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?
1 Answer
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!"
