Computing the determinant of a matrix over a field

Let A be a square matrix over a field F.

  1. Compute the determinant of each type of elementary matrix.
  2. Prove that \mathsf{det}(A) is nonzero if and only if A is row-equivalent to the identity matrix. Suppose there is a sequence of row operations containing s row interchanges and t row multiplications, using the nonzero constants u_1, \ldots, u_t. Prove that \mathsf{det}(A) = (-1)^s (u_1 \cdots u_t)^{-1}.

We discussed elementary matrices in this previous exercise.

Note that if \sigma \in S_n is not the identity, then there must exist an element i such that \sigma(i) > i. (Otherwise, we can show by induction that \sigma = 1.) Now recalling the naive formula for computing \mathsf{det}(A), note that if A is an upper- or lower-triangular matrix, then \mathsf{det}(A) is merely the product of the diagonal entries.

In particular, we have \mathsf{det}(E_{s,s}(\alpha)) = \alpha, and \mathsf{det}(E_{s,t}(\alpha)) = 1 if s \neq t. Note also that the row operation of interchanging two rows is equivalent to three operations which add a multiple of one row to another and one scalar row multiplication by -1. Since determinants are multiplicative, the determinant of the elementary matrix achieving this operation is -1.

Suppose \mathsf{det}(A) \neq 0. In this previous exercise, we saw that the columns of A are linearly independent. In particular, all columns in the reduced row echelon form of A are pivotal, and so A is row equivalent to the identity matrix. Conversely, suppose A is row equivalent to the identity matrix. Then the columns of A are linearly independent, and so \mathsf{det}(A) \neq 0.

Suppose now that PA = I, where P is a product of elementary matrices. Since determinants are multiplicative, it is clear that \mathsf{det}(A) has the given form.

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