SATHEE: Matrices and Determinants 2 Question 1

Matrices and Determinants 2 Question 1

1. A value of $θ \in (0, π / 3)$ , for which

$| \begin{matrix} 1 + \cos^{2} θ & \sin^{2} θ & 4 \cos 6 θ \\ \cos^{2} θ & 1 + \sin^{2} θ & 4 \cos 6 θ \\ \cos^{2} θ & \sin^{2} θ & 1 + 4 \cos 6 θ \end{matrix} |$

(a) $\frac{π}{9}$

(b) $\frac{π}{18}$

(d) $\frac{7 π}{36}$

2019 Main, 12 April II

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Answer:

Correct Answer: 1. (a)

Solution:

Let $Δ = | \begin{matrix} 1 + \cos^{2} θ & \sin^{2} θ & 4 \cos 6 θ \\ \cos^{2} θ & 1 + \sin^{2} θ & 4 \cos 6 θ \\ \cos^{2} θ & \sin^{2} θ & 1 + 4 \cos 6 θ \end{matrix} | = 0$

Applying $C_{1} \to C_{1} + C_{2}$ , we get

$Δ = | \begin{matrix} 2 & \sin^{2} θ & 4 \cos 6 θ \\ 2 & 1 + \sin^{2} θ & 4 \cos 6 θ \\ 1 & \sin^{2} θ & 1 + 4 \cos 6 θ \end{matrix} | = 0$

Applying $R_{1} \to R_{1} - 2 R_{3}$ and $R_{2} \to R_{2} - 2 R_{3}$ , we get

$Δ = | \begin{matrix} 0 & - \sin^{2} θ & - 2 - 4 \cos 6 θ \\ 0 & 1 - \sin^{2} θ & - 2 - 4 \cos 6 θ \\ 1 & \sin^{2} θ & 1 + 4 \cos 6 θ \end{matrix} | = 0$

On expanding w.r.t. $C_{1}$ , we get

$\Rightarrow \sin^{2} θ (2 + 4 \cos 6 θ) + (2 + 4 \cos 6 θ) (1 - \sin^{2} θ) = 0$

$\Rightarrow 2 + 4 \cos 6 θ = 0 \Rightarrow \cos 6 θ = - \frac{1}{2} = \cos \frac{2 π}{3}$

$\Rightarrow 6 θ = \frac{2 π}{3} \Rightarrow θ = \frac{π}{9}$

$∵ θ \in 0, \frac{π}{3}$

Matrices and Determinants 2 Question 1

1. A value of $θ \in (0, π / 3)$ , for which

Answer:

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Matrices and Determinants 2 Question 1

1. A value of θ∈(0,π/3), for which

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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1. A value of $θ \in (0, π / 3)$ , for which