Exercise 4.1

Question 1:

Evaluate the determinants in Exercises 1 and 2.

Answer

 = 2(−1) − 4(−5) = − 2 + 20 = 18

Question 2:

Evaluate the determinants in Exercises 1 and 2.

(i)  (ii)

Answer

(i)   = (cos θ)(cos θ) − (−sin θ)(sin θ) = 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  , then show that Answer

The given matrix is                   .

Question 4:

If                        , then show that  Answer

The given matrix is

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

From equations (i) and (ii), we have:

Hence, the given result is proved.

Question 5:

Evaluate the determinants

(i)                            (iii)

(ii)                           (iv)

Answer

(i)  Let

.

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

(ii)  Let

.

By expanding along the first row, we have:

(iii)  Let

By expanding along the first row, we have:

(iv)  Let

By expanding along the first column, we have:

Question 6:

If                          , find  .

Answer

Let

By expanding along the first row, we have:

Question 7:

Find values of x, if

(i)  (ii)

Answer

(ii)

Question 8:

If                         , then x is equal to

(A) 6 (B) ±6 (C) −6 (D) 0

Answer

Answer: B

Hence, the correct answer is B.

Exercise 4.2

Question 1:

Using the property of determinants and without expanding, prove that:

Answer

Question 2:

Using the property of determinants and without expanding, prove that:

Answer

Here, the two rows R₁ and R₃ are identical.

∆ = 0.

Question 3:

Using the property of determinants and without expanding, prove that:

Answer

Question 4:

Using the property of determinants and without expanding, prove that:

Answer

By applying C₃ → C₃ + C_2, we have:

Here, two columns C₁ and C₃ are proportional.

∆ = 0.

Question 5:

Using the property of determinants and without expanding, prove that:

Answer

Applying R₂ → R₂ − R₃, we have:

Applying R₁ ↔R₃ and R₂ ↔R₃, we have:

Applying R₁ → R₁ − R₃, we have:

Applying R₁ ↔R₂ and R₂ ↔R₃, we have:

From (1), (2), and (3), we have:

Hence, the given result is proved.

Question 6:

By using properties of determinants, show that:

Answer We have,

Here, the two rows R₁ and R₃ are identical.

∴∆ = 0.

Question 7:

By using properties of determinants, show that:

Answer

Applying R₂ → R₂ + R₁ and R₃ → R₃ + R₁, we have:

Question 8:

By using properties of determinants, show that:

(i)

(ii)

Answer

(i)

Applying R₁ → R₁ − R₃ and R₂ → R₂ − R₃, we have:

Applying R₁ → R₁ + R₂, we have:

Expanding along C₁, we have:

Hence, the given result is proved.

(ii) Let

.

Applying C₁ → C₁ − C₃ and C₂ → C₂ − C₃, we have:

Applying C₁ → C₁ + C₂, we have:

Expanding along C₁, we have:

Hence, the given result is proved.

Question 9:

By using properties of determinants, show that:

Answer

Applying R₂ → R₂ − R₁ and R₃ → R₃ − R₁, we have:

Applying R₃ → R₃ + R₂, we have:

Expanding along R₃, we have:

Hence, the given result is proved.

Question 10:

By using properties of determinants, show that:

(i)

(ii)

Answer

(i)

Applying R₁ → R₁ + R₂ + R₃, we have:

Applying C₂ → C₂ − C₁, C₃ → C₃ − C₁, we have:

Expanding along C₃, we have:

Hence, the given result is proved.

(ii)

Applying R₁ → R₁ + R₂ + R₃, we have:

Applying C₂ → C₂ − C₁ and C₃ → C₃ − C₁, we have:

Expanding along C₃, we have:

Hence, the given result is proved.

Question 11:

By using properties of determinants, show that:

(i)

(ii)

Answer

(i)

Applying R₁ → R₁ + R₂ + R₃, we have:

Applying C₂ → C₂ − C₁, C₃ → C₃ − C₁, we have:

Expanding along C₃, we have:

Hence, the given result is proved.

(ii)

Applying C₁ → C₁ + C₂ + C₃, we have:

Applying R₂ → R₂ − R₁ and R₃ → R₃ − R₁, we have:

Expanding along R₃, we have:

Hence, the given result is proved.

Question 12:

By using properties of determinants, show that:

Answer

Applying R₁ → R₁ + R₂ + R₃, we have:

Applying C₂ → C₂ − C₁ and C₃ → C₃ − C₁, we have:

Expanding along R₁, we have:

Hence, the given result is proved.

Question 13:

By using properties of determinants, show that:

Answer

Applying R₁ → R₁ + bR₃ and R₂ → R₂ − aR₃, we have:

Expanding along R₁, we have:

Question 14:

By using properties of determinants, show that:

Answer

Taking out common factors a, b, and c from R₁, R₂, and R₃ respectively, we have:

Applying R₂ → R₂ − R₁ and R₃ → R₃ − R₁, we have:

Applying C₁ → aC₁, C₂ → bC_2, and C₃ → cC₃, we have:

Expanding along R₃, we have:

Hence, the given result is proved.

Question 15:

Choose the correct answer.

Let A be a square matrix of order 3 × 3, then   is equal to

A.            B.     C.   D.

Answer

Answer: C

A is a square matrix of order 3 × 3.

Hence, the correct answer is C.

Question 16:

Which of the following is correct?

A.  Determinant is a square matrix.

B.  Determinant is a number associated to a matrix.

C.  Determinant is a number associated to a square matrix.

D.  None of these Answer Answer: C

We know that to every square matrix,  of order n. We can associate a number

called the determinant of square matrix A, where  element of A. Thus, the determinant is a number associated to a square matrix.

Hence, the correct answer is C.

Exercise 4.3

Question 1:

Find area of the triangle with vertices at the point 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)

Answer

(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) is given by the relation,

Hence, the area of the triangle is                                   .

Question 2:

Show that points

are collinear

Answer

Area of ∆ABC is given by the relation,

Thus, the area of the triangle formed by points A, B, and C is zero.

Hence, the points A, B, and C are collinear.

Question 3:

Find values of k if area of triangle is 4 square units and vertices are (i) (k, 0), (4, 0), (0, 2) (ii) (−2, 0), (0, 4), (0, k)

Answer

It is given that the area of triangle is 4 square units.

∴∆ = ± 4.

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

∆ =

∴−k + 4 = ± 4

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.

Question 4:

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

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

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

Answer

Answer: D

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:

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.

Exercise 4.4

Question 1:

Write Minors and Cofactors of the elements of following determinants:

(i)  (ii)

Answer

(i)  The given determinant is . 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)  The given determinant is   . 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:

(i)                  (ii)

Answer

(i)  The given determinant is

.

By the definition of minors and cofactors, we have:

M₁₁ = minor of a₁₁=   M₁₂ = minor of a₁₂=

M₁₃ = minor of a₁₃ =

M₂₁ = minor of a₂₁ =   M₂₂ = minor of a₂₂ =   M₂₃ = minor of a₂₃ =   M₃₁ = minor of a₃₁=   M₃₂ = minor of a₃₂ =

M₃₃ = minor of a₃₃ =

A₁₁ = cofactor of a₁₁= (−1)¹⁺¹ M₁₁ = 1 A₁₂ = cofactor of a₁₂ = (−1)¹⁺² M₁₂ = 0 A₁₃ = cofactor of a₁₃ = (−1)¹⁺³ M₁₃ = 0 A₂₁ = cofactor of a₂₁ = (−1)²⁺¹ M₂₁ = 0 A₂₂ = cofactor of a₂₂ = (−1)²⁺² M₂₂ = 1 A₂₃ = cofactor of a₂₃ = (−1)²⁺³ M₂₃ = 0 A₃₁ = cofactor of a₃₁ = (−1)³⁺¹ M₃₁ = 0 A₃₂ = cofactor of a₃₂ = (−1)³⁺² M₃₂ = 0 A₃₃ = cofactor of a₃₃ = (−1)³⁺³ M₃₃ = 1

(ii)  The given determinant is

.

By definition of minors and cofactors, we have:

M₁₁ = minor of a₁₁=

M₁₂ = minor of a₁₂=

M₁₃ = minor of a₁₃ =

M21 = minor of a21 =

M₂₂ = minor of a₂₂ =   M₂₃ = minor of a₂₃ =

M₃₁ = minor of a₃₁=

M₃₂ = minor of a₃₂ =

M₃₃ = minor of a₃₃ =

A₁₁ = cofactor of a₁₁= (−1)¹⁺¹ M₁₁ = 11 A₁₂ = cofactor of a₁₂ = (−1)¹⁺² M₁₂ = −6 A₁₃ = cofactor of a₁₃ = (−1)¹⁺³ M₁₃ = 3 A₂₁ = cofactor of a₂₁ = (−1)²⁺¹ M₂₁ = 4 A₂₂ = cofactor of a₂₂ = (−1)²⁺² M₂₂ = 2 A₂₃ = cofactor of a₂₃ = (−1)²⁺³ M₂₃ = −1 A₃₁ = cofactor of a₃₁ = (−1)³⁺¹ M₃₁ = −20 A₃₂ = cofactor of a₃₂ = (−1)³⁺² M₃₂ = 13 A₃₃ = cofactor of a₃₃ = (−1)³⁺³ M₃₃ = 5

Question 3:

Using Cofactors of elements of second row, evaluate                      . Answer

The given determinant is               . We have:

M21 =

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

M₂₂ =

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

M23 =

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

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

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

Question 4:

Using Cofactors of elements of third column, evaluate Answer

The given determinant is

.

We have:

M₁₃ =

M₂₃ =   M₃₃ =

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

A₂₃ = cofactor of a₂₃ = (−1)²⁺³ M₂₃ = − (z − x) = (x − z) A₃₃ = cofactor of a₃₃ = (−1)³⁺³ M₃₃ = (y − x)

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

Hence,

Question 5:

For the matrices A and B, verify that (AB)′ =    where

(i)

(ii)

Answer

(i)

(ii)

Exercise 4.5

Question 1:

Find adjoint of each of the matrices.

Answer

Question 2:

Find adjoint of each of the matrices.

Answer

Question 3:

Answer

Question 4:

Answer

Question 6:

Find the inverse of each of the matrices (if it exists).

Answer

Question 7:

Find the inverse of each of the matrices (if it exists).

Answer

Question 8:

Find the inverse of each of the matrices (if it exists).

Answer

Question 9:

Find the inverse of each of the matrices (if it exists).

Answer

Question 10:

Find the inverse of each of the matrices (if it exists).

.

Answer

Question 11:

Find the inverse of each of the matrices (if it exists).

Answer

Question 12:

From (1) and (2), we have:

(AB)⁻¹ = B⁻¹A⁻¹

Hence, the given result is proved.

Question 13:

If , show that                           . Hence find      . Answer

Question 14:

For the matrix                         , find the numbers a and b such that A² + aA + bI = O. Answer

We have:

Comparing the corresponding elements of the two matrices, we have:

Hence, −4 and 1 are the required values of a and b respectively.

Question 15:

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

A−1.

Answer

From equation (1), we have:

Question 16:

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

Answer

From equation (1), we have:

Question 17:

Let A be a nonsingular square matrix of order 3 × 3. Then  is equal to

A.     B.    C.    D.

Answer B

We know that,

Hence, the correct answer is B.

Question 18:

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

A. det (A) B.              C. 1 D. 0 Answer

Since A is an invertible matrix,

Hence, the correct answer is B.

Exercise 4.6

Question 1:

Examine the consistency of the system of equations.

x + 2y = 2 2x + 3y = 3

Answer

The given system of equations is:

x + 2y = 2 2x + 3y = 3

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

∴ A is non-singular.

Therefore, A⁻¹ exists.

Hence, the given system of equations is consistent.

Question 2:

Examine the consistency of the system of equations. 2x − y = 5

x + y = 4 Answer

The given system of equations is:

2x − y = 5

x + y = 4

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

∴ A is non-singular.

Therefore, A⁻¹ exists.

Hence, the given system of equations is consistent.

Question 3:

Examine the consistency of the system of equations.

x + 3y = 5 2x + 6y = 8

Answer

The given system of equations is:

x + 3y = 5 2x + 6y = 8

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

∴ A is a singular matrix.

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

Question 4:

Examine the consistency of the system of equations.

x + y + z = 1

2x + 3y + 2z = 2 ax + ay + 2az = 4 Answer

The given system of equations is:

x + y + z = 1

2x + 3y + 2z = 2

ax + ay + 2az = 4

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

∴ A is non-singular.

Therefore, A⁻¹ exists.

Hence, the given system of equations is consistent.

Question 5:

Examine the consistency of the system of equations. 3x − y − 2z = 2

2y − z = −1

3x − 5y = 3

Answer

The given system of equations is:

3x − y − 2z = 2 2y − z = −1

3x − 5y = 3

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

∴ A is a singular matrix.

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

Question 6:

Examine the consistency of the system of equations. 5x − y + 4z = 5

2x + 3y + 5z = 2 5x − 2y + 6z = −1

Answer

The given system of equations is:

5x − y + 4z = 5 2x + 3y + 5z = 2 5x − 2y + 6z = −1

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

∴ A is non-singular.

Therefore, A⁻¹ exists.

Hence, the given system of equations is consistent.

Question 7:

Solve system of linear equations, using matrix method.

Answer

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

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

Question 8:

Solve system of linear equations, using matrix method.

Answer

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

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

Question 9:

Solve system of linear equations, using matrix method.

Answer

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

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

Question 10:

Solve system of linear equations, using matrix method. 5x + 2y = 3

3x + 2y = 5

Answer

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

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

Question 11:

Solve system of linear equations, using matrix method.

Answer

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

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

Question 12:

Solve system of linear equations, using matrix method.

x − y + z = 4 2x + y − 3z = 0

x + y + z = 2 Answer

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

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

Question 13:

Solve system of linear equations, using matrix method.

2x + 3y + 3z = 5 x − 2y + z = −4 3x − y − 2z = 3 Answer

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

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

Question 14:

Solve system of linear equations, using matrix method.

x − y + 2z = 7

3x + 4y − 5z = −5 2x − y + 3z = 12 Answer

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

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

Question 15:

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

Answer

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

Question 16:

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

Answer

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

Now,

X = A⁻¹ B

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.

Miscellaneous Solutions

Question 1:

Hence, ∆ is independent of θ.

Question 2:

Without expanding the determinant, prove that

Answer

Hence, the given result is proved.

Question 3:

Evaluate

Answer

Expanding along C₃, we have:

Question 4:

If a, b and c are real numbers, and

Expanding along R₁, we have:

Hence, if ∆ = 0, then either a + b + c = 0 or a = b = c.

Question 5:

Answer

Question 6:

Prove that

Answer

Expanding along R₃, we have:

Hence, the given result is proved.

Question 8:

Let                                     verify that

(i)

(ii)

Answer

(i)

We have,

(ii)

Question 9:

Evaluate

Answer

Expanding along R₁, we have:

Question 10:

Evaluate

Answer

Expanding along C₁, we have:

Question 11:

Using properties of determinants, prove that:

Answer

Expanding along R₃, we have:

Hence, the given result is proved.

Question 12:

Using properties of determinants, prove that:

Answer

Expanding along R₃, we have:

Hence, the given result is proved.

Question 13:

Using properties of determinants, prove that:

Answer

Expanding along C₁, we have:

Hence, the given result is proved.

Question 14:

Using properties of determinants, prove that:

Answer

Expanding along C₁, we have: