NCERT Exemplar - Determinants(Part-2) | Mathematics (Maths) Class 12 - JEE PDF Download

Objective Type Questions (M.C.Q.)
Q.24. Ifthen value of x is
(a) 3 
(b) ± 3 
(c) ± 6 
(d) 6
Ans. (c)
Solution.
Given that
⇒
⇒ 2x² – 40 = 18 + 14 ⇒ 2x² = 32 + 40
⇒ 2x² = 72 ⇒ x² = 36
∴ x = ± 6
Hence, the correct option is (c).

Q.25. The value of determinant
(a) a³ + b³ + c³ 
(b) 3bc 
(c) a³ + b³ + c³ – 3abc 
(d) None of these
Ans. (d)
Solution.
Here, we have
C₂ → C₂ + C₃


(Taking a + b + c common from C₂)
R₁ → R₁ - R₂, R₂ → R₂ - R₃

Expanding along C₂

⇒ (a + b + c) (a - b) (b - c) (- 1)
⇒ (a + b + c) (a - b) (c - b)
Hence, the correct option is (d).

Q.26. The area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is 9 sq. units. The value of k will be
(a) 9 
(b) 3 
(c) – 9 
(d) 6
Ans. (b)
Solution.
Area of triangle with vertices (x₁ y₁), (x₂ y₂) and (x₃, y₃) will be:

⇒
⇒
⇒
3k = 9 ⇒ k = 3
Hence, the correct option is (b).

Q.27. The determinantequals 
(a) abc (b – c) (c – a) (a – b)
(b) (b – c) (c – a) (a – b)
(c) (a + b + c) (b – c) (c – a) (a – b)
(d) None of these
Ans. (d)
Solution.
Let

(Taking (b – a) common from C₁ and C₃)

C₁ → C₁ - C₃

(C₁ and C₂ are identical columns.)
= (a – b)² × 0
= 0
Hence, the correct option is (d).

Q.28. The number of distinct real roots of in the interval
(a) 0 
(b) 2 
(c) 1 
(d) 3
Ans. (c)
Solution.
Given that

C₁ → C₁ + C₂ + C₃
⇒
Taking 2 cos x + sin x common from C₁
⇒
R₁ → R₁ - R₂, R₂ → R₂ - R₃

⇒ (2 cos x + sin x) (cos x – sin x)² = 0

Hence, the correct option is (c).

Q.29. If A, B and C are angles of a triangle, then the determinant
is equal to
(a) 0 
(b) – 1 
(c) 1 
(d) None of these
Ans. (a)
Solution.
Let
C₁ → aC₁ + bC₂ + cC₃



Hence, the correct option is (a).

Q.30. Let is equal to
(a) 0 
(b) – 1 
(c) 2 
(d) 3
Ans. (a)
Solution.
We have f(t) =
Expanding along R₁

= cos t(t² - 2t² ) - t( 2t sint - 2t sint )+ ( 2t sint- t sint )
= -t² cos t+ t sin t
∴
⇒
⇒
Hence, the correct option is (a).

Q.31. The maximum value ofis (θ  is real number)
(a) 1/2
(b) √3/2
(c) √2
(d) 2√3/4
Ans. (a)
Solution.
Given that:
C₁ → C₁ - C₂, C₂ → C₂ - C₃

Expanding along R₁

⇒
but maximum value of sin 2θ = 1
Hence, the correct option is (a).

Q.32. If then
(a) f (a) = 0 
(b) f (b) = 0 
(c) f (0) = 0 
(d) f (1) = 0
Ans. (c)
Solution.
Given that:

Expanding along R₁ =
= (a – b) [2a(a + c)] = (a – b) × 2a × (a + c) ≠ 0

Expanding along R₁

= - (b - a) [( - 2b) (b - c)] = 2b(b – a) (b – c) ≠ 0

Expanding along R₁ =
= a(bc) – b(ac)= abc – abc = 0
Hence, the correct option is (c). 

Q.33. If, then A^–1 exists if
(a) λ = 2 
(b) λ ≠ 2 
(c) λ ≠ – 2 
(d) None of these
Ans. (d)
Solution.
We have,

Expanding along R₁ =
= 2(6 – 5) – λ(0 – 5) – 3(0 – 2)
= 2 + 5λ + 6 = 8 + 5λ
If A^–1 exists then |A| ≠ 0
∴ 8 + 5λ ≠ 0  so λ ≠
Hence, the correct option is (d).

Q.34. If A and B are invertible matrices, then which of the following is not correct?
(a) adj A = |A| × A^-1 
(b) det (A)^-1 = [det(A)]^-1 
(c) (AB)^-1 = B^-1A^-1 
(d) (A + B)^-1 = B^-1 + A^-1
Ans. (d)
Solution.
If A and B are two invertible matrices then
(a) adj A = |A| × A^-1 is correct
(b) det (A)^–1 = [det(A)]^–1 = is correct
(c) Also, (AB)^–1 = B^–1A^–1 is correct
(d) (A + B)^-1adj (A + B)
∴ (A + B)^-1 ≠ B^-1+ A^-1
Hence, the correct option is (d).

Q.35. If x, y, z are all different from zero and then value of x^–1 + y^–1 + z^–1 is
(a) xyz 
(b) x^–1y^–1z^–1 
(c) – x – y – z 
(d) –1
Ans. (d)
Solution.
Given that

Taking x, y and z common from R₁, R₂ and R₃ respectively.

R₁ → R₁ + R₂ + R₃

Takingcommon from R₁

C₁ → C₁ - C₂, C₂ → C₂ - C₃

Expanding along R₁

⇒
⇒+ 1 = 0 and xyz ≠ 0 (x ≠ y ≠ z ≠ 0)

∴ x^–1 + y^–1 + z^–1 = – 1
Hence, the correct option is (d).

Q.36. The value of the determinant
(a) 9x²(x + y) 
(b) 9y²(x + y) 
(c) 3y²(x + y) 
(d) 7x²(x + y)
Ans. (b)
Solution.
Let
C₁ → C₁ + C₂ + C₃

[Taking (3x + 3y) common from C₁]  
R₁ → R₁ - R₂, R₂ → R₂ - R₃

Expanding along C₁

⇒ 3(x + y) (y² + 2y²) ⇒ 3(x + y) (3y²) ⇒ 9y²(x + y)
Hence, the correct option is (b).

Q.37. There are two values of a which makes determinant,
then sum of these number is
(a) 4 
(b) 5 
(c) – 4 
(d) 9
Ans. (c)
Solution.
Given that,
Expanding along C₁

⇒ (2a² + 4) – 2(– 4a – 20) = 86 

⇒ 2a² + 4 + 8a + 40 = 86
⇒ 2a² + 8a + 4 + 40 – 86 = 0 

⇒ 2a² + 8a – 42 = 0

⇒ a² + 4a – 21 = 0

⇒ a² + 7a – 3a – 21 = 0
⇒ a(a + 7) – 3(a + 7) = 0
⇒ (a – 3) (a + 7) = 0
∴ a = 3, – 7
Required sum of the two numbers = 3 – 7 = – 4.
Hence, the correct option is (c).

Fill in the blanks
Q.38. If A is a matrix of order 3 × 3, then |3A| = _______ .
Ans.
We know that for a matrix of order 3 × 3,
|KA| = K³ |A|
∴

Q.39. If A is invertible matrix of order 3 × 3, then |A ^–1 |  _______ .
Ans.
We know that for an invertible matrix A of any order,


Q.40. If x, y, z ∈ R, then the value of determinant
is equal to  _______.
Ans.
We have,
C₁ → C₁ - C₂

[applying (a + b)² – (a – b)² = 4ab]

(Taking 4 common from C₁)
⇒ 4 × 0 = 0 (∵ C₁ and C₃ are identical columns)

Q.41. If cos 2θ = 0,=_______ .
Ans.
Given that: cos 2θ = 0
⇒
∴
The determinant can be written as

Expanding along C₁,


Q.42. If A is a matrix of order 3 × 3, then (A ²)^–1  = ________.
Ans.
For any square matrix A, (A²)^–1 = (A^–1)².

Q.43. If A is a matrix of order 3 × 3, then number of minors in determinant of A are ________.
Ans.
The order of a matrix is 3 × 3
∴ Total number of elements = 3 × 3 = 9
Hence, the number of minors in the determinant is 9.

Q.44. The sum of the products of elements of any row with the co-factors of corresponding elements is equal to _________.
Ans.
The sum of the products of elements of any row with the co-factors of corresponding elements is equal to the value of the determinant of the given matrix.
Let
Expanding along R₁

⇒ a₁₁M₁₁ + a₁₂M₁₂ + a₁₃M₁₃
(where M₁₁, M₁₂ and M₁₃ are the minors of the corresponding elements)

Q.45. If x = – 9 is a root ofthen other two roots are __________.
Ans.
We have,
Expanding along R₁
⇒
⇒ x(x² – 12) – 3(2x – 14) + 7(12 – 7x) = 0
⇒ x³ – 12x –6x + 42 + 84 – 49x = 0
 x³ – 67x + 126 = 0 ...(1)
The roots of the equation may be the factors of 126 i.e., 2 × 7 × 9
9 is given the root of the determinant put x = 2 in eq. (1)
(2)³ – 67 × 2 + 126 ⇒ 8 - 134 + 126 = 0
Hence, x = 2 is  the other root.
Now, put x = 7 in eq. (1)
(7)³ – 67(7) + 126 ⇒ 343 - 469 + 126 = 0
Hence, x = 7 is  also the other root of the determinant.

Q.46. = _______.
Ans.
Let
C₁ → C₁ – C₃

Taking (z – x) common from C₁

R₁ → R₁ - R₂, R₂ → R₂ - R₃

Taking (y – z) common from R₂


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

Q.47. If f(x)  == A + Bx + Cx² + ..., then A =  ________.
Ans.
Given that

Taking (1 + x)¹⁷, (1 + x)²³ and (1 + x)⁴¹ common from R₁, R₂ and R₃ respectively

⇒ (1 + x)¹⁷ . (1+x)²³ . (1+x)⁴¹ . 0 (R₁ and R₃ are identical)
∴ 0 = A + Bx + Cx² + &
By comparing the like terms, we get A = 0.

State True or False for the statements of the following Exercises:
Q.48. ( A³ ) ^–1 = ( A ⁻¹ )³ , where A is a square matrix and |A| ≠ 0.
Ans.
Since (A^K)^–1 = (A^–1)^K where K ∈ N
So, (A³)^-1 = (A^-1)³ is true

Q.49. , where a is any real number and A is a square matrix.
Ans.
If A is a non-singular square matrix, then for any non-zero
scalar ‘a‘, aA is invertible.
∴
So, (aA) is inverse of
is true.

Q.50. |A^–1| ≠ |A|^–1 , where A is non-singular matrix.
Ans.
False.
Since = for a non-singular matrix.

Q.51. If A and B are matrices of order 3 and |A| = 5 , |B| = 3 , then |3AB|  = 27 × 5 × 3 = 405.
Ans.
True.


Q.52. If the value of a third order determinant is 12, then the value of the determinant formed by replacing each element by its co-factor will be 144.
Ans.
True.
Since
If A is a square matrix of order n
then
∴

Q.53. , where a, b, c are in A.P.
Ans.
True.
Let
R₂ → 2R₂ - (R₁ + R₃)

a, b, c are in A.P.
∴ b – a = c – b ⇒ 2b = a + c


Q.54. |adj. A| = |A|² , where A is a square matrix of order two.
Ans.
False.
Sincewhere n is the order of the square matrix.

Q.55. The determinantis equal to zero.
Ans.
True.
Let
Splitting up C₃

[∵ C₁ and C₃ are identical]

[Taking cos A and cos B common from C₂ and C₃ respectively]
= cos A cos B (0) [∵ C₂ and C₃ are identical]
= 0

Q.56. If the determinantsplits into exactly K determinants of order 3, each element of which contains only one term, then the value of K is 8.
Ans.
True.
Let
Splitting up C₁

Splitting up C₂ in both determinants

Similarly by splitting C₃ in each determinant, we will get 8 determinants.

Q.57. Let then
Ans. True.
Given that:
L.H.S.
C₁ → C₁ + C₂ + C₃

[Taking 2 common from C₁]
C₁ → C₁ - C₂  =
C₃ → C₃ - C₁^d  =
Splitting up C₂

= 2 × 16 = 32

Q.58. The maximum value of
Ans.
True.
Let
C₁ → C₁ - C₂, C₂ → C₂ - C₃

Expanding along C₃
= sin θ cos θ – 0 = sin θ cos θ

[Maximum value of sin 2θ = 1]