Let a ∈ R and A be a matrix of order 3 × 3 such that det(A)=-4 and A+I=[1, a, 1; 2, 1, 0; a, 1, 2], where I is the identity matrix of order 3 × 3. If det((a+1) adj((a-1) A)) is 2ᵐ 3ⁿ, m, n ∈{0,1,2, …, 20}, then m+n is equal to :

Solution

Maths sir · 5 steps

Peel off I to get A, find a from det A, then let det(kB) and det(adj B) do the work.

1. Find A. Subtract I from the given matrix: A = (0, a, 1; 2, 0, 0; a, 1, 1).
2. Find a. Expand along row 2, which has only one non-zero entry: det A = -2 (a · 1 - 1 · 1) = -2(a-1). Setting this equal to -4 gives a - 1 = 2, so a = 3, a + 1 = 4 and a - 1 = 2.
3. Two rules. For a 3 × 3 matrix B, det(kB) = k³ det B and det(adj B) = (det B)².

   det(4 adj(2A)) = 4³(det(2A))² = 4³(2³ × (-4))²

4. Put in numbers. det(2A) = 8 × (-4) = -32 = -2⁵, so its square is 2¹⁰. Then 4³ × 2¹⁰ = 2⁶ × 2¹⁰ = 2¹⁶, which is 2¹⁶ × 3⁰.
5. So. m = 16 and n = 0, so m + n = 16, option (d).

The idea to keep. For an n × n matrix, a scalar comes out of a determinant as kⁿ, and det(adj B) = (det B)ⁿ⁻¹. Apply them from the outside in, one layer at a time.

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