Let A = bmatrix x & & & 2 & & & 3 - x bmatrix , where (0, 2 ) and x R . If | adj A| = 25 , then the sum of the squares of all possible real values of x is equal to

Step-by-step solution

Given A = bmatrix x & & & 2 & & & 3 - x bmatrix Expanding the determinant of A along the first row: |A| = x(2(3 - x) - ) - ( (3 - x) - ) + ( - 2 ) Using trigonometric identities = 1 , = , and = : |A| = x(6 - 2x - 1) - ( (3 - x) - ) + ( - 2 ) |A| = x(5 - 2x) - (2 - x) + (- ) |A| = 5x - 2x^2 - (2 - x) - 1 |A| = -2x^2 + 6x - 3 We are given | adj A| = 25 . For a 3 3 matrix, | adj A| = |A|^2 . Thus, |A|^2 = 25 |A| = 5 or |A| = -5 . Case 1: |A| = 5 -2x^2 + 6x - 3 = 5 2x^2 - 6x + 8 = 0 x^2 - 3x + 4 = 0 Discriminant D = 9