Algebra past year questions, CUET Mathematics 2025 22 May Shift 1 > Matrices & Determinants past year questions, Algebra, CUET Mathematics 2025 22 May Shift 1Mediumcommon3 of 85If A is a square matrix and I is an identity matrix of same order such that A2=AA^2 = AA2=A, then (I+A)3−8I(I + A)^3 - 8I(I+A)3−8I is equal toAA - I7(A - I)ISolution✅ Correct Option: 3Given A2=AA^2 = AA2=A and need to find (I+A)3−8I(I + A)^3 - 8I(I+A)3−8I. (I+A)2=(I+A)(I+A)(I + A)^2 = (I + A)(I + A)(I+A)2=(I+A)(I+A) =I+A+A+A2= I + A + A + A^2=I+A+A+A2 =I+2A+A2= I + 2A + A^2=I+2A+A2 =I+2A+A= I + 2A + A=I+2A+A =I+3A= I + 3A=I+3A (I+A)3=(I+A)2×(I+A)(I + A)^3 = (I + A)^2 \times (I + A)(I+A)3=(I+A)2×(I+A) =(I+3A)(I+A)= (I + 3A)(I + A)=(I+3A)(I+A) =I+A+3A+3A2= I + A + 3A + 3A^2=I+A+3A+3A2 =I+4A+3A2= I + 4A + 3A^2=I+4A+3A2 =I+4A+3A= I + 4A + 3A=I+4A+3A =I+7A= I + 7A=I+7A (I+A)3−8I=(I+7A)−8I(I + A)^3 - 8I = (I + 7A) - 8I(I+A)3−8I=(I+7A)−8I =I+7A−8I= I + 7A - 8I=I+7A−8I =7A−7I= 7A - 7I=7A−7I =7(A−I)= 7(A - I)=7(A−I) Therefore, (I+A)3−8I=7(A−I)(I + A)^3 - 8I = 7(A - I)(I+A)3−8I=7(A−I)Related questions:2025: 16 May Shift 1Let A = [aᵢⱼ]ₙₓₙ and B = [bᵢⱼ]ₙₓₙ. Then which of the following is/are true? (A) AB = BA (B) (AB)⁻¹ = B⁻¹ A⁻¹ (C) (AB)T=BTAT(AB)^T = B^T A^T(AB)T=BTAT (D) AB = 0 ⇒ A = 0 or B = 0 Choose the correct answer from the options given below:2025: 13 May Shift 2Match List-I with List-II List-IList-II(Matrix)(Determinant)(A) [17−35]\begin{bmatrix} 1 & 7 \\ -3 & 5 \end{bmatrix}[1−375](I) 24(B) [−25−3−3]\begin{bmatrix} -2 & 5 \\ -3 & -3 \end{bmatrix}[−2−35−3](II) 32(C) [−128−168]\begin{bmatrix} -12 & 8 \\ -16 & 8 \end{bmatrix}[−12−1688](III) 21(D) [159−21−11]\begin{bmatrix} 15 & 9 \\ -21 & -11 \end{bmatrix}[15−219−11](IV) 26 Choose the correct answer from the options given below:2022: 10 Aug Shift 1If [2x−1−3633y−24−6−44z−3]\begin{bmatrix} 2x-1 & -3 & 6 \\ 3 & 3y-2 & 4 \\ -6 & -4 & 4z-3 \end{bmatrix}2x−13−6−33y−2−4644z−3 is skew symmetric matrix, then xyzxyzxyz is equal to2026: 18 May Shift 2If A=[0−tanα/2tanα/20]A=\begin{bmatrix} 0 & -\tan\alpha/2 \\ \tan\alpha/2 & 0 \end{bmatrix}A=[0tanα/2−tanα/20], then the value of [cosα−sinαsinαcosα]\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}[cosαsinα−sinαcosα] for all α∈R\alpha \in \mathbb{R}α∈R(set of real numbers), is (I is identity matrix of order 2)2025: 13 May Shift 2For the system [2−3−46][xy]=[5−10]\begin{bmatrix} 2 & -3 \\ -4 & 6 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 \\ -10 \end{bmatrix}[2−4−36][xy]=[5−10] which of the following statements are correct? (A) The system has no solution. (B) The system is consistent. (C) It has infinitely many solutions. (D) It has a unique solution. Choose the correct answer from the options given below:2025: 13 May Shift 1Let AAA and BBB are square matrices of order 3 such that A+B=[2−2−4−1341−2−3]A + B = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}A+B=2−11−23−2−44−3. If AAA is a symmetric matrix, then the value of ∣B∣|B|∣B∣ is