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Matrices and Determinants question

2019 · 12 Apr · Shift 2 · Q33
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  5. /2019 · 12 Apr · Shift 2 · Q33

Matrices and Determinants question

2019 · 12 Apr · Shift 2 · Q33

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
A value of θ∈(0,π3)\theta \in \left( {0,{\pi \over 3}} \right)θ∈(0,3π​), for which ∣1+cos⁡2θsin⁡2θ4cos⁡6θcos⁡2θ1+sin⁡2θ4cos⁡6θcos⁡2θsin⁡2θ1+4cos⁡6θ∣=0\left| {\begin{matrix} {1 + {{\cos }^2}\theta } & {{{\sin }^2}\theta } & {4\cos 6\theta } \\ {{{\cos }^2}\theta } & {1 + {{\sin }^2}\theta } & {4\cos 6\theta } \\ {{{\cos }^2}\theta } & {{{\sin }^2}\theta } & {1 + 4\cos 6\theta } \\ \end{matrix} } \right| = 0​1+cos2θcos2θcos2θ​sin2θ1+sin2θsin2θ​4cos6θ4cos6θ1+4cos6θ​​=0, is :
  1. A
    π18{\pi \over {18}}18π​
  2. B
    π9{\pi \over {9}}9π​
  3. C
    7π24{{7\pi } \over {24}}247π​
  4. D
    7π36{{7\pi } \over {36}}367π​
View written solutionFree

Correct answer: B

  1. Let c2=cos⁡2θ,s2=sin⁡2θ,x=4cos⁡6θ.c^2=\cos^2\theta,\qquad s^2=\sin^2\theta,\qquad x=4\cos 6\theta.c2=cos2θ,s2=sin2θ,x=4cos6θ. Then the determinant becomes
1+c^2 & s^2 & x\\ c^2 & 1+s^2 & x\\ c^2 & s^2 & 1+x \end{vmatrix}.$$ Since $$c^2+s^2=1,$$ we simplify the entries: $$1+c^2=1+\cos^2\theta,\qquad 1+s^2=1+\sin^2\theta.$$ 2. Now apply row operations that do not change the determinant: - $R_1 \to R_1-R_3$ - $R_2 \to R_2-R_3$ Then $$R_1=(1,0,-1),\qquad R_2=(0,1,-1),\qquad R_3=(c^2,s^2,1+x).$$ So $$D=\begin{vmatrix} 1 & 0 & -1\\ 0 & 1 & -1\\ c^2 & s^2 & 1+x \end{vmatrix}.$$ 3. Expand along the first row: $$D=1\begin{vmatrix}1 & -1\\ s^2 & 1+x\end{vmatrix}+(-1)\begin{vmatrix}0 & 1\\ c^2 & s^2\end{vmatrix}.$$ Compute each minor: $$\begin{vmatrix}1 & -1\\ s^2 & 1+x\end{vmatrix}=1(1+x)-(-1)s^2=1+x+s^2,$$ $$\begin{vmatrix}0 & 1\\ c^2 & s^2\end{vmatrix}=0\cdot s^2-1\cdot c^2=-c^2.$$ Hence $$D=(1+x+s^2)+(-1)(-c^2)=1+x+s^2+c^2=1+x+1=x+2.$$ Using $x=4\cos 6\theta$, $$D=2+4\cos 6\theta.$$ 4. Given $D=0$, $$2+4\cos 6\theta=0$$ $$\cos 6\theta=-\frac12.$$ 5. Solve for $\theta\in\left(0,\frac\pi3\right)$. Since $$6\theta\in(0,2\pi),$$ and in this interval $$\cos y=-\frac12\quad \Rightarrow \quad y=\frac{2\pi}{3},\ \frac{4\pi}{3}.$$ Thus $$6\theta=\frac{2\pi}{3}\quad \text{or}\quad 6\theta=\frac{4\pi}{3}.$$ So $$\theta=\frac{\pi}{9}\quad \text{or}\quad \theta=\frac{2\pi}{9}.$$ 6. Check the options: - A: $\frac\pi{18}$ gives $6\theta=\frac\pi3$, so $\cos 6\theta=\frac12$ ❌ - B: $\frac\pi9$ gives $6\theta=\frac{2\pi}3$, so $\cos 6\theta=-\frac12$ ✅ - C: $\frac{7\pi}{24}>\frac\pi3$ ❌ - D: $\frac{7\pi}{36}$ gives $6\theta=\frac{7\pi}{6}$, so $\cos 6\theta=-\frac{\sqrt3}{2}$ ❌ Therefore the correct option is $$\boxed{\frac\pi9}.$$
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