If you assume the matrix $A$ is invertible, then $A^{-1} \cdot A = I$. Is there an assumption for invertibility in Mathematica 9? How can one make the following evaluate to the identity matrix $I_3$?

$Assumptions = {Element[A, Matrices[{3, 3}]]}
  • 1
    $\begingroup$ Look at the Tensor* functions. Specifically, pass your result to TensorReduce or TensorExpand. $\endgroup$ Feb 20, 2013 at 14:54

1 Answer 1


This gives what you want! Adding the axiom of matrix power $A^0=I$ to Simplify gives us

Assuming[Element[A, Matrices[{n, n}]] && Det[A] != 0,TensorExpand[Inverse[A].A]], 
ForAll[{A}, MatrixPower[A, 0] == IdentityMatrix[n]]]


Use n=3 to validate your case.

  • $\begingroup$ Why is the axiom of matrix power not assumed by default? $\endgroup$
    – masterxilo
    Feb 10, 2017 at 20:02
  • $\begingroup$ Perhaps because 0^0 is an indeterminate expression that can be both 0 and 1 with a sensible definition. One has 0^x = 0 but also x^0 = 1. With 0^0, they contradict each other, one has to make a call. Perhaps Mathematica does not make that for the user. $\endgroup$ Mar 8, 2017 at 9:01

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