Let A be a 3 × 3 matrix of non-negative real elements such that A[1; 1; 1]=3[1; 1; 1]. Then the maximum value of det(A) is _________.

Solution

Maths sir · 4 steps

Each row of A adds up to 3, so bound every term of the determinant by the product of the row sums.

1. Read the condition. Multiplying A by the column of ones adds up each row. So the condition says every row of A sums to 3, with all entries aᵢⱼ ≥ 0.
2. Expand the determinant. det A is a sum of 6 terms ± a₁ᵢa₂ⱼa₃ₖ, one from each row and each column. Each term is non-negative before its sign, so det A is at most the sum of these 6 products.

   det A ≤ Σ₆ₜₑᵣₘₛ a₁ᵢa₂ⱼa₃ₖ ≤ (a₁₁+a₁₂+a₁₃)(a₂₁+a₂₂+a₂₃)(a₃₁+a₃₂+a₃₃) = 3 × 3 × 3

3. Why the bound holds. Expanding the product of the three row sums gives all 27 products a₁ᵢa₂ⱼa₃ₖ, and none of them is negative. The 6 determinant terms are among them, so their sum cannot exceed 27.
4. So. A = 3I meets the condition, since each row sums to 3, and det(3I) = 27. The bound is reached, so the maximum value is 27.

The common slip: 3. Taking the given 3 as the determinant itself gives 3. The 3 is the sum of each row, and the determinant can be as large as 3³.

The idea to keep. When a matrix times the ones column appears, read it as row sums. With non-negative entries, the determinant can never exceed the product of the row sums.

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