Let A = bmatrix 2 & -1 & 1 -1 & 2 & -1 1 & -1 & 2 bmatrix and P be an invertible matrix of order 3 . The value of 1 16 | adj (2 P⁻¹ A P - 6I)| is equal to

Step-by-step solution

Let M = 2 P⁻¹ A P - 6I . We can rewrite M by factoring out 2P⁻¹ on the left and P on the right: M = 2 P⁻¹ (A - 3I) P Using the multiplicative property of determinants, |M| = |2 P⁻¹ (A - 3I) P| . Since P is a 3 3 matrix, the scalar 2 comes out of the determinant as 2^3 : |M| = 2^3 |P⁻¹| |A - 3I| |P| = 8 |A - 3I| Now, compute A - 3I : A - 3I = bmatrix 2 & -1 & 1 -1 & 2 & -1 1 & -1 & 2 bmatrix - bmatrix 3 & 0 & 0 0 & 3 & 0 0 & 0 & 3 bmatrix = bmatrix -1 & -1 & 1 -1 & -1 & -1 1 & -1 & -1 bmatrix Expanding the determina