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​Solve the Following Determinant Equation: ∣ ∣ ∣ ∣ ∣ 3 − 2 Sin ( 3 θ ) − 7 8 Cos ( 2 θ ) − 11 14 2 ∣ ∣ ∣ ∣ ∣ = 0 - Mathematics

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प्रश्न

​Solve the following determinant equation:

\[\begin{vmatrix}3 & - 2 & \sin\left( 3\theta \right) \\ - 7 & 8 & \cos\left( 2\theta \right) \\ - 11 & 14 & 2\end{vmatrix} = 0\]

 

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उत्तर

Let Δ `=|(3,-2,sin(3theta)),(-7,8,cos(2theta)),(-11,14,2)|=0`

`=>|(1,-2,sin(3theta)),(1,8,cos(2theta)),(3,14,2)|=0`      `["Applying"  C_1->C_1+C_2]`

`=>|(1,-2,sin(3theta)),(0,10,cos(2theta)-sin(3theta)),(0,20,2-3sin(3theta))|=0`    `["Applying"  R_2->R_2-R_1  "and"  R_3->R_3-3R_1]`

`=>10(2-3sin(3theta))-20(cos(2theta))-sin(3theta)=0`

`=>20-10sin(3theta)-20cos(2theta)=0`

`=>sin(3theta)+2cos(2theta)-2=0`

`=>3sintheta-4sin^3theta+2-4sin^2theta-2=0`

`=>-sintheta(4sin^2theta+4sintheta-3)=0`

`=>sintheta=0    or    4sin^2theta+4sintheta-3=0`

`=>theta=npi     or   (2sintheta+3)(2sintheta-1)=0`

`=>theta=npi     or   sintheta=-3/2   or   sintheta=1/2`

`=>theta=npi    or    theta=npi+(-1)^npi/6,ninZ`

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अध्याय 6: Determinants - Exercise 6.2 [पृष्ठ ६१]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 6 Determinants
Exercise 6.2 | Q 52.9 | पृष्ठ ६१

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