Let A = [aᵢⱼ] be a real matrix of order 3 × 3, such that aᵢ₁ + aᵢ₂ + aᵢ₃ = 1, for i = 1, 2, 3. Then, the sum of all the entries of the matrix A³ is equal to :

Solution

Maths sir · 3 steps

Each row adding to 1 means A times the all-ones column gives that column back. Use that three times.

1. Row sums as product. Let e = (1; 1; 1). The i-th entry of Ae is aᵢ₁ + aᵢ₂ + aᵢ₃, which is 1 for every row. So Ae = e.

   A³ e = A² (Ae) = A² e = A(Ae) = Ae = e

2. Read off the rows. The i-th entry of A³ e is the sum of row i of A³. Since A³ e = e, every row of A³ also adds up to 1, and there are 3 rows.
3. So. The sum of all entries of A³ is 1 + 1 + 1 = 3, which is option (c).

The idea to keep. Row sums live in Ae, where e is the all-ones column. If every row of A adds to k, then Ae = ke, so every row of Aⁿ adds to kⁿ.

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