Given the matrices A = bmatrix 1 & 0 & 1 0 & 1 & 0 1 & 0 & 2 bmatrix and B = bmatrix 2 & 1 & 0 1 & 1 & 2 0 & 2 & 1 bmatrix , then the minor M₂₃ of the matrix (A B⁻¹)⁻¹ is:

Step-by-step solution

Let C = (A B⁻¹)⁻¹ . Using the reversal law for matrix inverses, we have: C = (B⁻¹)⁻¹ A⁻¹ = B A⁻¹ First, we find the inverse of matrix A . The determinant of A is: |A| = 1(2 - 0) - 0 + 1(0 - 1) = 1 The adjugate matrix of A is the transpose of its cofactor matrix. Since A is symmetric, adj (A) is also symmetric: adj (A) = bmatrix 2 & 0 & -1 0 & 1 & 0 -1 & 0 & 1 bmatrix Thus, A⁻¹ = 1 |A| adj (A) = bmatrix 2 & 0 & -1 0 & 1 & 0 -1 & 0 & 1 bmatrix Now, we compute C = B A⁻¹ : C = bmatrix 2 & 1 & 0 1 & 1 & 2 0 & 2 & 1 bmat