Let A = [aᵢⱼ] be a square matrix of order 3 such that aᵢⱼ = 2ʲ⁻ⁱ, for all i, j = 1, 2, 3. Then, the matrix A² + A³ + ...... + A¹⁰ is equal to :

Solution

Maths sir · 4 steps

Every row of A is a multiple of (1, 2, 4), so A² is just 3A and the powers form a GP.

1. Write out A. Since aᵢⱼ = 2ʲ⁻ⁱ = 1/2ⁱ × 2ʲ, each entry is a number from row i times a number from column j. So A is a column times a row, and every row is a multiple of (1, 2, 4).

   A = (1, 2, 4; 1/2, 1, 2; 1/4, 1/2, 1) = (1; 1/2; 1/4) (1, 2, 4) = uvᵀ

2. Square it. A² = u (vᵀu) vᵀ, and vᵀu = 1 + 1 + 1 = 3 is just a number. So A² = 3A, then A³ = 3A² = 9A, and in general Aⁿ = 3ⁿ⁻¹A.
3. Add the powers. A² + A³ + … + A¹⁰ = (3 + 3² + … + 3⁹)A. This is a GP with first term 3, ratio 3 and 9 terms, so the sum is 3(3⁹ - 1)/(3 - 1) = (3¹⁰ - 3)/2.
4. So. The sum equals ((3¹⁰ - 3)/2)A, which is option (a).

The idea to keep. If every row of a matrix is a multiple of one row (rank 1), then A² = (trace of A) A. Here the trace is 3, so every power of A is a multiple of A.

Answer from the NTA final answer key. Solution by PrepWiser.