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JEE MATHEMATICS MCQ: MATRICES

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Application +4 | -1
cos⁡θ[cos⁡θ        sin⁡θ−sin⁡θ          cos⁡θ]+sin⁡θ[sin⁡θ−cos⁡θcos⁡θ        sin⁡θ]\cos \theta \left[ {\begin{array}{cccccccccccccccccccc} {\cos \theta \;\;\;\;\sin \theta }\\ { - \sin \theta \;\;\;\;\;\cos \theta } \end{array}} \right] + \sin \theta \left[ {\begin{array}{cccccccccccccccccccc} {\sin \theta - \cos \theta }\\ {\cos \theta \;\;\;\;\sin \theta } \end{array}} \right] is equal to
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Correct option: D
[1001]\left[ {\begin{array}{cccccccccccccccccccc} 1&0\\ 0&1 \end{array}} \right]
Option D \left[ {\begin{{array}}{{*{20}{c}}}1&0\\0&1\end{{array}}} \right]\left[ {\begin{{array}}{{*{20}{c}}}1&0\\0&1\end{{array}}} \right] is correct because after performing the matrix multiplication and addition, and using the trigonometric identity cos⁡2θ+sin⁡2θ=1{{\cos }^2}\theta + {{\sin }^2}\theta = 1, the result is the identity matrix.

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If ∣ab−cb+cc+abc−aa−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 x2−x3=1,  −x1+2x3=−2,  x1−2,  x1−2x2=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+y−2=02x + 3y + 1 = 0,3x + y - 2 = 0 and ax+2y−b=0ax + 2y - b = 0 are consistent, then The value of the determinant ∣1(α2x−α−2x)2(α2x+α−2x)21(β2x−β−2x)2(β2x+β−2x)21(γ2x−γ−2x)2(γ2x+γ−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 ω\omega is the cube root of unity, then ∣1ωω2ωω21ω21ω∣\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 2×22 \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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