Inverse Matrix

If matrix B is the inverse matrix of A then

$AB = I$

Assume that A is the matrix to cause a rotation of angle theta around the z-axis in a 3D vector space.

$A= \begin{bmatrix} cos (\theta) && - sin(\theta) && 0 \\ sin(\theta) && cos(\theta) && 0 \\ 0 && 0 && 1 \end{bmatrix}$

If the claim is true, we can create the inverse matrix, $A^-1$ by building the matrix that will cause a rotation of angle negative theta around the z-axis in a 3D vector space.

$A^{-1}= \begin{bmatrix} cos (- \theta) && sin(\theta) && 0 \\ sin(- \theta) && cos(- \theta) && 0 \\ 0 && 0 && 1 \end{bmatrix}$

Because cosine is an Even Function, cos(x) = cos(-x)
Because sine is an Odd Function, sin(x) = – sin(-x)

We use these two rules on the above matrix to get the following:

$A^{-1}= \begin{bmatrix} cos ( \theta) && sin(\theta) && 0 \\ - sin( \theta) && cos( \theta) && 0 \\ 0 && 0 && 1 \end{bmatrix}$

We multiply these two matrices together:

$A \: A^{-1}= \begin{bmatrix} cos (\theta) && - sin(\theta) && 0 \\ sin(\theta) && cos(\theta) && 0 \\ 0 && 0 && 1 \end{bmatrix} \begin{bmatrix} cos ( \theta) && sin(\theta) && 0 \\ - sin( \theta) && cos( \theta) && 0 \\ 0 && 0 && 1 \end{bmatrix}$

$A \: A^{-1} = \begin{bmatrix} cos^2 (\theta) + sin^2(\theta) && cos(\theta)sin(\theta) - sin(\theta)cos(\theta) && 0 \\ sin( \theta)cos( \theta) - cos( \theta)sin( \theta) && sin^2 (\theta) + cos^2(\theta) && 0 \\ 0 && 0 && 1 \end{bmatrix}$

$A \: A^{-1} = \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && 1 \end{bmatrix}$

This confirms that the matrix we wrote for $A^{-1}$ and the matrix we wrote for A are inverse matrices, one to the other.

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