If cos2θ = 0, then `|(0, costheta, sin theta),(cos theta, sin theta,0),(sin theta, 0, cos theta)|^2` = ______.

If cos2θ = 0, then `|(0, costheta, sin theta),(cos theta, sin theta,0),(sin theta, 0, cos theta)|^2` = `- 1/sqrt(2)`. Explanation: Δ = `|(0, costheta, sin theta),(cos theta, sin theta,0),(sin theta, 0, cos theta)|^2` = `0 - cos theta(costheta) + sintheta(0- sin^2theta)` = `-(cos^3theta + sin^2theta)` cos2θ = 0 ⇒ 2θ = `pi/2` ⇒ θ = `pi/4` &there4; Δ = `-(cos^3 pi/4 + sin^3 pi/4)` = `-((1/sqrt(2))^3 +(1/sqrt(2))^3)` =`- 1/sqrt(2)`

If cos2θ = 0, then |0cosθsinθcosθsinθ0sinθ0cosθ|2 = ______. - Mathematics

Question

If cos2θ = 0, then ${| \begin{matrix} 0 & \cos θ & \sin θ \\ \cos θ & \sin θ & 0 \\ \sin θ & 0 & \cos θ \end{matrix} |}^{2}$ = ______.

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Solution

If cos2θ = 0, then ${| \begin{matrix} 0 & \cos θ & \sin θ \\ \cos θ & \sin θ & 0 \\ \sin θ & 0 & \cos θ \end{matrix} |}^{2}$ = $- \frac{1}{\sqrt{2}}$ .

Explanation:

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

= $0 - \cos θ (\cos θ) + \sin θ (0 - \sin^{2} θ)$

= $- (\cos^{3} θ + \sin^{2} θ)$

cos2θ = 0

⇒ 2θ = $\frac{π}{2}$

⇒ θ = $\frac{π}{4}$

∴ Δ = $- (\cos^{3} \frac{π}{4} + \sin^{3} \frac{π}{4})$

= $- ({(\frac{1}{\sqrt{2}})}^{3} + {(\frac{1}{\sqrt{2}})}^{3})$

= $- \frac{1}{\sqrt{2}}$

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Properties of Determinants

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Q 41 Q 40 Q 42

Chapter 4: Determinants - Exercise [Page 83]

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NCERT Exemplar Mathematics [English] Class 12

Chapter 4 Determinants
Exercise | Q 41 | Page 83

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