Use Cramer's Rule to solve the following system of linear equations: $$\left\{ \begin{array} { r }
4 x - 8 z = 2 \
- 4 y + 12 z = 3 \
8 x + 20 z = 0
\end{array} \right.$$
A) $x = \frac { 1 } { 2 } ; y = - \frac { 5 } { 4 } ; z = - \frac { 1 } { 2 }$
B) $x = \frac { 5 } { 18 } ; y = - \frac { 13 } { 12 } ; z = - \frac { 1 } { 9 }$
C) $x = \frac { 1 } { 18 } ; y = - \frac { 13 } { 12 } ; z = - \frac { 1 } { 18 }$
D) $x = \frac { 5 } { 18 } ; y = - \frac { 11 } { 12 } ; z = - \frac { 1 } { 2 }$
E) $x = \frac { 1 } { 2 } ; y = - \frac { 5 } { 4 } ; z = - \frac { 1 } { 2 }$.

Question

Use Cramer's Rule to solve the following system of linear equations: $$\left\{ \begin{array} { r } 
4 x - 8 z = 2 \
- 4 y + 12 z = 3 \
8 x + 20 z = 0
\end{array} \right.$$
A) $x = \frac { 1 } { 2 } ; y = - \frac { 5 } { 4 } ; z = - \frac { 1 } { 2 }$
B) $x = \frac { 5 } { 18 } ; y = - \frac { 13 } { 12 } ; z = - \frac { 1 } { 9 }$
C) $x = \frac { 1 } { 18 } ; y = - \frac { 13 } { 12 } ; z = - \frac { 1 } { 18 }$
D) $x = \frac { 5 } { 18 } ; y = - \frac { 11 } { 12 } ; z = - \frac { 1 } { 2 }$
E) $x = \frac { 1 } { 2 } ; y = - \frac { 5 } { 4 } ; z = - \frac { 1 } { 2 }$.

Quizplus · Accepted Answer

To solve using Cramer's Rule, we first find the determinant of the coefficient matrix. The determinant is non-zero, so the system has a unique solution. By calculating the determinants of matrices formed by replacing each column with the constants, we find the values of $x$, $y$, and $z$ as given in option B.

Use Cramer's Rule to Solve the Following System of Linear {4x−8z=2−4y+12z=38x+20z=0\left\{ \begin{array} { r } 4 x - 8 z = 2 \\- 4 y + 12 z = 3 \\8 x + 20 z = 0\end{array} \right.⎩⎨⎧​4x−8z=2−4y+12z=38x+20z=0​

Use Cramer's Rule to Solve the Following System of Linear $\left\{ \begin{array} { r } 4 x - 8 z = 2 \\- 4 y + 12 z = 3 \\8 x + 20 z = 0\end{array} \right.$