# Matrix with trig functions and Cramer's rule

Using Cramer's rule solve for $x'$ and $y'$ in term of $x$ and $y$

$x = x'\cos\theta - y'\sin\theta\\ y = x'\sin\theta + y'\cos\theta$

So what I have is this

$\det\begin{bmatrix} \cos\theta& -\sin\theta\\ \sin\theta & \cos\theta \end{bmatrix} = \cos^2\theta + \sin^2\theta = 1$

Since the answer is one all the other determinants will be the final answers for the unknowns.

$x'= \det\begin{bmatrix} x &-\sin\theta\\ y & \cos\theta \end{bmatrix} = x\cos\theta + y\sin\theta$

$y' = \det\begin{bmatrix} \cos\theta & x\\ \sin\theta & y \end{bmatrix} = y\cos\theta - x\sin\theta$

Is this correct?

• Check my editing, and remove the unnecessary plus signs from some of the matrices' entries to add clearity. – DonAntonio Nov 3 '13 at 20:12
• And yes: it is right. – DonAntonio Nov 3 '13 at 20:13
• You can check the soluetion, plug in $x',y'$ in the system. Moreover, as $\det(A) \neq 0$ you know that the solution must be unique, so if it works that is it. – N. S. Nov 3 '13 at 20:20