Let A be a 3 × 3 matrix such that Xᵀ AX=O for all nonzero 3 × 1 matrices X=[x; y; z]. If A[1; 1; 1]=[1; 4; -5], A[1; 2; 1]=[0; 4; -8], and det(adj(2( A+I)))=2^α 3^β 5^γ, α, β, γ ∈ N, then α²+β²+γ² is

Solution

Maths sir · 5 steps

X^T A X = O for every X tells you A is skew-symmetric. After that only three numbers remain to find.

1. Skew-symmetric. XᵀAX is a 1 × 1 number, so it equals its own transpose XᵀAᵀX. That makes it equal to Xᵀ(A + Aᵀ)/2X, and the only symmetric matrix giving zero for every X is O. So Aᵀ = -A, and the diagonal of A is zero.

   A = (0, a, b; -a, 0, c; -b, -c, 0)

2. Find a, b, c. The first condition gives a + b = 1, -a + c = 4, -b - c = -5. The second gives 2a + b = 0. Subtracting, a = -1, then b = 2 and c = 3. Check with the last row of the second condition: -b - 2c = -2 - 6 = -8.
3. Determinant of 2(A+I). For this skew-symmetric form, det(A + I) = 1 + a² + b² + c² = 1 + 1 + 4 + 9 = 15. Pulling the 2 out of all 3 rows gives det(2(A+I)) = 2³ × 15 = 120.
4. Adjoint. For an n × n matrix M, det(adj M) = (det M)ⁿ⁻¹. With n = 3 this is 120² = 14400 = 2⁶ × 3² × 5².
5. So. α = 6, β = 2, γ = 2, so α² + β² + γ² = 36 + 4 + 4 = 44.

The common slip: 12. Taking det(2(A+I)) as 2 × 15 = 30 instead of 2³ × 15 gives 900 = 2² × 3² × 5² and a final answer of 12.

The idea to keep. If XᵀAX = 0 for every X, then A is skew-symmetric. Also remember det(kM) = kⁿ det M and det(adj M) = (det M)ⁿ⁻¹.

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