1.Let A=[aij]A = [a_{ij}]A=[aij] be a square matrix of order 3 such that aij=1a_{ij} = 1aij=1 for all i,j=1,2,3i, j = 1, 2, 3i,j=1,2,3. Then, the matrix A2+A3+…+A10A^2 + A^3 + \ldots + A^{10}A2+A3+…+A10 is equal to:a.(310−1)A(3^{10} - 1) A(310−1)Ab.(310+1)A(3^{10} + 1) A(310+1)Ac.310−32⋅3⋅A\dfrac{3^{10} - 3}{2} \cdot 3 \cdot A2310−3⋅3⋅Ad.310−32A\dfrac{3^{10} - 3}{2} A2310−3ALogin to continueOnly logged in users canattempt or see the solution.
