Use the Gauss-Jordan method to solve the system of equations. If the system has infinitely many solutions, let the last
variable be the arbitrary variable.
-$$\begin{array} { l }
2 x - 2 y + 3 z = 3 \
x + 2 y - z = 7 \
2 y + z = 1
\end{array}$$
A) $\varnothing$
B) $\{ ( - 4,1,15 ) \}$
C) $\{ ( 10,8 , - 5 ) \}$
D) $\{ ( 4,1 , - 1 ) \}$

Question

Use the Gauss-Jordan method to solve the system of equations. If the system has infinitely many solutions, let the last
variable be the arbitrary variable.
-$$\begin{array} { l } 
2 x - 2 y + 3 z = 3 \
x + 2 y - z = 7 \
2 y + z = 1
\end{array}$$
A) $\varnothing$
B) $\{ ( - 4,1,15 ) \}$
C) $\{ ( 10,8 , - 5 ) \}$
D) $\{ ( 4,1 , - 1 ) \}$

Quizplus · Accepted Answer

To solve the system of equations using the Gauss-Jordan method, we first write the augmented matrix for the system and then use row operations to reduce it to reduced row echelon form (RREF).Given system:$$2x - 2y + 3z = 3$$$$x + 2y - z = 7$$$$2y + z = 1$$The augmented matrix is:$$\left[\begin{array}{ccc|c}2 & -2 & 3 & 3 \1 & 2 & -1 & 7 \0 & 2 & 1 & 1\end{array}\right]$$We perform row operations to simplify this matrix into RREF:1. $R_2 = R_2 - \frac{1}{2}R_1$ to eliminate $x$ from the second equation.2. Use $R_3$ to eliminate $x$ and $y$ from the third equation appropriately.3. Scale rows if necessary to get leading 1s.4. Use back substitution to solve for $x$, $y$, and $z$.After performing these operations (which are too lengthy to detail step-by-step here), we find the solution to be $x = 4$, $y = 1$, and $z = -1$, which corresponds to choice D.

Use the Gauss-Jordan Method to Solve the System of Equations 2x−2y+3z=3x+2y−z=72y+z=1\begin{array} { l } 2 x - 2 y + 3 z = 3 \\x + 2 y - z = 7 \\2 y + z = 1\end{array}2x−2y+3z=3x+2y−z=72y+z=1​

Use the Gauss-Jordan Method to Solve the System of Equations $\begin{array} { l } 2 x - 2 y + 3 z = 3 \\x + 2 y - z = 7 \\2 y + z = 1\end{array}$