Let every element of a nonzero square matrix of order n be equal to its algebraic cofactor, and prove that R (a) = n

Let every element of a nonzero square matrix of order n be equal to its algebraic cofactor, and prove that R (a) = n

Let adj (a) be the adjoint matrix of a, then the condition is written as a '= adj (a)
Using adj (a) * a = det (a) * I, DET (a) is nonzero, otherwise a'a = 0, a = 0, which is contradictory

Let a be a non-zero real square matrix of order n, and every element AIJ of a is equal to its algebraic cofactor, namely AIJ = AIJ, (I, j = 1,2,3 n) Prove that a is reversible

This problem can be proved in this way,
The element of the adjoint matrix A * (J, I) of a is AIJ algebraic cofactor AIJ. From this, we can see that the title you gave is that each element AIJ of a is equal to its algebraic cofactor, that is AIJ = AIJ, and a = (a *) '
Another way is a * = a 'where' means transpose. This connects the problem with adjoint matrix
The proof of adjoint matrix is fixed
To the contrary, if a is irreversible, that is R (a)

Let the determinant of n-order square matrix a be equal to 0, and there is an algebraic cofactor a (ij) not equal to 0. It is proved that the general solution of the system AX = 0 is
k（A（i1）,A（i2）,… Transpose of, a (in))

It is proved that because | a | = 0, so AA * = | a | = 0, so the column vectors of a * are solutions of AX = 0. And because | a | = 0, so r (a) = 1, so r (a) > = n-1, so r (a) = n-1