Let A=[2, 0, 1; 1, 1, 0; 1, 0, 1], B=[B₁, B₂, B₃], where B₁, B₂, B₃ are column matrics, and
AB₁=[1; 0; 0], AB₂=[2; 3; 0], AB₃=[3; 2; 1]
If α=|B| and β is the sum of all the diagonal elements of B, then α³+β³ is equal to ____________.

Solution

Maths sir · 4 steps

Stack the three equations into one, AB = C, then use |A| for α and A⁻¹ for β.

1. Stack the columns. Since B = [B₁, B₂, B₃], the product AB is [AB₁, AB₂, AB₃]. So the three given equations combine into one matrix equation.

   AB = C = (1, 2, 3; 0, 3, 2; 0, 0, 1)

2. Find alpha. |A| = 2(1 - 0) - 0 + 1(0 - 1) = 1. C is upper triangular, so |C| = 1 × 3 × 1 = 3. From |A||B| = |C| we get α = |B| = 3.
3. Find B. With |A| = 1, A⁻¹ = adj A = (1, 0, -1; -1, 1, 1; -1, 0, 2). Then B = A⁻¹C = (1, 2, 2; -1, 1, 0; -1, -2, -1), so β = 1 + 1 - 1 = 1.
4. So. α³ + β³ = 27 + 1 = 28.

The common slip: 152. Taking β as the trace of C = AB, which is 5, instead of the trace of B gives 27 + 125 = 152.

The idea to keep. When A times each column of B is given, write it as AB = C. Then |B| = |C|/|A| needs no inverse, and B = A⁻¹C gives every entry you need.

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