Question 366418
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left|\frac{1}{4}\ \ \ \frac{1}{4}\ \ 2\cr1\ -1\ \ 4\right|]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4R1\ \rightarrow\ R1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left|1\ \ \ 1\ \ 8\cr1\ -1\ \ 4\right|]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -R1\ +\ R2\ \rightarrow\ R2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left|1\ \ \ 1\ \ \ \,8\cr0\ -2\ -4\right|]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -R2/2\ \rightarrow\ R2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left|1\ \ 1\ \ 8\cr0\ \ 1\ \ 2\right|]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -R2\ +\ R1\ \rightarrow\ R1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left|1\ \ 0\ \ 6\cr0\ \ 1\ \ 2\right|]


So the solution set is *[tex \Large \left\{\left(6,2\right)\right\}]






John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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