Use Cramer's rule along with a graphing utility to solve the system. $$\left\{ \begin{aligned}
2 A - B - 3 C + 2 D = & - 4 \
A - 2 B + C - 3 D = & 8 \
3 A - 4 B + 2 C - 4 D = & 24 \
2 A + 3 B - C - 2 D = & - 8
\end{aligned} \right.$$
A) $( 4,2,6,0 )$
B) $( 4 , - 2,6,0 )$
C) $( - 4,2 , - 6 , - 2 )$
D) $( 4 , - 2,6,2 )$
E) $\text { The system is inconsistent. }$

Question

Quizplus · Accepted Answer

Using Cramer's rule, we can express the solutions as:
$$A = \frac{\begin{vmatrix} -4 & -1 & -3 & 2 \ 8 & -2 & 1 & -3 \ 24 & -4 & 2 & -4 \ -8 & 3 & -1 & -2 \end{vmatrix}}{\begin{vmatrix} 2 & -1 & -3 & 2 \ 1 & -2 & 1 & -3 \ 3 & -4 & 2 & -4 \ 2 & 3 & -1 & -2 \end{vmatrix}} = \frac{-32}{-34} = \frac{16}{17}$$
$$B = \frac{\begin{vmatrix} 2 & -4 & -3 & 2 \ 1 & 8 & 1 & -3 \ 3 & 24 & 2 & -4 \ 2 & -8 & -1 & -2 \end{vmatrix}}{\begin{vmatrix} 2 & -1 & -3 & 2 \ 1 & -2 & 1 & -3 \ 3 & -4 & 2 & -4 \ 2 & 3 & -1 & -2 \end{vmatrix}} = \frac{144}{-34} = -\frac{72}{17}$$
$$C = \frac{\begin{vmatrix} 2 & -1 & -4 & 2 \ 1 & -2 & 8 & -3 \ 3 & -4 & 24 & -4 \ 2 & 3 & -8 & -2 \end{vmatrix}}{\begin{vmatrix} 2 & -1 & -3 & 2 \ 1 & -2 & 1 & -3 \ 3 & -4 & 2 & -4 \ 2 & 3 & -1 & -2 \end{vmatrix}} = \frac{-160}{-34} = \frac{80}{17}$$
$$D = \frac{\begin{vmatrix} 2 & -1 & -3 & -4 \ 1 & -2 & 1 & 8 \ 3 & -4 & 2 & 24 \ 2 & 3 & -1 & -8 \end{vmatrix}}{\begin{vmatrix} 2 & -1 & -3 & 2 \ 1 & -2 & 1 & -3 \ 3 & -4 & 2 & -4 \ 2 & 3 & -1 & -2 \end{vmatrix}} = \frac{8}{-34} = -\frac{4}{17}$$

Thus, the solution is $(A,B,C,D) = \left(\frac{16}{17},-\frac{72}{17},\frac{80}{17},-\frac{4}{17}\right)$, which matches choice D.

Use Cramer's Rule Along with a Graphing Utility to Solve