Let A=[2, 1, 2; 6, 2, 11; 3, 3, 2] and P=[1, 2, 0; 5, 0, 2; 7, 1, 5]. The sum of the prime factors of |P⁻¹ A P-2 I| is equal to

Solution

Maths sir · 3 steps

Write 2I as P⁻¹(2I)P. Then P cancels out of the determinant, and only |A - 2I| is left.

1. Pull out P. Since P⁻¹(2I)P = 2I, we get P⁻¹AP - 2I = P⁻¹(A - 2I)P. Using |XY| = |X||Y| and |P⁻¹| = 1/|P|, the factors of |P| cancel.

   |P⁻¹AP - 2I| = 1/|P| |A - 2I| |P| = |A - 2I|

2. Subtract 2 on diagonal. A - 2I = (0, 1, 2; 6, 0, 11; 3, 3, 0). Expand along the first row: 0 - 1(0 - 33) + 2(18 - 0) = 33 + 36 = 69.
3. So. 69 = 3 × 23, so the sum of the prime factors is 3 + 23 = 26, which is option (d).

The idea to keep. A similar matrix P⁻¹AP always has the same determinant as A. Write kI as P⁻¹(kI)P and the whole expression collapses to |A - kI|.

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