Question 1137268
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Check out <a href="http://www.mathcentre.ac.uk/resources/uploaded/sigma-matrices7-2009-1.pdf">this article</a> to see how to find the inverse of a 2x2 matrix. 


Part of the steps will have us compute the determinant. The determinant of the 2x2 matrix is found through this formula
E = determinant
E = a*d - b*c
where a,b,c,d are arranged like this in the matrix
*[Tex \Large \begin{bmatrix}a & b\\c & d\end{bmatrix}]
In this case,
a = cos(theta)
b = -sin(theta)
c = sin(theta)
d = cos(theta)
which we can see through this comparison
*[Tex \Large \begin{bmatrix}a & b\\c & d\end{bmatrix} = \begin{bmatrix}\cos(\theta) & -\sin(\theta)\\\sin(\theta) & \cos(\theta)\end{bmatrix}]



So the determinant is,
E = a*d - b*c
E = cos(theta)*cos(theta) - (-sin(theta))*sin(theta)
E = cos(theta)*cos(theta) + sin(theta)*sin(theta)
E = cos^2(theta) + sin^2(theta)
E = 1
For that last step, you use the pythagorean trig identity. 


Therefore, the expression {{{1/(a*d - b*c)}}} is equal to 1 {{{1/(a*d - b*c) = 1/1 = 1}}}, meaning that this portion does not affect the result. This is because multiplying by 1 does not change the answer. We can ignore this piece because the determinant is 1.


In that same article I posted, note how the original matrix looks like this
*[Tex \Large \begin{bmatrix}a & b\\c & d\end{bmatrix}]
and it becomes this
*[Tex \Large \begin{bmatrix}d & -b\\-c & a\end{bmatrix}]
You swap a and d; also change the signs of b and c but keep them in the same spot.


Since 
a = cos(theta)
b = -sin(theta)
c = sin(theta)
d = cos(theta)
we can say...
*[Tex \Large \begin{bmatrix}d & -b\\-c & a\end{bmatrix} = \begin{bmatrix}\cos(\theta) & -(-\sin(\theta))\\-\sin(\theta) & \cos(\theta)\end{bmatrix}]


*[Tex \Large \begin{bmatrix}d & -b\\-c & a\end{bmatrix} = \begin{bmatrix}\cos(\theta) & \sin(\theta)\\-\sin(\theta) & \cos(\theta)\end{bmatrix}]
which is the inverse of the original matrix. I'll let you confirm this by computing A*B and B*A, where A is the original matrix and B is the inverse matrix. You should get AB = I and BA = I. Recall that the matrix I is the identity matrix.
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