Algebra past year questions, CUET Mathematics 2025 19 May Shift 1 > Matrices & Determinants past year questions, Algebra, CUET Mathematics 2025 19 May Shift 1Mediumcommon1 of 85If A=[1101]A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}A=[1011], then the value of A20A^{20}A20 is:[1101]\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}[1011][2020020]\begin{bmatrix} 20 & 20 \\ 0 & 20 \end{bmatrix}[2002020][12001]\begin{bmatrix} 1 & 20 \\ 0 & 1 \end{bmatrix}[10201][1001]\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}[1001]Solution✅ Correct Option: 3Given A=[1101]A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}A=[1011] Calculate A2A^2A2: A2=A×AA^2 = A \times AA2=A×A =[1101]×[1101]= \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=[1011]×[1011] =[1201]= \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}=[1021] Calculate A3A^3A3: A3=A2×AA^3 = A^2 \times AA3=A2×A =[1201]×[1101]= \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=[1021]×[1011] =[1301]= \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}=[1031] The pattern emerges: A1=[1101]A^1 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}A1=[1011] A2=[1201]A^2 = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}A2=[1021] A3=[1301]A^3 = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}A3=[1031] The general form is An=[1n01]A^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}An=[10n1] For n=20n = 20n=20: A20=[12001]A^{20} = \begin{bmatrix} 1 & 20 \\ 0 & 1 \end{bmatrix}A20=[10201]Related questions:2022: 30 Aug Shift 1The value of the expression x2+y2x−y\frac{x^2 + y^2}{x - y}x−yx2+y2 is:2026: 26 May Shift 2If A and B are symmetric matrices of the same order then which of the following statements are correct? A. AB is symmetric matrix B. AB - BA is skew symmetric matrix C. BTABB^TABBTAB is symmetric matrix D. A−ATA - A^TA−AT is skew symmetric matrix Choose the correct answer from the options given below:2025: 26 May Shift 2If y=−4y = -4y=−4 is a root of ∣y231y132y∣=0\begin{vmatrix} y & 2 & 3 \\ 1 & y & 1 \\ 3 & 2 & y \end{vmatrix} = 0y132y231y=0, then the product of the other two roots is2025: 22 May Shift 1Let A=[15210531492535210]A = \begin{bmatrix} 152 & 105 & 3 \\ 149 & 25 & 35 \\ 2 & 1 & 0 \end{bmatrix}A=15214921052513350. If AijA_{ij}Aij denotes the co-factor of an element aija_{ij}aij of the matrix A, then the value of a11A21+a12A22+a13A23a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23}a11A21+a12A22+a13A23 is equal to2022: 17 Aug Shift 2Match List - I with List - II. List - IList - II(A) [0−5950−3−930]\begin{bmatrix}0 & -5 & 9 \\ 5 & 0 & -3 \\ -9 & 3 & 0\end{bmatrix}05−9−5039−30(I) Scalar matrix(B) [500050005]\begin{bmatrix}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{bmatrix}500050005(II) Diagonal matrix(C) [1000−50007]\begin{bmatrix}1 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 7\end{bmatrix}1000−50007(III) Symmetric matrix(D) [3−21−2−56160]\begin{bmatrix}3 & -2 & 1 \\ -2 & -5 & 6 \\ 1 & 6 & 0\end{bmatrix}3−21−2−56160(IV) Skew-symmetric matrix Choose the correct answer from the options given below :2025: 14 May Shift 2If A=[x−343y−5−4z0]A = \begin{bmatrix} x & -3 & 4 \\ 3 & y & -5\\-4&z&0 \end{bmatrix}A=x3−4−3yz4−50 is a Skew-Symmetric matrix and adj A=[aij]3×3adj \ A = [a_{ij}]_{3 \times3}adj A=[aij]3×3, then a11+a22+a33a_{11} + a_{22} + a_{33}a11+a22+a33 is equal to