a square matrix has an inverse iff
the determinant is nonzero
There are different ways to state the conditions for a square matrix to have an inverse, but one common way is the following:
A square matrix has an inverse if and only if its determinant is nonzero.
In other words, if A is a n-by-n square matrix, denoted as A = [a_ij], then A has an inverse if and only if det(A) ≠ 0, where det(A) is the determinant of A.
This condition is equivalent to saying that A is non-singular, i.e., its row echelon form (REF) has no zero rows. This can be seen from the fact that computing the REF of a matrix is equivalent to performing elementary row operations, which do not affect the determinant. If the REF of A has a zero row, then det(A) = 0, and hence A is not invertible. Conversely, if det(A) ≠ 0, then the REF of A can be reduced to the identity matrix by elementary row operations, which means that A is invertible by definition.
Another way to state the condition for a matrix to be invertible is in terms of its columns. Specifically, a matrix A is invertible if and only if its columns form a linearly independent set. This condition is equivalent to saying that the nullspace of A is trivial, i.e., {x | Ax = 0} = {0}. To see this, suppose that the columns of A are linearly dependent. Then there exist scalars x_1, x_2, …, x_n, not all zero, such that Ax = 0, where x = [x_1, x_2, …, x_n]^T. This means that A cannot have an inverse, since A times any nonzero vector in the nullspace would also be zero. Conversely, if the columns of A are linearly independent, then the system Ax = b has a unique solution for any given b, and hence A is invertible.
Therefore, the condition for a square matrix to have an inverse is equivalent to any of the following:
– its determinant is nonzero
– its row echelon form has no zero rows
– its columns form a linearly independent set
– its nullspace is trivial
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