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Q: Is Inverse of the inverse matrix the original matrix?

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If an identity matrix is the answer to a problem under matrix multiplication, then each of the two matrices is an inverse matrix of the other.

if A then B (original) if not A then not B (inverse)

of course it does

A matrix A is orthogonal if itstranspose is equal to it inverse. So AT is the transpose of A and A-1 is the inverse. We have AT=A-1 So we have : AAT= I, the identity matrix Since it is MUCH easier to find a transpose than an inverse, these matrices are easy to compute with. Furthermore, rotation matrices are orthogonal. The inverse of an orthogonal matrix is also orthogonal which can be easily proved directly from the definition.

The original function's RANGE becomes the inverse function's domain.

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That is called an inverse matrix

I could do that if you gave me the original matrix.

(I-A)-1 is the Leontief inverse matrix of matrix A (nxn; non-singular).

No. A square matrix has an inverse if and only if its determinant is nonzero.

The inverse of a rotation matrix represents a rotation in the opposite direction, by the same angle, about the same axis. Since M-1M = I, M-1(Mv) = v. Thus, any matrix inverse will "undo" the transformation of the original matrix.

it doesnt have an inverse since only square matrices have an inverse

it is used to find the inverse of the matrix. inverse(A)= (adj A)/ mod det A

To find the original matrix of an inverted matrix, simply invert it again. Consider A^-1^-1 = A^1 = A

If an identity matrix is the answer to a problem under matrix multiplication, then each of the two matrices is an inverse matrix of the other.

The fact that the matrix does not have an inverse does not necessarily mean that none of the variables can be found.

From Wolfram MathWorld: The inverse of a square matrix A, sometimes called a reciprocal matrix, is a matrix A-1 such that AA-1=I where I is the identity matrix.

if A then B (original) if not A then not B (inverse)

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