NCERT Solutions Class 12 Mathematics Chapter 4 Determinants

Exercise 4.1

1. Evaluate the determinants in Exercises 1 and 2.

Solution

2. Evaluate the determinants in Exercises 1 and 2. 

Solution

3. If A

Solution

4. If A

Solution

5. Evaluate the determinants 

Solution

6. If A

Solution

7. Find values of x, if 

Solution

8. If

(A) 6
(B) ± 6
(C) -6
(D) 0 
Solution

⇒ x² - 36 = 36 - 36 
⇒ x² - 36 = 0 
⇒ x² = 36 
⇒ x = ± 6 
Hence, the correct answer is B.

Exercise 4.2

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

Solution

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

Solution

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

Solution

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

Solution

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

Solution

6. By using properties of determinants, show that: 

Solution 

We have, 

7. By using properties of determinants, show that : 

Solution

8. By using properties of determinants, show that : 

Solution

= (a - b)(b - c)(c - a)(a + b + c)
Hence, the given result is proved. 

9. By using properties of determinants, show that : 

Solution

= (x - y)(z - x)(z- y)[(-xz - yz) + (-x² – xy + x² )]
= -(x - y)(z -x)(z - y)(xy + yz + zx)
= (x - y)(y - z)(z - x)(xy + yz + zx)
Hence, the given result is proved.

10. By using properties of determinants, show that : 

Solution

11. By using properties of determinants, show that : 

Solution

Hance, the guven result is proved

12. By using properties of determinants show that : 

Solution

Hance, the guven result is proved

13. By using properties of determinants show that : 

Solution 

14. By using properties of determinants show that : 

Solution

15. Let A be a square matrix of order 3 × 3, then | kA| is equal to
(A) k|A|
(B) k² | A |
(C) k³ | A |
(D) 3k | A |

Solution

6. 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 the above.
Solution
We know that to every square matrix, A = [a_ij] of order n , we can associate a number called the determinant of square matrix A, where a_ij = (i, j)^th element of A. Thus, the determinant is a number associated to a square matrix. Hence, the correct option is C.

Exercise 4.3

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

= (1/2)[-2(2 + 8) + 3(3 + 1) + 1(-24 + 2)] 
= (1/2)[-2(10) + 3(4) + 1(-22)] 
= (1/2)[-20 + 12 - 22] 
= (1/2) [-30]
= -30/2 = -15 
Hence, area of the triangle is 15 square units.

2. Show that the points A(a, b + c), B(b, c + a), C(c, a +  b) are collinear. 
Solution
The area of the triangle with vertices A(a, b + c), B(b, c + a), C(c, a +  b) is given by the absolute value of 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.

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) 
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 

= (1/2) [-2(4 - k)
= k - 4 
∴  k - 4 = ± 4
When k - 4 = - 4 , k = 0 
When k - 4 = 4, k = 8 
Hence, k = 0, 8

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. 
Solution
(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. 

⇒ 1/2 [ 1(6 - y) - 2(3 - x) + 1(3y - 6x)] = 0 
⇒ 6 - y - 6 + 2x + 3y - 6x = 0 
⇒ 2y - 4x = 0 
⇒ y = 2x 
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. 

⇒ 1/2 [3(3 - y) - 1(9 - x) + 1(9y - 3x)] = 0 
⇒ 9 - 3y - 9 + x + 9y - 3x = 0 
⇒ 6y - 2x = 0 
⇒ x - 3y = 0 
Hence, the equation of the line joining the given points is x - 3y = 0.

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 , 

= 1/2 [2(4 - 4) + 6(5 - k) + 1(20 -4k)] 
= 1/2 [30 - 6k + 20 - 4k ] 
= 1/2 [50 - 10k] 
= 25 - 5k
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.

Exercise 4.4

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

Solution
(i) The given determinant is

Minor of 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

2. Write Minors and Cofactors of the elements of 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

3. Using Cofactors of elements of second row, evaluate Δ =

Solution

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.
Δ = a₂₁ A₂₁ + a₂₂ A₂₂ + a₂₃ A₂₃ = 2(7) + 0(7) + 1(-7) = 14 - 7 = 7

4. sing Cofactors of elements of third column, evaluate Δ =

Solution

A₁₃ = cofactor of a₁₃ = (-1)¹⁺³ M₁₃ = (z- y)
A₂₃ = cofactor of a₂₃ = (-1)²⁺³ M₂₃ = -(z - x) = (x - z) 
A₃₃ = cofactor of a₃₃ = (-1)³⁺³ M₃₃ = (y - x)
We know that Δ is equal to the sum of the product of the elements of the second row with their corresponding cofactors. 
∴ Δ = a₂₁ A₂₁ + a₂₂ A₂₂ + a₂₃ A₂₃ 
= yz(z - y) + zx(x - z) + xy(y - x) 
= yz² - y² z  +x² z - xz² + xy² - x² y
= (x² z - y² z) + (yz² - xz² ) + (xy² - x² y)
= z(x²  - y² ) + z² (y -x) + xy(y -x)
= z(x - y)(x + y) + z² (y -x) + xy(y - x) 
= (x - y)[zx - zy - z² - xy]
= (x - y) [z(x -z) + y(z - x)] 
= (x - y) (z - x)[-z + y]
= (x - y)(y - z)(z - x)
Hence, Δ = (x - y)(y - z)(z - x) .

5. If ∆ = 

and A_ij  is Cofactors of a_ij , then value of Δ is given by
(A) a₁₁ A₃₁+ a₁₂ A₃₂ + a₁₃ A₃₃
(B) a₁₁ A₁₁+ a₁₂ A₂₁ + a₁₃ A₃₁
(C) a₂₁ A₁₁+ a₂₂ A₁₂ + a₂₃ A₁₃
(D) a₁₁ A₁₁+ a₂₁ A₂₁ + a₃₁ A₃₁
Solution 
We know that :
∆ = Sum of the product of the elements of a column (or a row) with their corresponding cofactors
∆ = a₂₁ A₂₁ + a₂₂ A₂₂ + a₂₃ A₂₃
Hence, the value of Δ is given by the expression given in alternative D.
The correct answer is D.

Exercise 4.5

1. Find the adjoint of the matrix

Solution

2. Find adjoint of each of the matrices.

Solution

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

Solution

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

Solution

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

Solution

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

Solution

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

Solution

We have,  
|A| = 1(10 - 0) - 2(0 - 0) + 3(0 - 0) = 10 
Now, 
A₁₁ = 10 - 0 = 10 , A₁₂ = -(0 - 0) = 0, A₁₃ = 0 - 0 = 0,
A₂₁ = -(10 - 0) = -10, A₂₂ = 5 - 0 = 5, A₂₃ = -(0 - 0) = 0 
A₃₁ = 8 - 6= 2 , A₃₂ = -(4 - 0) = -4, A₃₃ = 2 - 0 = 2 

8. Find the inverse of each of the matrix (if it exists). 

Solution

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

Solution

We have, 
|A| = 2(-1 - 0) - 1(4 - 0) + 3(8 - 7) 
= 2(-1) - 1(4) + 3(1)
= -2 - 4 + 3
= - 3 
Now, 
A₁₁ = -1 - 0 = -1 , A₁₂ = -(4 - 0) = -4, A₁₃ = 8 - 7 = 1
A₂₁ = -(1 - 6) = 5, A₂₂ = 2 + 21 = 23, A₂₃ = -(4 + 7) = -11 
A₃₁ = 0 - 3 = 3 , A₃₂ = -(0 - 12) = 12, A₃₃ = -2 - 4 = -6

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

Solution

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

Solution

12. Let A =

Solution

From (1) and (2), we have : 
(AB)^-1 = B^-1 A^-1 
Hence, the given result is proved.

13. If A =

Solution

Hence, A² - 5A + 7I = 0.
∴ A . A - 5A = -7I
⇒ A.A(A^-1 ) - 5AA^-1 = -7IA^-1 [Post-multiplying by A^-1 as  |A| ≠ 0]
⇒ A(AA^-1 ) - 5I = -7A^-1 
⇒ AI -5I = -7A^-1 
⇒ A^-1 = - (1/7)[A - 5I]
⇒ A^-1 = (1/7)[5I - A] 

14. For the matrix A = 

find the numbers a and b such that A²  + aA + bI = 0. 
Solution

15. For the matrix A =

show that A³ - 6A² + 5A + 11I = 0 . Hence, find A^-1 . 
Solution

A³ - 6A² + 5A + 11I = 0 
⇒ (AAA)A^-1  - 6(AA)A^-3 + 5AA^-1 + 11IA^-1  = 0  [Post-multiplying by A^-1 as  |A| ≠ 0]
⇒ AA(AA^-1 ) - 6A(AA^-1 ) + 5(AA^-1 ) = 11(IA^-1)
⇒ A² - 6A + 5I = -11A^-1   
⇒ A + aI = -bA^-1 
⇒ A^-1 = -(1/11)[A² - 6A + 5I]  ...(1)
Now, 
A² - 6A + 5I 

16. If A = 

verify that A³ - 6A² + 9A - 4I = 0 and hence find A^-1.
Solution

∴ A³ - 6A² + 9A + 4I = 0 
Now, 
A³ - 6A² + 9A + 4I = 0 
⇒ (AAA)A^-1  - 6(AA)A^-3 + 9AA^-1 + 4IA^-1  = 0   [Post-multiplying by A^-1 as  |A| ≠ 0]
⇒ AA(AA^-1 ) - 6A(AA^-1 ) + 9(AA^-1 ) = 4(IA^-1  )
⇒ AAI - 6AI + 9I = 4A^-1 
⇒ A² - 6A + 9I = 4A^-1   
⇒ A + aI = -bA^-1 
⇒ A^-1 = (1/4)[A² - 6A + 9I]  ...(1)
A² - 6A + 9I 

17. Let A be a nonsingular square matrix of order 3 × 3. Then |adj A| is equal to
(A) |A |
(B) |A|²
(C) |A|³
(D) 3|A|
Solution
We know that, 

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
Since A is an invertible matrix, A^-1 exists and A^-1 = (1/|A|) adj A. 

Exercise 4.6

 

1. Examine the consistency of the system of equations.
x + 2y = 2
2x + 3y = 3
Solution
The given system of equation is : 
x + 2y = 2 
2x + 3y = 3 
The given system of equations can be written in the form of AX = B, where 

Now, 
|A| = 1(3) - 2(2) = 3 - 4 = -1 ≠ 0 
∴ A is non - singular. 
Therefore, A^-1 exists.
Hence, the given system of equations is consistent.

2. Examine the consistency of the system of equations.
2x − y = 5
x + y = 4
Solution
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 

Now, 
|A| = 2|1| - (-1)(1) = 2 + 1 = 3 ≠ 0 
∴ A is non - singular. 
Therefore, A^-1 exists. 
Hence, the given system of equations is consistent.

3. Examine the consistency of the system of equations.
x + 3y = 5
2x + 6y = 8
Solution
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 

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

4. Examine the consistency of the system of equations.
x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
Solution
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 

Now,  
|A| = 1(6a - 2a) - 1(4a - 2a) + 1(2a - 3a)
= 4a - 2a - a = 4a - 3a = a ≠ 0 
∴ A is non - singular. 
Therefore, A^-1 exists. 
Hence, the given system of equations is consistent.

5. Examine the consistency of the system of equations.
3x − y − 2z = 2
2y − z = −1
3x − 5y = 3
Solution 
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 

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

6. Examine the consistency of the system of equations.
5x − y + 4z = 5
2x + 3y + 5z = 2
5x − 2y + 6z = −1
Solution
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 

Now, 
|A| = 5(18 + 10) + 1(12 - 25) + 4(- 4 - 15) 
= 5(28) + 1(-13) + 4(-19)
= 140 - 13 - 76 
= 51 ≠ 0 
A is non  - singular. 
Therefore, A^-1 exists. 
Hence, the given system of equations is consistent.

7. Solve system of linear equations, using matrix method.
5x + 2y = 4
7x + 3y = 5
Solution
The given system of equations can be written in the form of AX = B, where 

8. Solve system of linear equations, using matrix method.
2x – y = –2
3x + 4y = 3
Solution
The given system of equations can be written in the form of AX = B, where 

9. Solve system of linear equations, using matrix method.
4x – 3y = 3
3x – 5y = 7
Solution
The given system of equation can be written in the form of AX = B, where 

10.Solve system of linear equations, using matrix method.
5x + 2y = 3
3x + 2y = 5
Solution
The given system of equations can be written in the form of AX = B, where

11. Solve system of linear equations, using matrix method.
2x + y + z = 1
x – 2y – z = 3/2
3y – 5z = 9
Solution
The given system of equations can be written in the form of AX = B, where 

Now, 
|A| = 2(10 + 3) - 1(-5 - 3) + 0 = 2(13) - 1(-8) = 26 + 8 = 34 ≠ 0 
Thus, A is non - singular. Therefore, its inverse exists. 
Now,
A₁₁ = 13, A₁₂ = 5, A₁₃ = 3
A₂₁ = 8, A₂₂ = -10 , A₂₃ = -6
A₃₁ = 1, A₃₂ = 3, A₃₃ = -5

12. Solve system of linear equations, using matrix method.
x − y + z = 4
2x + y − 3z = 0
x + y + z = 2
Solution
The given system of equations can be written in the form of AX= B, where 

Now, 
|A| = 1(1 + 3) + 1(2 + 3) + 1(2 - 1) = 4 + 5 + 1= 10 ≠ 0 
Thus, A is non - singular. Therefore, its inverse exists. 
Now, 
A₁₁ = 4, A₁₂ = -5, A₁₃ = 1
A₂₁ = 2, A₂₂ = 0 , A₂₃ = -2
A₃₁ = 2, A₃₂ = 5, A₃₃ = 3

13. Solve system of linear equations, using matrix method.
2x + 3y + 3z = 5
x − 2y + z = −4
3x − y − 2z = 3
Solution
The given system of equations can be written in the form AX = B, where 

Now, 
|A| = 2(4 + 1) - 3(-2 - 3) + 3(-1 + 6)
= 2(5) - 3(-5) + 3(5) 
= 10 + 15 + 15
= 40 ≠ 0 
Thus, A is non - singular. Therefore, its inverse exists. 
Now,  
A₁₁ = 5, A₁₂ = 5, A₁₃ = 5
A₂₁ = 3, A₂₂ = -13 , A₂₃ = 11
A₃₁ = 9, A₃₂ = 1, A₃₃ = -7

14. Solve system of linear equations, using matrix method.
x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12
Solution
The given system of equations can be written in the form of AX = B, where 

Now, 
|A| = 1(12 - 5) + 1(9 + 10) + 2(-3 - 8)
= 7 + 19 - 22 
= 4 ≠ 0 
Thus, A is non - singular. Therefore, its inverse exists. 
Now,  
A₁₁ = 7, A₁₂ = -19, A₁₃ = -11
A₂₁ = 1, A₂₂ = -1 , A₂₃ = -1
A₃₁ = -3, A₃₂ = 11, A₃₃ = 7

15. If A = 

find A⁻¹. Using A⁻¹ solve the system of equations
2x – 3y + 5z = 11
3x + 2y – 4z = – 5
x + y – 2z = – 3
Solution

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.
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 

x = 5, y = 8 and z = 8 
Hence, the cost of onion per kg is Rs. 5, the cost of wheat per kg is Rs.  8 and the cost of rice per kg is Rs. 8.

Miscellaneous Solutions

1. Prove that the determinant 

is independent of θ is independent of θ. 
Solution

= x(–x² – 1) – sinθ(– x sinθ – cos θ) + cos θ(– sinθ + x cosθ)
= –x³ – x + xsin² θ + sinθ cosθ – sinθ cosθ + x cos²θ
= –x³ – x + x(sin²θ + cos²θ)
= –x³ – x + x
= –x³ (Independent of θ)
Hence, Δ is independent of θ.

2. Without expanding the determinant, prove that

Solution

3. Evaluate 

Solution  Δ = 

Expanding along C₃ , we have :
∆ = − sin α(−sin α sin²β – cos² β sinα) + cos α(cos α cos² β + cos α sin² β)
= sin² α(sin² β + cos² β) + cos² α(cos² β + sin² β)
= sin² α(1) + cos² α(1)
= 1

4. If a, b and c are real numbers, and triangle  = 

Show that either a + b + c = 0 or a = b = c. 

Solution

Expanding along R₁ , we have : 
Δ = 2 (a + b + c)(1)[(b - c)(c - b) - (b - a)(c - a)] 
= 2(a + b + c)[-b² - c² + 2bc - bc + ba + ac - a² ]
= 2(a + b + c)[ab + bc +  ca - a²  - b²  - c² ]

It is given that Δ = 0. 
(a + b + c)[ab + bc + ca - a² - b² - c²] = 0 
⇒ Either a + b + c = 0, or ab + bc + ca - a² - b² - c² = 0 .
Now,  
ab + bc + ca - a² - b² - c² = 0 
⇒ -2ab - 2bc - 2ca + 2a² + 2b² + 2c²  = 0 
⇒ (a - b)² + (b - c)² + (c - a)² = 0 
⇒ (a - b)² = (b - c)² = (c - a)²  = 0 [(a-b)², (b-c)², (c-a)² are non-negative] 
⇒ (a - b) = (b - c) = (c - a) = 0 
⇒ a = b = c
Hence, if Δ = 0 then either a + b + c = 0 or a = b = c. 

5. Solve the equations

= 0, a ≠ 0 
Solution

Expanding along R₁ , we have : 
(3x + a)[1 × a² ]= 0 
⇒ a² (3x + a) = 0 
But a ≠ 0. 
Therefore, we have : 
3x + a = 0 
⇒ x = -a/3 

6. Prove that 

= 4a² b² c² 
Solution

Expanding along R₃ , we have : 
Δ = 2ab² c[a(c - a) + a(a + c)]
= 2ab² c[ac - a² + a² + ac]
= 2ab² c(2ac)
= 4a² b² c² 
Hence, the given result is proved.

7. If A ^(-1) = 

Solution
We know that (AB)^-1 = B^-1 A^-1 .

8. Let A = 

verify that 
(i) [adj A]^-1 = adj(A^-1) 
(ii) (A^-1)^-1 = A 
Solution

(i) |adj A| = 14(-4 - 9) - 11(-11 - 15) - 5(-33 + 20)
= 14(-13) - 11(-26) - 5(-13) 
= -182 + 286 + 65 = 169 
we have , 

9. Evaluate

Solution

10. Evaluate 

Solution

11. Using properties of determinants, prove that : 

Solution

Expanding along R₃ , we have : 
Δ =  (β - α) (γ - α)[- (γ - β)( - α - β - γ)]
= (β - α) (γ - α) (γ - β)(α + β + γ)
= (α - β) (β - γ) (γ - α)(α + β + γ)
Hence, the given result is proved. 

12. Using properties of determinants, prove that : 

(1 + pxyz)(x - y)(y - z)(z -x), where p is any scalar. 
Solution

13. Using properties of determinants, prove that :

Solution

Expanding along C₁ , we have : 
Δ = (a + b + c) [(2b + a)(2c + a) - (a - b)(a - c)]
= (a + b +c)[4abc + 2ab + 2ac + a² - a² + ac + ba - bc]
= (a + b + c)(3ab + 3bc + 3ac)
= 3(a + b + c)(ab + bc + ca)
Hence, the given result is proved. 

14. Using properties of determinants, prove that : 

Solution

15 . Solve the system of the following equations 
2/x + 3/y + 10/z = 4 
4/x - 6/y + 5/z = 1 
6/x + 9/y - 20/x = 2 
Solution
Let 1/x = p, 1/y = q, 1/z = r. 
Then the given system of equations is as follows : 
2p + 3q + 10r = 4 
4p - 6q + 5r = 1 
6p + 9q - 20r = 2 
This system can be written in the form of AX = B, where 

Now, 
|A| = 2(120 - 45) - 3 (-80 - 30) + 10(36 + 36)
= 150 + 330 + 720 
=1200 
Thus, A is non - singular. Therefore, its inverse exists. 
Now, 
A₁₁ = 75, A₁₂ = 110, A₁₃ = 72 
A₂₁ = 150, A₂₂ = -100, A₂₃ = 0 
A₃₁ = 75, A₃₂ = 30, A₃₃ = -24 
∴ A^-1 = 1/|A| [adj A]

16. Choose the correct answer. 
If a, b, c, are in A.P., then the determinant 

A. 0
B. 1 
C. x 
D. 2x 
Solution 

Here, all the elements of the first row (R₁ ) are zero. 
Hence, we have Δ = 0.
The correct answer is A.

18. If x, y, z are nonzero real numbers, then the inverse of matrix A =

Solution

19. Choose the correct answer . 

A. Det (A) = 0
B. Det (A) ∈ (2, ∞)
C. Det (A) ∈ (2, 4)
D. Det (A)∈ [2, 4]
Solution

∴ |A| = 1(1 + sin² θ) - sinθ( - sinθ + sinθ) + 1 (sin² θ + 1)
= 1 + sin² θ + sin² θ + 1 
= 2 + 2sin² θ
= 2(1 + sin² θ) 
Now, 0 ≤ θ ≤ 2π 
⇒ 0 ≤ sinθ  ≤ 1 
⇒ 0 ≤ sin² θ  ≤ 1 
⇒ 1 ≤ 1 + sin² θ  ≤ 2
⇒ 2 ≤ 2(1 + sin² θ)  ≤ 2
∴ Det (A) ∊ [2, 4]
The correct answer is D.