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JEE MATHEMATICS MCQ: MATRICES

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Application +4 | -1
The product of the matrices
A=[cos⁑2θcos⁑θsin⁑θcos⁑θsin⁑θsin⁑2θ]A=\left[\begin{array}{cc}\cos ^2 \theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & \sin ^2 \theta\end{array}\right] and
B=[cos⁑2Ο•cos⁑ϕsin⁑ϕcos⁑ϕsin⁑ϕsin⁑2Ο•]B=\left[\begin{array}{cc}\cos ^2 \phi & \cos \phi \sin \phi \\ \cos \phi \sin \phi & \sin ^2 \phi\end{array}\right] is a null matrix if ΞΈβˆ’Ο•=\theta-\phi=
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Correct option: A
(2n+1)Ο€2(2 n+1) \frac{\pi}{2}
Option A ((2n+1)Ο€2(2 n+1) \frac{\pi}{2}) is correct because the product of the matrices is a null matrix only when cos⁑(ΞΈβˆ’Ο•)=0\cos(\theta - \phi) = 0, which occurs when ΞΈβˆ’Ο•\theta - \phi is an odd multiple of Ο€2\frac{\pi}{2}.

More Questions from MATRICES

Practice All Chapter MCQs β†’
If ∣abβˆ’cb+cc+abcβˆ’aaβˆ’ba+bc∣=0\left|\begin{array}{ccc}a & b-c & b+c \\ c+a & b & c-a \\ a-b & a+b & c\end{array}\right|=0 and a,b,ca, b, c are real numbers then the line ax+by+c=0a x+b y+c=0 passes through the fixed point is The number of solutions of the system of equations x2βˆ’x3=1,β€…β€Šβˆ’x1+2x3=βˆ’2,β€…β€Šx1βˆ’2,β€…β€Šx1βˆ’2x2=3{x_2} - {x_3} = 1,\; - {x_1} + 2{x_3} = - 2,\;{x_1} - 2,\;{x_1} - 2{x_2} = 3 is If the equations 2x+3y+1=0,3x+yβˆ’2=02x + 3y + 1 = 0,3x + y - 2 = 0 and ax+2yβˆ’b=0ax + 2y - b = 0 are consistent, then The value of the determinant ∣1(Ξ±2xβˆ’Ξ±βˆ’2x)2(Ξ±2x+Ξ±βˆ’2x)21(Ξ²2xβˆ’Ξ²βˆ’2x)2(Ξ²2x+Ξ²βˆ’2x)21(Ξ³2xβˆ’Ξ³βˆ’2x)2(Ξ³2x+Ξ³βˆ’2x)2∣\left|\begin{array}{ccc}1 & \left(\alpha^{2 x}-\alpha^{-2 x}\right)^2 & \left(\alpha^{2 x}+\alpha^{-2 x}\right)^2 \\ 1 & \left(\beta^{2 x}-\beta^{-2 x}\right)^2 & \left(\beta^{2 x}+\beta^{-2 x}\right)^2 \\ 1 & \left(\gamma^{2 x}-\gamma^{-2 x}\right)^2 & \left(\gamma^{2 x}+\gamma^{-2 x}\right)^2\end{array}\right|, is If Ο‰\omega is the cube root of unity, then ∣1ωω2ωω21Ο‰21Ο‰βˆ£\left| {\begin{array}{cccccccccccccccccccc} 1&\omega &{{\omega ^2}}\\ \omega &{{\omega ^2}}&1\\ {{\omega ^2}}&1&\omega \end{array}} \right| is equal to For any matrix AA of order 2Γ—22 \times 2, if A(adj⁑A)=[100010]A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right], then ∣A∣|A| is equal to
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