Use Cramer's Rule to solve the following system of linear equations: $$\left\{ \begin{aligned}
3 x - 6 z & = 8 \
- 6 y + 13 z & = 3 \
6 x + 15 z & = 0
\end{aligned} \right.$$
A) x = $\frac { 40 } { 27 }$ , y = $- \frac { 16 } { 27 }$ , z = $- \frac { 289 } { 162 }$
B) x = $\frac { 40 } { 27 }$ , y = $- \frac { 289 } { 162 }$ , z = $- \frac { 16 } { 27 }$
C) x = $\frac { 27 } { 40 }$ , y = $- \frac { 162 } { 289 }$ , z = $- \frac { 27 } { 16 }$
D) x = $\frac { 27 } { 40 }$ , y = $- \frac { 289 } { 162 }$ , z = $- \frac { 16 } { 27 }$
E) x = $\frac { 40 } { 27 }$ , y = $- \frac { 162 } { 289 }$ , z = $- \frac { 16 } { 27 }$

Question

Use Cramer's Rule to solve the following system of linear equations:   $$\left\{ \begin{aligned}
3 x - 6 z & = 8 \
- 6 y + 13 z & = 3 \
6 x + 15 z & = 0
\end{aligned} \right.$$
A) x = $\frac { 40 } { 27 }$ , y = $- \frac { 16 } { 27 }$ , z = $- \frac { 289 } { 162 }$
B) x = $\frac { 40 } { 27 }$ , y = $- \frac { 289 } { 162 }$ , z = $- \frac { 16 } { 27 }$  
C) x = $\frac { 27 } { 40 }$ , y = $- \frac { 162 } { 289 }$ , z = $- \frac { 27 } { 16 }$
D) x = $\frac { 27 } { 40 }$ , y = $- \frac { 289 } { 162 }$ , z = $- \frac { 16 } { 27 }$
E) x = $\frac { 40 } { 27 }$ , y = $- \frac { 162 } { 289 }$ , z = $- \frac { 16 } { 27 }$

Quizplus · Accepted Answer

Cramer's Rule involves calculating determinants. Solving the system using determinants gives the correct values for x, y, and z as in choice B.

Use Cramer's Rule to Solve the Following System of Linear {3x−6z=8−6y+13z=36x+15z=0\left\{ \begin{aligned}3 x - 6 z & = 8 \\- 6 y + 13 z & = 3 \\6 x + 15 z & = 0\end{aligned} \right.⎩⎨⎧​3x−6z−6y+13z6x+15z​=8=3=0​

Use Cramer's Rule to Solve the Following System of Linear $\left\{ \begin{aligned}3 x - 6 z & = 8 \\- 6 y + 13 z & = 3 \\6 x + 15 z & = 0\end{aligned} \right.$