Solve the system, if possible, by using Cramer's rule. If Cramer's rule does not apply, solve the system by
using another method.
-$$\begin{array} { l }
6 x - 9 y + 9 z = 3 \
- 4 x + 6 y - 6 z = - 2 \
2 x - 3 y + 3 z = 2
\end{array}$$
A) $\{ ( - 1,1,2 ) \}$
B) { }
C) $\{ ( 1 , - 1 , - 2 ) \}$
D) $\left\{ \left( 1 + \frac { 3 } { 2 } y - \frac { 3 } { 2 } z \right) \mid y \right.$ and $z$ are any real numbers $\}$

Question

Solve the system, if possible, by using Cramer's rule. If Cramer's rule does not apply, solve the system by
using another method.
-$$\begin{array} { l } 
6 x - 9 y + 9 z = 3 \
- 4 x + 6 y - 6 z = - 2 \
2 x - 3 y + 3 z = 2
\end{array}$$
A) $\{ ( - 1,1,2 ) \}$
B) {    }
C) $\{ ( 1 , - 1 , - 2 ) \}$
D) $\left\{ \left( 1 + \frac { 3 } { 2 } y - \frac { 3 } { 2 } z \right) \mid y \right.$ and $z$ are any real numbers $\}$

Quizplus · Accepted Answer

The system of equations is dependent, meaning that all three equations represent the same plane in three-dimensional space. This can be observed by noting that the second and third equations are multiples of the first equation. Therefore, Cramer's rule does not apply because the determinant of the coefficient matrix is zero, indicating an infinite number of solutions or no solution. However, in this case, since the equations are dependent, it means there are infinitely many solutions, but they cannot be represented by a single point or parameterized as given in the options A, C, or D.

Solve the System, If Possible, by Using Cramer's Rule 6x−9y+9z=3−4x+6y−6z=−22x−3y+3z=2\begin{array} { l } 6 x - 9 y + 9 z = 3 \\- 4 x + 6 y - 6 z = - 2 \\2 x - 3 y + 3 z = 2\end{array}6x−9y+9z=3−4x+6y−6z=−22x−3y+3z=2​

Solve the System, If Possible, by Using Cramer's Rule $\begin{array} { l } 6 x - 9 y + 9 z = 3 \\- 4 x + 6 y - 6 z = - 2 \\2 x - 3 y + 3 z = 2\end{array}$