TITLE: Determine if a matrix is unimodular
QUESTION [8 upvotes]: Is deciding if an integer square matrix has determinant $\pm 1$ faster that calculating the determinant of the matrix?

REPLY [2 votes]: From Santha and Tan, "Verifying the determinant in parallel":

"for the determinant, there doesn't seem to be a simpler way to verify it than the computation itself."

Other notes: 
Matrix multiplication, inversion, and determinant-finding are all equally complex. I will quote from Kaltofen and Villard, "Computing the sign or the value of the determinant of an integer matrix, a complexity survey":

In algebraic complexity—i.e. when counting the number of operations in
  an abstract domain—we refer to Strassen [52] and Bunch & Hopcroft [13]
  for the reduction of the problem of computing the determinant to
  matrix multiplication. Conversely, Strassen [53] and Bunch & Hopcroft
  [13] reduce matrix multiplication to matrix inversion, and Baur &
  Strassen reduce matrix inversion to computing the determinant [7]

Since unimodularity amounts to invertibility, your question is equivalent to "is determining invertibility easier than actually computing the inverse (if it exists)?"