SATHEE: Matrices and Determinants 2 Question 3

Matrices and Determinants 2 Question 3

3. If $Δ_{1} = | \begin{matrix} x & \sin θ & \cos θ \\ - \sin θ & - x & 1 \\ \cos θ & 1 & x \end{matrix} |$ and $Δ_{2} = | \begin{matrix} x & \sin 2 θ & \cos 2 θ \\ - \sin 2 θ & - x & 1 \\ \cos 2 θ & 1 & x \end{matrix} |, x \neq 0$ , then for all $θ \in 0, \frac{π}{2}$

(a) $Δ_{1} + Δ_{2} = - 2 (x^{3} + x - 1)$

(b) $Δ_{1} - Δ_{2} = - 2 x^{3}$

(d) $Δ_{1} - Δ_{2} = x (\cos 2 θ - \cos 4 θ)$

(2019 Main, 10 April I)

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Answer:

Correct Answer: 3. (c)

Solution:

Given determinants are

$Δ_{1} = | \begin{matrix} x & \sin θ & \cos θ \\ - \sin θ & - x & 1 \\ \cos θ & 1 & x \end{matrix} |$

$\begin{aligned} = - x^{3} + \sin θ \cos θ - \sin θ \cos θ + x \cos^{2} θ - x + x \sin^{2} θ \\ = - x^{3} \\ and Δ_{2} = | \begin{array}{c} x & \sin 2 θ & \cos 2 θ \\ - \sin 2 θ & - x & 1 \\ \cos 2 θ & 1 & x \end{array} |, x \neq 0 \\ = - x^{3} (similarly as Δ_{1}) \end{aligned}$

So, according to options, we get $Δ_{1} + Δ_{2} = - 2 x^{3}$

Matrices and Determinants 2 Question 3

3. If $Δ_{1} = | \begin{matrix} x & \sin θ & \cos θ \\ - \sin θ & - x & 1 \\ \cos θ & 1 & x \end{matrix} |$ and $Δ_{2} = | \begin{matrix} x & \sin 2 θ & \cos 2 θ \\ - \sin 2 θ & - x & 1 \\ \cos 2 θ & 1 & x \end{matrix} |, x \neq 0$ , then for all $θ \in 0, \frac{π}{2}$

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Matrices and Determinants 2 Question 3

3. If Δ1=|xsin⁡θcos⁡θ−sin⁡θ−x1cos⁡θ1x| and Δ2=|xsin⁡2θcos⁡2θ−sin⁡2θ−x1cos⁡2θ1x|,x≠0, then for all θ∈0,π2

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3. If $Δ_{1} = | \begin{matrix} x & \sin θ & \cos θ \\ - \sin θ & - x & 1 \\ \cos θ & 1 & x \end{matrix} |$ and $Δ_{2} = | \begin{matrix} x & \sin 2 θ & \cos 2 θ \\ - \sin 2 θ & - x & 1 \\ \cos 2 θ & 1 & x \end{matrix} |, x \neq 0$ , then for all $θ \in 0, \frac{π}{2}$