If A = [ 0, - tan ( θ/2 ); tan ( θ/2 ), 0 ] and
(I₂ + A)(I₂ - A)⁻¹ = [ a, - b; b, a ], then 13(a² + b²) is equal to

Solution

Maths sir · 4 steps

Take determinants on both sides: a² + b² falls out without multiplying a single matrix.

1. Write both matrices. Let t = tan θ/2. Then I₂ + A = (1, -t; t, 1) and I₂ - A = (1, t; -t, 1). Both have determinant 1 + t², which is never zero, so I₂ - A is invertible.
2. Use determinants. The determinant of a product is the product of determinants, and det(M⁻¹) = 1/(det M). The right side has determinant a² + b².

   a² + b² = det(I₂ + A)/det(I₂ - A) = (1 + t²)/(1 + t²) = 1

3. Check by multiplying. Working it out fully gives 1/(1+t²)(1-t², -2t; 2t, 1-t²), so a = cos θ and b = sin θ. Again a² + b² = 1.
4. So. 13(a² + b²) = 13 × 1 = 13.

The common slip: 1. Stopping at a² + b² = 1 gives 1. The question asks for 13 times that.

The idea to keep. When the answer only needs a² + b² from a matrix of the form (a, -b; b, a), take determinants first. It is often 1 without any multiplication.

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