = −1

3x − 5y = 3

Solution
Matrix form of given equations is AX = B

Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.

Question6.

5x − y + 4z = 5

2x + 3y + 5z = 2

5x − 2y + 6z = −1

Solution
Matrix form of given equations is AX = B

∴ A is non-singular.

Therefore, A⁻¹ exists.

Hence, the given system of equations is consistent.

Solve the system of linear equations, using matrix method, in Exercise 7 to 10.

Question7.

5x + 2y =4

7x + 3y = 5

Solution
Matrix form of given equations is AX = B

Question8.

2x – y = – 2

3x + 4y = 3

Solution
Matrix form of given equations is AX = B

Question9.

4x – 3y = 3

3x – 5y = 7

Solution
Matrix form of given equations is AX = B

Question10.

5x + 2y = 3

3x + 2y = 5

Solution
Matrix form of given equations is AX = B

Thus, A is non-singular. Therefore, its inverse exists.

Solve the system of linear equations, using matrix method, in Exercise 11 to 14.

Question11.

Solution
Matrix form of given equations is AX = B

Question12.

x − y + z = 4

2x + y − 3z = 0

x + y + z = 2

Solution
Matrix form of given equations is AX = B

Question13.

2x + 3y + 3z = 5

x − 2y + z = −4

3x − y − 2z = 3

Solution
Matrix form of given equations is AX = B

Question14.

x − y + 2z = 7

3x + 4y − 5z = −5

2x − y + 3z = 12

Solution
Matrix form of given equations is AX = B

Question15. If A =  find A⁻¹. Using A⁻¹ solve the system of equations

Solution

Question16. The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ` 60. The cost of 2 kg onion, 4 kg wheat and 2 kg rice is ` 90. The cost of 6 kg onion, 2 k wheat and 3 kg rice is ` 70. Find cost of each item per kg by matrix method.

Solution :

Let the cost of onions, wheat, and rice per kg be Rs x, Rs y,and Rs z respectively.

Then, the given situation can be represented by a system of equations as:

4x + 3y + 2z = 60

2x + 4y + 6z = 90

6x + 2y + 3z = 70

This system of equations can be written in the form of AX = B, where

Hence, the cost of onions is Rs 5 per kg, the cost of wheat is Rs 8 per kg, and the cost of rice is Rs 8 per kg.

Chapter 4 Miscellaneous

 

 

 

1. Prove that the determinant  is independent of  

Ans. Let 

Expanding along first row,

 

 

  =  =  which is independent of 

2. Without expanding the determinants, prove that: 

Ans. L.H.S. = 

Multiplying R₁ by  R₂ by  and R₃ by ,  = 

=  =  [Interchanging C₁ and C₃]

=  =  [Interchanging C₂ and C₃]

Proved.

 

 

2. Evaluate:  

Ans. Let 

Expanding along first row,

=

= 

= 

= 

= 1

 

3. If  and B =  find  

Ans. Given:  and B = 

Since,   [Reversal law] ……….(i)

Now 

=  = 

Therefore,  exists.

  and  and 

 adj. B =  = 

 

From eq. (i), 

 

= 

---

4. Let A =  verify that:

(i)    

(ii)  

Ans. Given: Matrix A = 

 

  = 

Therefore,  exists.

  and 

and 

 adj. A =  = B (say)

  =            ………(i)

 

=  = 

Therefore,  exists.

  and 

and 

 adj. B =  = 

 

= 

=  ….(ii)

Now to find  (say), where

C = 

= 

C = 

C =  =  =

Therefore,  exists.

  and 

and 

 adj. A = 

=  ……….(iii)

Again 

= 

=  = A (given)

(i) 

 = 

[From eq. (ii) and (iii)]

(ii) 

  = 

5. Evaluate:  

Ans. Let 

=  

= 

=  

= 

= 

= 

= 

= 

= 

6. Evaluate:  

Ans. Let 

=  

=  =  = 

 

7. Solve the system of the following equations: (Using matrices):

 

Ans. Putting  and  in the given equations,

  

 the matrix form of given equations is  [AX= B]

Here,   A =  X =  and B = 

 

= 

= 

  exists and unique solution is  ……….(i)

Now     and 

and 

 adj. A =  = 

And 

 From eq. (i),

 

= 

= 

 

 

 

8. If  are non-zero real numbers, then the inverse of matrix A =  is:

(A) 

(B)  

(C) 

(D)  

Ans. Given: Matrix A = 

 

 

  exists and unique solution is  ……….(i)

Now     and  and 

 adj. A =  = 

And 

= 

= 

= 

Therefore, option (A) is correct.

 

9. Let A =  where  Then:

(A) Det (A) = 0 

(B) Det (A)   

(C) Det (A)  

(D) Det (A) 

Ans. Given: Matrix A = 

 

 

  ……….(i)

Since   

  [  cannot be negative]

 

 

 

Therefore, option (D) is correct.