This is not true!
numpy.linalg.inv(A) actually calls numpy.linalg.solve(A,I) , where I is the identifier and decides to use lapack LU factorization .
That is, in the end, it eliminates the Gauss, where orthogonality is not defined by default.
And I donโt think there is a shot in the dark to check something like A * AT = I , since the matrix time matrix is โโexpensive.
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