Determinants Class 12 Important Questions with Solutions Previous Year Questions
Expansion of Determinant
Question 1.
Find |AB|, if A =
Answer:
Question 2.
Find the maximum value of (Delhi 2016)
Answer:
= [(1 + sinθ)(1 + cosθ) – 1] – [1 + cosθ – 1] + [1 – 1 – sinθ]
= 1 + cosθ + sinθ + sinθcosθ – 1 – cosθ – sinθ
= sinθcosθ
= |
We know that, maximum value of sin 2θ is 1.
∴ Δmax =
Question 3.
If
Answer:
We have,
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 =
Answer:
Given, A =
Now, Cofactors of a21 = (-1)
= -(18 – 21) = 3
Question 5.
If A =
Answer:
Clearly, |A| =
and |B| =
∴ |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
Answer:
Question 7.
If
Answer:
First, expand both determinants, which gives ” equation in x and then solve that equation to find the value of x.
Given,
⇒ 2x2 – 40 = 18 -(-14)
⇒ 2x2 – 40 = 32
⇒ 2x2 = 72
⇒ x2 = 36
∴ x = ± 6
Question 8.
If =
Answer:
Given,
x = -2
Question 9.
Write the value of the determinant (Delhi 2014C)
Answer:
Let Δ =
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
Answer:
Expand both determinants which gives equation in x and then solve that equation to find the value of x.
Given,
⇒ 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
Answer:
Given,
∴ x = 2
Question 12.
If Aij is the cofactor of the element a of the determinant
Answer:
Let Δ =
Here, a32 = 5
Given, Aij is the cofactor of the element aij of A.
∴ A32 = (-1)3+2
⇒ a32 . A32 = 5 × 32 = 110
Question 13.
If Δ =
Answer:
Cofactor of element a32
= (-1)3+2
Question 14.
If Δ =
Answer:
Minor of elements a22 =
Question 15.
If Δ =
Answer:
Minor of the elements a23 =
Question 16.
For what value of x, A =
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
∴
⇒ (2x + 2)(x – 2) – 2x2 = 0
⇒ 2x2 – 2x – 4 – 2x2 = 0
⇒ -2x = 4
∴ x = -2
Question 17.
For what value of x, the matrix
Answer:
Let, A =
If matrix A is singular, then
|A| = 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
Answer:
Given, A =
∴ x = 4
Question 19.
For what value of x, matrix
Answer:
Given A =
∴ x = 2
Question 20.
For what value of x, the matrix
Answer:
Given A =
∴ x = 3
Question 21.
Evaluate
Answer:
Let Δ =
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
Answer:
Given,
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
Answer:
Determinant can be easily expand along that row or column which have maximum zeroes.
Let Δ =
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
Answer:
Minor of elements a23 = 13
Question 25.
What positive value of x makes following pair of determinants equal? (All India 2010)