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.
-$$\begin{array} { l }
x - y + 2 z + w = 5 \
y + z = 1 \
z - w = 4
\end{array}$$
A) $\{ ( - 6 - 4 w , - 3 - w , 4 + w , w ) \}$
B) $\{ ( - 10 , - 4,5,1 ) \}$
C) $\varnothing$
D) $\{ ( - 6 , - 3,4 , \mathrm {~W} ) \}$

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.
-$$\begin{array} { l } 
x - y + 2 z + w = 5 \
y + z = 1 \
z - w = 4
\end{array}$$
A) $\{ ( - 6 - 4 w , - 3 - w , 4 + w , w ) \}$
B) $\{ ( - 10 , - 4,5,1 ) \}$
C) $\varnothing$
D) $\{ ( - 6 , - 3,4 , \mathrm {~W} ) \}$

Quizplus · Accepted Answer

To solve the system of equations using the Gauss-Jordan method, we first write the augmented matrix and then apply row operations to get it into reduced row echelon form (RREF). The system is:1. $x - y + 2z + w = 5$2. $y + z = 1$3. $z - w = 4$The augmented matrix is:$$\left[\begin{array}{cccc|c}1 & -1 & 2 & 1 & 5 \0 & 1 & 1 & 0 & 1 \0 & 0 & 1 & -1 & 4 \\end{array}\right]$$We can use the third equation to express $z$ in terms of $w$: $z = 4 + w$.Substituting $z$ into the second equation to find $y$: $y + (4 + w) = 1 \Rightarrow y = -3 - w$.Substituting $z$ and $y$ into the first equation to find $x$: $x - (-3 - w) + 2(4 + w) + w = 5 \Rightarrow x = -6 - 4w$.Thus, the solution set is $\{(-6 - 4w, -3 - w, 4 + w, w)\}$, where $w$ is an arbitrary variable.

Use the Gauss-Jordan Method to Solve the System of Equations x−y+2z+w=5y+z=1z−w=4\begin{array} { l } x - y + 2 z + w = 5 \\y + z = 1 \\z - w = 4\end{array}x−y+2z+w=5y+z=1z−w=4​

Use the Gauss-Jordan Method to Solve the System of Equations $\begin{array} { l } x - y + 2 z + w = 5 \\y + z = 1 \\z - w = 4\end{array}$