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I'm having some trouble with the following linear algebra question on Cramer's rule: An explanation for your choice of answer please.

Determine whether Cramer's Rule can be applied on the following system

$$ \begin{cases}x_1\cos(u) - x_2\sin(u) = 1\\x_1\sin(u)+x_2\cos(u)=-3\end{cases}$$

Thanks in advance.

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closed as unclear what you're asking by user228113, SchrodingersCat, vrugtehagel, martini, drhab Feb 18 '16 at 16:01

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Welcome to MSE! this site uses latex for math notations, please use it. Also, it is highly recommended to show some effort. What have you tried? $\endgroup$ – Galc127 Feb 18 '16 at 13:19
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    $\begingroup$ Well, what is the condition in Cramer's rule (i.e. what needs to be satisfied for you to be allowed to apply it?) $\endgroup$ – Tobias Kildetoft Feb 18 '16 at 13:20
  • $\begingroup$ It depends on which are the indeterminates and which are the parameters. $\endgroup$ – user228113 Feb 18 '16 at 13:25
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To use Cramer's rule, the square coefficient matrix $A$ (for the system $AX=B$ with as many equations as unknowns) has to have a nonzero determinant.

What is $A$ for your system? Check its determinant.

Note: a nonzero determinant guarantees a unique solution (given by Cramer's rule).

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    $\begingroup$ Why would the OP check that the determinant is $0$ when it is in fact not? $\endgroup$ – Tobias Kildetoft Feb 18 '16 at 13:34
  • $\begingroup$ Thanks, my mistake! (edited) $\endgroup$ – StackTD Feb 18 '16 at 13:36
  • $\begingroup$ Aaaah... Thanks... So finding the determinant of this coefficient matrix A by cofactor expansion along the first row, gives us (\cos\theta)(\cos\theta) + ((-\sin\theta)(-\sin\theta)) = \cos^2\theta + \sin^2\theta = 1 $\therefore determinant of A is nonzero and Cramer's rule can be used. True. p.s. I don't know how to use latex. Will check it out soon. $\endgroup$ – Samiera Ebrahim Feb 18 '16 at 20:19
  • $\begingroup$ That's correct! $\endgroup$ – StackTD Feb 18 '16 at 20:54

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