Problem 5. (1 point)
The linear system
\[
\begin{aligned}
2 x_{3} & =14 \\
-x_{2}+x_{3} & =-2 \\
7 x_{1}+7 x_{2}+4 x_{3} & =0
\end{aligned}
\]
is not in echelon form.
Begin by choosing which of the following statements are correct. If there is more than one reason why the system is not in echelon form, type the letters as a comma separated list.
A. The system is not in echelon form because a variable is the leading variable of two or more equations.
B. The system is not in echelon form because the system is not organized in a descending "stair step" pattern so that the index of the leading variables increases from the top to bottom.
C. The system is not in echelon form because not every equation has a leading variable.
Correct Letter(s): $A$
Now write the system in echelon form.
Equation 1: $x 1+0 x 2+0 x 3=-13$
Equation $2: 0 x 1+x 2+0 x 3=9$
Equation $3: 0 x 1+0 x 2+x 3=7$
Finally, solve the system. Use $\mathbf{x} 1, \mathbf{x} 2$, and $\mathbf{x} \mathbf{3}$ to enter the variables $x_{1}, x_{2}$, and $x_{3}$. If necessary, use $s 1, s 2$, etc to enter the free variables $s_{1}, s_{2}$, etc.
\[
\left(x_{1}, x_{2}, x_{3}\right)=(-13,9,7)
\]
Answer
Rewrite the system in echelon form and solve it: \(\begin{bmatrix} 7 & 7 & 4 \\ 0 & -1 & 1 \\ 0 & 0 & 2 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ -2 \\ 14 \end{bmatrix}\) which gives the solution \(\boxed{(x_{1}, x_{2}, x_{3}) = (-13, 9, 7)}\)
Step 1 ：Identify the reason the system is not in echelon form: The system is not in echelon form because it is not organized in a descending 'stair step' pattern so that the index of the leading variables increases from the top to bottom. \(\boxed{B}\)
Step 2 ：Rewrite the system in echelon form and solve it: \(\begin{bmatrix} 7 & 7 & 4 \\ 0 & -1 & 1 \\ 0 & 0 & 2 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ -2 \\ 14 \end{bmatrix}\) which gives the solution \(\boxed{(x_{1}, x_{2}, x_{3}) = (-13, 9, 7)}\)
