How is this proposition proven?

Question

If A is an nxn matrix, det(adj A)= ((det(A))^n-1.

Answer 1

As a result of Laplace's expansion for matrix determinants,

det(A).I = A.adj(A).

So, taking determinants of both sides, we see that

det(A)^n = det(det(A).I) = det(A.adj(A)) = det(A).det(adj(A)),

and so det(adj(A)) = det(A)^(n-1).

Answer 2

Suppose det A is nonzero. Then you have that A⁻¹ = 1/det (A) * adj (A). Clearly, det (A⁻¹) = 1/det (A), and you know that det (kA) = k^n det (A). So, 1/det (A) = det (1/det (A) * adj (A)) = 1/det (A)^n * det (adj (A)). Multiplying by det (A)^n yields det (A)^(n-1) = det (adj(A)).

If det (A) = 0, we still have the fact that A*adj(A) = det (A)*I_n, which in this case means A*adj(A) = 0. This means that ker (A*adj(A)) = R^n. If A≠0, then ker(A)≠R^n, but Im (adj (A)) ⊆ ker (A), so Im (adj (A)) ≠ R^n and thus rank (adj (A)) < n, so by the rank-nullity theorem nullity (adj (A)) = n-rank(adj (A)) > 0, so ker (adj (A))≠{0}, therefore adj (A) is singular and has determinant zero. Since det (A) = 0, this means that det (adj (A)) = 0 = det (A)^(n-1), as claimed.

Finally, if A=0, it can be verified by manual calculation that adj (A) = 0, so det (adj (A)) = 0 = det (A)^(n-1).

(of course all this assumes that n≠1. It's not entirely clear how to define the adjoint of a 1×1 matrix, since deleting any row or column leaves you with no numbers left to find a determinant. If we want to preserve the expansion by minors when considering the determinant of a hypothetical 0×0 matrix, then we would have to say the determinant of such a matrix is 1. In this case, we would define adj ([x]) = [1] regardless of the value of x. This makes sense, because it preserves the identity A*adj (A) = det (A) * I_n in the case of 1×1 matrices as well. In this case, we would also have that det (adj (A)) = 1 = det (A)^0 (recalling that 0^0=1), regardless of the value of A, so this identity would hold when n=1 as well.)