Let x,y,z > 1 and A = [ 1, log ₓy, log ₓz; log _yx, 2, log _yz; log _zx, log _zy, 3 ]. Then |adj (adj A²)| is equal to

Solution

Maths sir · 4 steps

Turn every log into a ratio of natural logs; the ratios cancel and leave a plain number matrix.

1. Change of base. Let a = ln x, b = ln y, c = ln z. All three are positive because x, y, z > 1. Then logₓ y = b/a, and every off-diagonal entry in row i, column j becomes (log of column variable) divided by (log of row variable).
2. Pull out factors. Take 1/a, 1/b, 1/c out of rows 1, 2, 3, and a, b, c out of columns 1, 2, 3. The factors multiply to 1, and what is left has every entry equal to 1 except the diagonal 1, 2, 3.

   |A| = |1, 1, 1; 1, 2, 1; 1, 1, 3| = 1(6-1) - 1(3-1) + 1(1-2) = 2

3. Adjoint twice. For an n × n matrix B, |adj B| = |B|ⁿ⁻¹, so |adj(adj B)| = |B|^((n-1)²). Here n = 3, so the exponent is 4. Take B = A², with |A²| = |A|² = 4.
4. So. |adj(adj A²)| = 4⁴ = 2⁸, which is option (b).

The idea to keep. When a matrix has entries like logₓ y, switch to natural logs and factor them out of rows and columns. For adjoint applied twice on an n × n matrix, the exponent is (n-1)².

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