Let A = [ a₁; a₂ ] and B = [ b₁; b₂ ] be two 2 × 1 matrices with real entries such that A = XB, where
X = 1/√3[ 1, - 1; 1, k ], and k∈R.
If a₁² + a₂² = 2/3(b₁² + b₂²) and (k² + 1) b₂² ≠ -2b₁b₂, then the value of k is __________.

Solution

Maths sir · 4 steps

Write a₁ and a₂ in terms of b₁, b₂, square and add, then factor out (k-1).

1. Multiply out XB. From A = XB we get a₁ = (b₁ - b₂)/√3 and a₂ = (b₁ + k b₂)/√3.
2. Square and add. a₁² + a₂² = 1/3[2b₁² + 2(k-1)b₁ b₂ + (k²+1)b₂²]. Set this equal to 2/3(b₁² + b₂²) and multiply both sides by 3. The 2b₁² cancels on both sides.

   2(k-1)b₁ b₂ + (k²-1)b₂² = 0 ⇒ (k-1)[2b₁ b₂ + (k+1)b₂²] = 0

3. Drop the bracket. The condition (k²+1)b₂² ≠ -2b₁ b₂ is there to rule out the second bracket being zero. At k = 1 it is the same statement, since k² + 1 = k + 1 = 2. That leaves only the factor k - 1 = 0.
4. So. k = 1.

The idea to keep. When a matrix condition must hold for all inputs, expand it into one polynomial equation and factor. The side condition in the question tells you which factor to throw away.

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