Let A = bmatrix & 3 & - bmatrix where , Z . It is given that (A) = -1 and - 2 = 4 . The sum of all elements of the matrix (A+I)¹⁰ , where I is the 2 2 identity matrix, is:

Step-by-step solution

Given A = bmatrix & 3 & - bmatrix , we have (A) = - ^2 - 3 = -1 . We are also given - 2 = 4 = 2 + 4 . Substituting into the determinant equation: -(2 + 4)^2 - 3 = -1 -(4 ^2 + 16 + 16) - 3 + 1 = 0 4 ^2 + 19 + 15 = 0 (4 + 15)( + 1) = 0 Since Z , we must have = -1 . Then = 2(-1) + 4 = 2 . Thus, A = bmatrix 2 & -1 3 & -2 bmatrix . Observe that tr (A) = 0 and (A) = -1 . By the Cayley-Hamilton theorem, A^2 - tr (A)A + (A)I = 0 A^2 - I = 0 A^2 = I . Now, consider (A+I)^2 = A^2 + 2A + I . Since A^2 = I , this becomes I + 2