Using determinant to find constraints on equation

[TheDemx27]

Basically I don't know how to get to the constraints from the system of equations. In class we used det to find constraints for homogenous equations, but we didn't go over this situation. Someone spell it out for me?

 

[mathman]

The determinant is $sin^2(\theta )(1+cos^2(\theta ))$.

 

[TheDemx27]

The determinant is $sin^2(\theta )(1+cos^2(\theta ))$.

How did you get that?
$det(A)=cos^2(\theta )sin^2(\theta )+sin^4(\theta )$ Which then simplifies into what they got: $=\frac{1}{2}\ (1-cos(\theta ))=sin^2(\theta )$

My question is how they proceed after that anyways.

 

[Stephen Tashi]

The solution for a particular unknown in a system of linear equations can be expressed as a ratio of two determinants. In your example, the determinant of the denominator is $\sin^2(q)$. The numerator is the determinant of a matrix formed by using $C_1, C_2$ in place of some entries in the coefficient matrix, depending on which unknown you are solving for. See if the standard method of solving equations in that manner ( e.g. https://www.cliffsnotes.com/study-g...tions-using-determinants-with-three-variables ) derives the equations the article ends up with.$\begin{pmatrix} \cos(\theta) \sin(\theta) & \sin^2(\theta) \\ -\sin^2(\theta) & \cos(\theta)\sin(\theta) \end{pmatrix} \begin{pmatrix} X \\ Y \end{pmatrix} =\begin{pmatrix} C_1 \\ C_2 \end{pmatrix}$

Solve for $X,Y$.

 

[mathman]

How did you get that?
$det(A)=cos^2(\theta )sin^2(\theta )+sin^4(\theta )$ Which then simplifies into what they got: $=\frac{1}{2}\ (1-cos(\theta ))=sin^2(\theta )$

My question is how they proceed after that anyways.

my mistake