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Solve the following system using Cramer's Rule.

$$2x + y = 1$$

$$x - 4y = 14$$

I haven't done Cramer's rule for 2x2 matrices, but I figured that the same rules applied as in a 3x3, here's what I did;

$$\det(A) = -8-1 = -9$$


$D_x$ = $ \begin{align} \begin{bmatrix} 4 & 1\\ 14 & -4 \end{bmatrix} \end{align} $ $ = -16-14 = -30 \quad \quad\quad D_y = \begin{align} \begin{bmatrix} 2 & 4\\ 1 & 14 \end{bmatrix} \end{align} = 28 - 4 = 24 $


$$x = \frac{-30}{-9} = \frac{10}{3}; \quad\quad y = \frac{24}{3} = 8$$

Now the solution comes up with;

$Dx$ = $ \begin{align} \begin{bmatrix} 1 & 1\\ 14 & -4 \end{bmatrix} \end{align} $

$Dy$ = $ \begin{align} \begin{bmatrix} 2 & 1\\ 1 & 14 \end{bmatrix} \end{align} $

Which then results in a completely different answer, it's more likely that I made a mistake, could anyone point out what I did wrong?

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    $\begingroup$ You appear to have $4$ in your $Dx$ and $Dy$ where you should have $1$. Perhaps a transcription error? $\endgroup$ – vadim123 May 28 '13 at 0:05
  • $\begingroup$ You're write, when I re-wrote the question in my textbook I made that mistake. Thank you for pointing it out for me. $\endgroup$ – Matthew Brzezinski May 28 '13 at 0:07
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I think your thought process is just fine: however, I suspect that you may have transposed $1$ and $4$ in the respective columns for $D_x$ and $D_y$, which would account for the "discrepancy".

Did you mis-transcribe the value $4$ (right-hand side of equation 1), having meant to write $1$?

And if so, perform your calculations using Cramer's rule, using the same process, and I think you'll be "good to go!"

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