Determinants Class 12 Important Questions with Solutions Previous Year Questions

Expansion of Determinant

Question 1.
Find |AB|, if A = [0012] and B = [3050]. (All India 2019)
Answer:
Determinants Class 12 Maths Important Questions Chapter 4 1

Question 2.
Find the maximum value of (Delhi 2016)
11111+sinθ1111+cosθ
Answer:
Determinants Class 12 Maths Important Questions Chapter 4 2
= [(1 + sinθ)(1 + cosθ) – 1] – [1 + cosθ – 1] + [1 – 1 – sinθ]
= 1 + cosθ + sinθ + sinθcosθ – 1 – cosθ – sinθ
= sinθcosθ
= |12|(2sinθcosθ) = 12sin2θ
We know that, maximum value of sin 2θ is 1.
∴ Δmax = 12 × 1 = 12

Question 3.
If xsinθcosθsinθx1cosθ1x = 8, write the value of x. (Foreign 2016)
Answer:
We have, xsinθcosθsinθx1cosθ1x = 8

On expanding along R1, we get
x (- x2 -1) – sinθ(- x sinθ – cosθ) + cosθ(- sinθ + x cosθ) = 8
⇒ – x3 – x + x sin2θ + sinθcosθ – sinθcosθ + x cos2θ = 8
⇒ – x3 – x + x (sin2θ + cos2θ) = 8
⇒ -x3 – x + x = 8 [∵ sin2θ + cos2θ = 1]
⇒ – x3 = 8 ⇒ x3 + 8 = 0 ⇒ x3 + 23 = 0
⇒ (x + 2)(x2 + 4 – 2x) = 0
⇒ x = -2 [∵ x2 – 2x + 4 = 0, gives imaginary values]

Question 4.
If A = 544637323, then write the cofactor of the element a21 of its 2nd row. (Foreign 2015)
Answer:
Given, A = 544637323

Now, Cofactors of a21 = (-1)6733
= -(18 – 21) = 3

Question 5.
If A = [1321] and B = [1131], write the value of |AB|. (Delhi 2015C)
Answer:
Clearly, |A| = 1321 = -1 – 6 = – 7
and |B| = 1131 = 1 + 3 = 4
∴ |AB| = |A|.|B| = (-7)(4) = -28

Question 6.
In the interval it π/2 < x < π, find the value of x for which the matrix [2sinx132sinx] is singular. (All India 2015C)
Answer:
Determinants Class 12 Maths Important Questions Chapter 4 3

Question 7.
If 2x85x=6723, then write the value of x. (Delhi 2014)
Answer:
First, expand both determinants, which gives ” equation in x and then solve that equation to find the value of x.
Given, 2x85x=6723
⇒ 2x2 – 40 = 18 -(-14)
⇒ 2x2 – 40 = 32
⇒ 2x2 = 72
⇒ x2 = 36
∴ x = ± 6

Question 8.
If = 3x274=8674, then find the value of x. (All India 2014)
Answer:
Given, 3x274=8674
x = -2

Question 9.
Write the value of the determinant (Delhi 2014C)
pp1p+1p
Answer:
Let Δ = pp1p+1p

On expanding, we get
Δ = p2 – (p – 1)(p + 1)
⇒ Δ = p2 – (p2 – 12) [∵ a2 – b2 =(a + b) (a – b)]
⇒ Δ = p2 – p2 + 1
∴ Δ = 1

Question 10.
If 2x2(x+1)x+3x+1=1353, then find the value of x. (Delhi 2013C)
Answer:
Expand both determinants which gives equation in x and then solve that equation to find the value of x.
Given, 2x2(x+1)x+3x+1=1353
⇒ 2x (x +1) – (x + 3) (2x + 2) = 3 – 15
⇒ 2x2 + 2x – (2x2 + 8x + 6) = -12
⇒ – 6x – 6 =-12 ⇒ 6x = 6
∴ x = 1

Question 11.
If x+1x3x1x+2=4113, then write the value of x. (Delhi 2013)
Answer:
Given, x+1x3x1x+2=4113
∴ x = 2

Question 12.
If Aij is the cofactor of the element a of the determinant 261305547, then write the value of a32 . A32. (All India 2013)
Answer:
Let Δ = 261305547
Here, a32 = 5
Given, Aij is the cofactor of the element aij of A.
∴ A32 = (-1)3+22654 = -1(8 – 30) = 32
⇒ a32 . A32 = 5 × 32 = 110

Question 13.
If Δ = 521302813, write the cofactor of a23. (Delhi 2012)
Answer:
Cofactor of element a32
= (-1)3+25281 = (-1)(5 – 16) = 11

Question 14.
If Δ = 125203318, write the minor of element a22. (Delhi 2012)
Answer:
Minor of elements a22 = 1538 = 8 – 15 = -7

Question 15.
If Δ = =521302813, then write the minor of the element a23. (Delhi 2012)
Answer:
Minor of the elements a23 = 5132 = 10 – 3 = 7

Question 16.
For what value of x, A = [2(x+1)x2xx2] is a singular matrix? (All India 2011C)
Answer:
For a singular matrix, |A| = 0. Use this relation and solve it.
We know that, a matrix A is said to be singular, if |A| = 0
∴ [2(x+1)x2xx2] = 0
⇒ (2x + 2)(x – 2) – 2x2 = 0
⇒ 2x2 – 2x – 4 – 2x2 = 0
⇒ -2x = 4
∴ x = -2

Question 17.
For what value of x, the matrix [2x+4x+543] is a singular matrix? (All India 2011C)
Answer:
Let, A = [2x+4x+543]
If matrix A is singular, then
|A| = 0
⇒ 2x+4x+543 = 0
⇒ (2x + 4) × 3 – (x + 5) × 4 = 0
⇒ 6x + 12 – 4x – 20 = 0 ⇒ 2x = 8
∴ x = 4

Question 18.
For what value of x, the matrix [2xx+243] is a singular matrix? (Delhi 2011C)
Answer:
Given, A = [2xx+243]
∴ x = 4

Question 19.
For what value of x, matrix [6x3x41] is a singular matrix? (Delhi 2011C)
Answer:
Given A = [6x3x41]
∴ x = 2

Question 20.
For what value of x, the matrix [5x2x+14] is a singular? (Delhi 2011)
Answer:
Given A = [5x2x+14]
∴ x = 3

Question 21.
Evaluate cos15sin75sin15cos75. (All India 2011)
Answer:
Let Δ = cos15sin75sin15cos75

On expanding, we get
A = (cos 15° cos 75° – sin 15° sin 75°)
= cos (15° + 75°) [∵ cos x cos y – sin x sin y = cos (x + y)]
= cos 90° = 0 [∵ cos 90° = 0]

Question 22.
If x1xx=3142, then write the positive value of x. (Foreign 2011; All India 2008C)
Answer:
Given, x1xx=3142
On expanding, we get
x2 – x = 6 – 4
⇒ x2 – x – 2 = 0
⇒ (x – 2)(x + 1) = 0
∴ x = 2 or -1
Hence, the positive value of x is 2.

Question 23.
What is the value of determinant 024235046 ? (Delhi 2010)
Answer:
Determinant can be easily expand along that row or column which have maximum zeroes.
Let Δ = 024235046
Then, Δ = -1(12 – 16) [expanding along R1]
= -2(-4) = 8

Question 24.
Find the minor of the element of second row and third column (a23) in the following determinant 261305547. (Delhi 2010)
Answer:
Minor of elements a23 = 13

Question 25.
What positive value of x makes following pair of determinants equal? (All India 2010)
2x53x