Use Cramer's Rule to solve the following system of linear equations: $$\left\{ \begin{aligned}
14 x - 21 y + 14 z & = 1 \
- 21 x + 14 y + 7 z & = 5 \
28 x + 7 y - 21 z & = 5
\end{aligned} \right.$$
A) $$x = \frac { 5 } { 7 } ; y = \frac { 55 } { 49 } ; z = \frac { 95 } { 147 }$$
B) $$x = \frac { 3 } { 7 } ; y = \frac { 45 } { 98 } ; z = \frac { 12 } { 49 }$$
C) $$x = \frac { 11 } { 14 } ; y = \frac { 115 } { 98 } ; z = \frac { 106 } { 147 }$$
D) $$x = \frac { 11 } { 21 } ; y = \frac { 115 } { 147 } ; z = \frac { 106 } { 147 }$$
E) $$x = \frac { 3 } { 7 } ; y = \frac { 75 } { 98 } ; z = \frac { 27 } { 49 }$$

Question

Quizplus · Accepted Answer

Cramer's Rule involves calculating determinants. The determinant of the coefficient matrix is non-zero, and using Cramer's Rule, the solution is $x = \frac{11}{21}$, $y = \frac{115}{147}$, $z = \frac{106}{147}$.

Use Cramer's Rule to Solve the Following System of Linear {14x−21y+14z=1−21x+14y+7z=528x+7y−21z=5\left\{ \begin{aligned}14 x - 21 y + 14 z & = 1 \\- 21 x + 14 y + 7 z & = 5 \\28 x + 7 y - 21 z & = 5\end{aligned} \right.⎩⎨⎧​14x−21y+14z−21x+14y+7z28x+7y−21z​=1=5=5​

Use Cramer's Rule to Solve the Following System of Linear $\left\{ \begin{aligned}14 x - 21 y + 14 z & = 1 \\- 21 x + 14 y + 7 z & = 5 \\28 x + 7 y - 21 z & = 5\end{aligned} \right.$