Write the system of linear equations represented by the augmented matrix.Then use back-substitution to solve.(Use variables x, y, and z.) $$\left[ \begin{array} { r r r r r }
1 & 9 & - 3 & \vdots&- 39 \
0 & 1 & 2 & \vdots & 0 \
0 & 0 & 1 & \vdots & 2
\end{array} \right]$$
A)x = 4, y = 2, z = -3
B)x = 3, y = -4, z = 2
C)x = 3, y = 4, z = -2
D)x = -18, y = -1, z = 4
E)x = -4, y = 3, z = 2

Question

Write the system of linear equations represented by the augmented matrix.Then use back-substitution to solve.(Use variables x, y, and z.) $$\left[ \begin{array} { r r r r r } 
1 & 9 & - 3 &  \vdots&- 39 \
0 & 1 & 2 & \vdots & 0 \
0 & 0 & 1 & \vdots & 2
\end{array} \right]$$
A)x = 4, y = 2, z = -3
B)x = 3, y = -4, z = 2
C)x = 3, y = 4, z = -2
D)x = -18, y = -1, z = 4
E)x = -4, y = 3, z = 2

Quizplus · Accepted Answer

The augmented matrix represents the system of equations:1. $x + 9y - 3z = -39$2. $y + 2z = 0$3. $z = 2$Starting from the bottom, we substitute $z = 2$ into the second equation to find $y$, and then use both $y$ and $z$ to find $x$ from the first equation.From the third equation, $z = 2$.Substituting $z = 2$ into the second equation gives $y + 2(2) = 0$, so $y = -4$.Finally, substituting $y = -4$ and $z = 2$ into the first equation gives $x + 9(-4) - 3(2) = -39$, solving for $x$ gives $x = 3$.Therefore, the solution is $x = 3$, $y = -4$, $z = 2$.

Write the System of Linear Equations Represented by the Augmented [19−3⋮−39012⋮0001⋮2]\left[ \begin{array} { r r r r r } 1 & 9 & - 3 & \vdots&- 39 \\0 & 1 & 2 & \vdots & 0 \\0 & 0 & 1 & \vdots & 2\end{array} \right]​100​910​−321​⋮⋮⋮​−3902​​

Write the System of Linear Equations Represented by the Augmented $\left[ \begin{array} { r r r r r } 1 & 9 & - 3 & \vdots&- 39 \\0 & 1 & 2 & \vdots & 0 \\0 & 0 & 1 & \vdots & 2\end{array} \right]$