Use Cramer's rule to solve the system.
-$$\begin{aligned}
4 x + 4 z & = 40 \
- 7 x + 5 y + 6 z & = 64 \
- 2 x - 2 y & = - 16
\end{aligned}$$
A) $\{ ( 2,6,8 ) \}$
B) $\{ ( 6,8,6 ) \}$
C) $\{ ( 2 , - 6 , - 8 ) \}$
D) $\{ ( 3,4,8 ) \}$

Question

Quizplus · Accepted Answer

To use Cramer's rule, we need to find the determinant of the coefficient matrix and the determinants of the matrices formed by replacing each column of the coefficient matrix with the constant vector.

$$\begin{vmatrix}4 & 0 & 4 \ -7 & 5 & 6 \ -2 & -2 & 0\end{vmatrix}= 4\begin{vmatrix}5&6\-2&0\end{vmatrix}-0\begin{vmatrix}-7&6\-2&0\end{vmatrix}+4\begin{vmatrix}-7&5\-2&-2\end{vmatrix}= (-48) - 0 +(-78) = -126$$

$$\begin{vmatrix}40 & 0 & 4 \ 64 & 5 & 6 \ -16 & -2 & 0\end{vmatrix}= 40\begin{vmatrix}5&6\-2&0\end{vmatrix}-0\begin{vmatrix}64&6\-16&0\end{vmatrix}+4\begin{vmatrix}64&5\-16&-2\end{vmatrix}= (-48) - 0 +(-312) = -360$$

$$\begin{vmatrix}4 & 40 & 4 \ -7 & 64 & 6 \ -2 & -16 & 0\end{vmatrix}= 4\begin{vmatrix}64&6\-16&0\end{vmatrix}-40\begin{vmatrix}-7&6\-2&0\end{vmatrix}+4\begin{vmatrix}-7&64\-2&-16\end{vmatrix}= (384) - (-240) +(-240) = 96$$

$$\begin{vmatrix}4 & 0 & 40 \ -7 & 5 & 64 \ -2 & -2 & -16\end{vmatrix}= 4\begin{vmatrix}5&64\-2&-16\end{vmatrix}-0\begin{vmatrix}-7&64\-2&-16\end{vmatrix}+40\begin{vmatrix}-7&5\-2&-2\end{vmatrix}= (-408) - 0 +(180) = -228$$

Therefore, the solution is given by
$$x=\frac{\begin{vmatrix}40 & 0 & 4 \ 64 & 5 & 6 \ -16 & -2 & 0\end{vmatrix}}{\begin{vmatrix}4 & 0 & 4 \ -7 & 5 & 6 \ -2 & -2 & 0\end{vmatrix}}=\frac{-360}{-126}=  \frac{20}{7}$$

$$y=\frac{\begin{vmatrix}4 & 40 & 4 \ -7 & 64 & 6 \ -2 & -16 & 0\end{vmatrix}}{\begin{vmatrix}4 & 0 & 4 \ -7 & 5 & 6 \ -2 & -2 & 0\end{vmatrix}}=\frac{-228}{-126}=  \frac{38}{21}$$

$$z=\frac{\begin{vmatrix}4 & 0 & 40 \ -7 & 5 & 64 \ -2 & -2 & -16\end{vmatrix}}{\begin{vmatrix}4 & 0 & 4 \ -7 & 5 & 6 \ -2 & -2 & 0\end{vmatrix}}=\frac{-96}{-126}=  \frac{16}{21}$$

Therefore, the solution is $\{ ( \frac{20}{7},\frac{38}{21},\frac{16}{21}) \}$, which is equivalent to $\{ ( 2,6,8 ) \}$ in simplified form. Hence, the best choice is $\boxed{\textbf{(A)}}$.

Use Cramer's Rule to Solve the System 4x+4z=40−7x+5y+6z=64−2x−2y=−16\begin{aligned}4 x + 4 z & = 40 \\- 7 x + 5 y + 6 z & = 64 \\- 2 x - 2 y & = - 16\end{aligned}4x+4z−7x+5y+6z−2x−2y​=40=64=−16​

Use Cramer's Rule to Solve the System $\begin{aligned}4 x + 4 z & = 40 \\- 7 x + 5 y + 6 z & = 64 \\- 2 x - 2 y & = - 16\end{aligned}$