CUET Mathematics 2022 6 Aug Shift 2Algebra > Hard111−1-1−1333−3-3−3✅ Correct Option: 4Related questions:27 May Shift 1If A is a square matrix such that A2=AA^2 = AA2=A and I is the identity matrix of same order as A, then the value of (A−2I)2−(2A+I)2+11A(A-2I)^2 - (2A + I)^2 + 11A(A−2I)2−(2A+I)2+11A is:16 July Shift 2If ∣(a−x)2(a−y)2(a−z)2(b−x)2(b−y)2(b−z)2(c−x)2(c−y)2(c−z)2∣=λ(a−b)(b−c)(c−a)⋅(x−y)(y−z)(z−x)\begin{vmatrix} (a-x)^2 & (a-y)^2 & (a-z)^2 \\ (b-x)^2 & (b-y)^2 & (b-z)^2 \\ (c-x)^2 & (c-y)^2 & (c-z)^2 \end{vmatrix} = \lambda(a-b)(b-c)(c-a) \cdot (x-y)(y-z)(z-x)(a−x)2(b−x)2(c−x)2(a−y)2(b−y)2(c−y)2(a−z)2(b−z)2(c−z)2=λ(a−b)(b−c)(c−a)⋅(x−y)(y−z)(z−x) then the value of λ\lambdaλ is:30 May Shift 1If A is a square matrix and I is the identity matrix of same order such that A2=IA^2 = IA2=I, then 3(A−I)3+3(A+I)3−15A3(A - I)^3 + 3(A + I)^3 - 15A3(A−I)3+3(A+I)3−15A is equal to