Let A = [ 0, - 2; 2, 0 ]. If M and N are two matrices given by M = Σₖ₌₁¹⁰ A²ᵏ and N = Σₖ₌₁¹⁰ A²ᵏ⁻¹ then MN² is :

Solution

Maths sir · 4 steps

Square A first: A² = -4I, so every power of A becomes a number times I or a number times A.

1. Square A. Multiplying out gives A² = (-4, 0; 0, -4) = -4I. So even powers are A²ᵏ = (-4)ᵏ I and odd powers are A²ᵏ⁻¹ = (-4)ᵏ⁻¹ A.

   M = (Σₖ₌₁¹⁰ (-4)ᵏ) I = 838860 I, N = (Σₖ₌₁¹⁰ (-4)ᵏ⁻¹) A = -209715 A

2. Sum the GPs. Both sums are geometric with ratio -4. The second is ((-4)¹⁰-1)/(-4-1) = 1048575/-5 = -209715, and the first is -4 times that, which is 838860 = 4 × 209715. So M is a scalar matrix and N is a multiple of A.
3. Build MN squared. N² = 209715² A² = -4 × 209715² I. Then MN² = 4 × 209715 × (-4 × 209715²) I = -16 × 209715³ I.
4. So. MN² is a negative multiple of I. It equals its own transpose, so it is symmetric. It is not I, because the scalar is not 1. The answer is (a).

The idea to keep. When a square of a matrix is a multiple of I, every power reduces to a number times I or a number times the matrix. Turn the matrix sums into ordinary GPs.

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