The question
Select the correct answer from the codes (A), (B), (C) and (D) as given below.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.
Assertion (A) : For matrix
A = [ 1 cos θ 1 − cos θ 1 cos θ − 1 − cos θ 1 ] , A = \begin{bmatrix} 1 & \cos\theta & 1 \\ -\cos\theta & 1 & \cos\theta \\ -1 & -\cos\theta & 1 \end{bmatrix}, A = 1 − cos θ − 1 cos θ 1 − cos θ 1 cos θ 1 , where θ ∈ [ 0 , 2 π ] \theta \in [0, 2\pi] θ ∈ [ 0 , 2 π ] ∣ A ∣ ∈ [ 2 , 4 ] |A| \in [2, 4] ∣ A ∣ ∈ [ 2 , 4 ]
Reason (R) : cos θ ∈ [ − 1 , 1 ] , ∀ θ ∈ [ 0 , 2 π ] \cos\theta \in [-1, 1],\ \forall\ \theta \in [0, 2\pi] cos θ ∈ [ − 1 , 1 ] , ∀ θ ∈ [ 0 , 2 π ]
Exam answer Show all explanations
Expanding along the first row:
∣ A ∣ = 1 ( 1 + cos 2 θ ) − cos θ ( − cos θ + cos θ ) + 1 ( cos 2 θ + 1 ) = 2 + 2 cos 2 θ \begin{aligned} |A| &= 1(1 + \cos^2\theta) \\ &\quad - \cos\theta(-\cos\theta + \cos\theta) \\ &\quad + 1(\cos^2\theta + 1) \\ &= 2 + 2\cos^2\theta \end{aligned} ∣ A ∣ = 1 ( 1 + cos 2 θ ) − cos θ ( − cos θ + cos θ ) + 1 ( cos 2 θ + 1 ) = 2 + 2 cos 2 θ Why?
Expanding along the first row uses the signs + + + − - − + + +
∣ A ∣ = a 11 M 11 − a 12 M 12 + a 13 M 13 \begin{aligned} |A| &= a_{11}M_{11} - a_{12}M_{12} \\ &\quad + a_{13}M_{13} \end{aligned} ∣ A ∣ = a 11 M 11 − a 12 M 12 + a 13 M 13 Each minor is the 2 × 2 2 \times 2 2 × 2
M 11 = 1 + cos 2 θ M 12 = − cos θ + cos θ M 13 = cos 2 θ + 1 \begin{aligned} M_{11} &= 1 + \cos^2\theta \\ M_{12} &= -\cos\theta + \cos\theta \\ M_{13} &= \cos^2\theta + 1 \end{aligned} M 11 M 12 M 13 = 1 + cos 2 θ = − cos θ + cos θ = cos 2 θ + 1 For example, deleting row 1 and column 3 leaves
M 13 = ( − cos θ ) ( − cos θ ) − ( 1 ) ( − 1 ) \begin{aligned} M_{13} &= (-\cos\theta)(-\cos\theta) \\ &\quad - (1)(-1) \end{aligned} M 13 = ( − cos θ ) ( − cos θ ) − ( 1 ) ( − 1 ) All three in full. Each 2 × 2 2 \times 2 2 × 2
M 11 = ∣ 1 cos θ − cos θ 1 ∣ = ( 1 ) ( 1 ) − ( cos θ ) ( − cos θ ) = 1 + cos 2 θ M 12 = ∣ − cos θ cos θ − 1 1 ∣ = ( − cos θ ) ( 1 ) − ( cos θ ) ( − 1 ) = − cos θ + cos θ = 0 M 13 = ∣ − cos θ 1 − 1 − cos θ ∣ = ( − cos θ ) ( − cos θ ) − ( 1 ) ( − 1 ) = cos 2 θ + 1 \begin{aligned} M_{11} &= \begin{vmatrix} 1 & \cos\theta \\ -\cos\theta & 1 \end{vmatrix} \\ &= (1)(1) - (\cos\theta)(-\cos\theta) \\ &= 1 + \cos^2\theta \\[4pt] M_{12} &= \begin{vmatrix} -\cos\theta & \cos\theta \\ -1 & 1 \end{vmatrix} \\ &= (-\cos\theta)(1) - (\cos\theta)(-1) \\ &= -\cos\theta + \cos\theta = 0 \\[4pt] M_{13} &= \begin{vmatrix} -\cos\theta & 1 \\ -1 & -\cos\theta \end{vmatrix} \\ &= (-\cos\theta)(-\cos\theta) - (1)(-1) \\ &= \cos^2\theta + 1 \end{aligned} M 11 M 12 M 13 = 1 − cos θ cos θ 1 = ( 1 ) ( 1 ) − ( cos θ ) ( − cos θ ) = 1 + cos 2 θ = − cos θ − 1 cos θ 1 = ( − cos θ ) ( 1 ) − ( cos θ ) ( − 1 ) = − cos θ + cos θ = 0 = − cos θ − 1 1 − cos θ = ( − cos θ ) ( − cos θ ) − ( 1 ) ( − 1 ) = cos 2 θ + 1 The middle minor is 0 0 0
Now multiply by the entries with their signs and add:
1 ( 1 + cos 2 θ ) − cos θ ( 0 ) + 1 ( cos 2 θ + 1 ) = 1 + cos 2 θ − 0 + cos 2 θ + 1 = ( 1 + 1 ) + ( cos 2 θ + cos 2 θ ) = 2 + 2 cos 2 θ \begin{aligned} &1(1 + \cos^2\theta) - \cos\theta(0) \\ &\quad + 1(\cos^2\theta + 1) \\ &= 1 + \cos^2\theta - 0 \\ &\quad + \cos^2\theta + 1 \\ &= (1 + 1) + (\cos^2\theta + \cos^2\theta) \\ &= 2 + 2\cos^2\theta \end{aligned} 1 ( 1 + cos 2 θ ) − cos θ ( 0 ) + 1 ( cos 2 θ + 1 ) = 1 + cos 2 θ − 0 + cos 2 θ + 1 = ( 1 + 1 ) + ( cos 2 θ + cos 2 θ ) = 2 + 2 cos 2 θ Reason (R) is true. Since cos θ ∈ [ − 1 , 1 ] \cos\theta \in [-1, 1] cos θ ∈ [ − 1 , 1 ] 0 ≤ cos 2 θ ≤ 1 0 \le \cos^2\theta \le 1 0 ≤ cos 2 θ ≤ 1
0 ≤ 2 cos 2 θ ≤ 2 ⇒ 2 ≤ 2 + 2 cos 2 θ ≤ 4 \begin{aligned} &0 \le 2\cos^2\theta \le 2 \\ \Rightarrow\ &2 \le 2 + 2\cos^2\theta \le 4 \end{aligned} ⇒ 0 ≤ 2 cos 2 θ ≤ 2 2 ≤ 2 + 2 cos 2 θ ≤ 4 Why?
The cosine of any angle lies between − 1 -1 − 1 1 1 1
Watch out when squaring: both end points − 1 -1 − 1 1 1 1 1 1 1 cos θ = 0 \cos\theta = 0 cos θ = 0 0 0 0 cos 2 θ \cos^2\theta cos 2 θ 0 0 0 1 1 1
Next, multiply every part by 2 2 2 2 2 2
0 ≤ cos 2 θ ≤ 1 2 ( 0 ) ≤ 2 cos 2 θ ≤ 2 ( 1 ) 0 ≤ 2 cos 2 θ ≤ 2 0 + 2 ≤ 2 + 2 cos 2 θ ≤ 2 + 2 2 ≤ 2 + 2 cos 2 θ ≤ 4 \begin{aligned} 0 &\le \cos^2\theta \le 1 \\ 2(0) &\le 2\cos^2\theta \le 2(1) \\ 0 &\le 2\cos^2\theta \le 2 \\ 0 + 2 &\le 2 + 2\cos^2\theta \le 2 + 2 \\ 2 &\le 2 + 2\cos^2\theta \le 4 \end{aligned} 0 2 ( 0 ) 0 0 + 2 2 ≤ cos 2 θ ≤ 1 ≤ 2 cos 2 θ ≤ 2 ( 1 ) ≤ 2 cos 2 θ ≤ 2 ≤ 2 + 2 cos 2 θ ≤ 2 + 2 ≤ 2 + 2 cos 2 θ ≤ 4 Both ends are reached: θ = π / 2 \theta = \pi/2 θ = π /2 ∣ A ∣ = 2 |A| = 2 ∣ A ∣ = 2 θ = 0 \theta = 0 θ = 0 ∣ A ∣ = 4 |A| = 4 ∣ A ∣ = 4
θ = π 2 : 2 + 2 ( 0 ) 2 = 2 θ = 0 : 2 + 2 ( 1 ) 2 = 2 + 2 = 4 \begin{aligned} \theta = \tfrac{\pi}{2}&: 2 + 2(0)^2 = 2 \\ \theta = 0&: 2 + 2(1)^2 = 2 + 2 = 4 \end{aligned} θ = 2 π θ = 0 : 2 + 2 ( 0 ) 2 = 2 : 2 + 2 ( 1 ) 2 = 2 + 2 = 4 So ∣ A ∣ ∈ [ 2 , 4 ] |A| \in [2, 4] ∣ A ∣ ∈ [ 2 , 4 ]
Why?
To get from ∣ A ∣ = 2 + 2 cos 2 θ |A| = 2 + 2\cos^2\theta ∣ A ∣ = 2 + 2 cos 2 θ [ 2 , 4 ] [2, 4] [ 2 , 4 ] cos θ \cos\theta cos θ
Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
Where this comes from Chapter 4, Determinants : Value of a 3 × 3 determinant.
Question type: Evaluate a determinant
How to spot it: click to see Type 1. Evaluate a determinant: cross for 2 × 2 2\times2 2 × 2 3 × 3 3\times3 3 × 3
Spot it: "evaluate" and a square block of numbers between straight bars.
Here: The Assertion rests on a 3 × 3 3 \times 3 3 × 3 ∣ A ∣ |A| ∣ A ∣ R 1 R_1 R 1 2 + 2 cos 2 θ 2 + 2\cos^2\theta 2 + 2 cos 2 θ 0 ≤ cos 2 θ ≤ 1 0 \le \cos^2\theta \le 1 0 ≤ cos 2 θ ≤ 1
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