NCERT Exemplar - Determinants(Part-2)
Objective Type Questions (M.C.Q.)
Q.24. If
then 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 determinant
equals
(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 of
is (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)-1
adj (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
Taking
common 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 of
then 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 (AK)-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.
Since
where n is the order of the square matrix.
Q.55. The determinant
is 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 determinant
splits 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 - C1d =
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]
FAQs on NCERT Exemplar - Determinants(Part-2)
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