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JEE MATHEMATICS MCQ: MATRICES

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If ∣a1b1c1a2b2c2a3b3c3∣=5,\left| {\begin{array}{cccccccccccccccccccc} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = 5, then the value of ∣b2c3−b3c2c2a3−c3a2a2b3−c3b2b3c1−b1c3c3a1−c1a3a3b1−a1b3b1c2−b2c1c1a2−c2a1a1b2−a2b1∣\left| {\begin{array}{cccccccccccccccccccc} {{b_2}{c_3} - {b_3}{c_2}}&{{c_2}{a_3} - {c_3}{a_2}}&{{a_2}{b_3} - {c_3}{b_2}}\\ {{b_3}{c_1} - {b_1}{c_3}}&{{c_3}{a_1} - {c_1}{a_3}}&{{a_3}{b_1} - {a_1}{b_3}}\\ {{b_1}{c_2} - {b_2}{c_1}}&{{c_1}{a_2} - {c_2}{a_1}}&{{a_1}{b_2} - {a_2}{b_1}} \end{array}} \right| is
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Correct option: B
25
Option B (25) is correct because the given determinant is the adjoint of A, and |adj(A)| = |A|^(n-1) where n is the order of the 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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