91Ó°ÊÓ

To prove this, we first recall that the ith row of C is \((C_{1i}, C_{2i}, ..., C_{ni})\) and the jth column of A is \((a_{j1},a_{j2},...,a_{jn})\). Then \((AC)_{ji} = \sum_{k=1}^n a_{jk}C_{ki}\). We will now analyze the matrix product AC and determine its determinant. We know that det(AC) = det(A) * det(C). The matrix AC will be a diagonal matrix with elements (det(A), det(A), ..., det(A)), since the product of the ith row of A with the jth column of C will give the determinant of A if i = j, and 0 otherwise. Therefore, the determinant of AC = (det(A))^n. Now, we have det(AC) = det(A) * det(C) = (det(A))^n. Divide both sides by det(A), det(C) = [det(A)]^(n-1).

First recall how the classical adjoint is constructed. If B is the matrix of cofactors associated with A, C = B^t. Now we want to show that the adjoint of A^t is the transpose of the adjoint of A (C^t). Consider the cofactor of element (i, j) in A^t, denoted as M_{ij}^{(A^t)}. By the definition of the transpose, this cofactor is the same as the cofactor of element (j, i) in A, i.e., M_{ji}^{(A)}. Then, the classical adjoint of A^t is formed by transposing the matrix of cofactors M_{ji}^{(A)}, which is the same as the transpose of the classical adjoint of A (C^t).

Recall that the inverse of A can be computed as A^{-1} = (1/det(A)) * C. We will first show that both A^{-1} and C preserve the upper triangular property, and then discuss why the invertible condition is necessary. First, consider A^{-1}. If A is upper triangular, its diagonal elements, a_{ii}, are all nonzero. Since det(A) is the product of its diagonal elements, det(A) is nonzero. Now, we form A^{-1} by multiplying each element of C by the scalar 1/det(A). This multiplication does not change the position of the nonzero elements, so A^{-1} remains upper triangular. Next, consider C. The elements of C are found by transposing the matrix of cofactors of A. Since A is upper triangular, every cofactor in A is determined by the product of elements in its upper triangular part. Therefore, the cofactors and, subsequently, C, remain upper triangular. A quick note on invertibility: If A were not invertible, it would have a determinant of 0, which would prevent us from computing A^{-1} using the formula A^{-1} = (1/det(A)) * C.