91影视

To show this, we will use two properties of rank and invertibility. Firstly, if A is invertible, then the product of A and its inverse A鈦宦� is the identity matrix I of size n, that is, AA鈦宦� = I. Secondly, the rank of a matrix is equal to the rank of its row space, and rank of a matrix is unchanged by elementary row operations. Now, let's perform elementary row operations on A to reduce it to row-echelon form, which we will call B. Since A is invertible, there must be n nonzero rows in B with n leading ones (pivots). This implies that the rank of B, and therefore the rank of A, is equal to n.

Now we will assume that the rank of the matrix A is n. This means there are n linearly independent rows in A. As a consequence, when performing Gaussian elimination or row reduction on A, we will not eliminate any rows to the zero row. Continue to perform elementary row operations until we reach a row-echelon form B and ultimately the reduced row-echelon form (also called the row-reduced echelon form), which we will call C. Since we did not eliminate any rows, C has n leading ones (pivots) and is a square matrix of size nxn. Now, the matrix C is the identity matrix I of size n because it contains n leading ones and the other elements in the matrix are zero. Since we can convert matrix A to the identity matrix I by elementary row operations, it means A is row-equivalent to I. By the definition of invertibility, if A can be converted to the identity matrix by a sequence of elementary row operations, then A is invertible. Thus, we have proved that A is invertible if and only if the rank of A is n.