Let A = [aᵢⱼ] = [log₅ 128, log₄ 5; log₅ 8, log₄ 25] . If Aᵢⱼ is the cofactor of aᵢⱼ , Cᵢⱼ = Σₖ₌₁² aᵢₖ Aⱼₖ , 1 ≤ i, j ≤ 2 , and C=[Cᵢⱼ] , then 8|C| is equal to :

Solution

Maths sir · 3 steps

That sum is just a row of A times a row of cofactors, so C is A times adj A, which is |A| I.

1. Read the sum. Cᵢⱼ = Σₖ aᵢₖAⱼₖ multiplies row i of A by the cofactors of row j. When i = j this is the expansion of |A| along row i. When i ≠ j it uses the cofactors of the wrong row, and that sum is always 0.

   C = A (adj A) = |A| I₂ ⇒ |C| = |A|²

2. Find |A|. log₅ 128 × log₄ 25 = 7 log₅ 2 × log₂ 5 = 7. Also log₄ 5 × log₅ 8 = 1/2 log₂ 5 × 3 log₅ 2 = 3/2. So |A| = 7 - 3/2 = 11/2.
3. So. |C| = (11/2)² = 121/4, and 8|C| = 242. The answer is (d).

The idea to keep. Row i of a matrix times the cofactors of row j gives |A| when i = j and 0 otherwise. For an n × n matrix that means A(adj A) = |A| I, so its determinant is |A|ⁿ.

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