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

Cramer's Rule involves calculating determinants. The determinant of the coefficient matrix is non-zero, allowing us to use Cramer's Rule. Solving for each variable using the determinants of matrices formed by replacing columns with the constant terms gives the solution: $x = \frac { 5 } { 18 } ; y = - \frac { 13 } { 12 } ; z = - \frac { 1 } { 9 }$.

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 $\left\{ \begin{array} { r } 4 x~~~ - 8 z = 2 \\- 4 y + 12 z = 3 \\8 x~ + 20 z = 0\end{array} \right.$