# The row echelon form of a system of linear equations is given. (a) Write the system of equations corresponding to the given matrix. Use x, y, or x, y,

The row echelon form of a system of linear equations is given.
(a) Write the system of equations corresponding to the given matrix.
Use x, y, or x, y, z, or ${x}_{1},{x}_{2},{x}_{3},{x}_{4}$
(b) Determine whether the system is consistent. If it is consistent, give the solution.
$\begin{array}{ccccc}1& 0& 3& 0& 1\\ 0& 1& 4& 3& 2\\ 0& 0& 1& 2& 3\\ 0& 0& 0& 0& 0\end{array}$
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Neelam Wainwright
a. The system represented by the matrix is
$\left\{\begin{array}{l}{x}_{1}+3{x}_{3}=1\\ {x}_{2}+4{x}_{3}+3{x}_{4}=2\\ {x}_{3}+2{x}_{4}=3\\ 0=0\end{array}$
b. This is a consistent system (because the last equation is true for all ordered quadruplets).
The system will not have a single solution, as it is, in effect, a system of THREE equations in FOUR variables.
It is a dependent system.
To solve, we take ${x}_{4}$ to be any real number $t\in \mathbb{R}$, and back-substitute into the other two equations:
Back substitute into (3): $\left[\begin{array}{c}{x}_{3}+2\left(t\right)=0\\ {x}_{3}=3-2t\end{array}\right]$
Back substitute into (2): $\left[\begin{array}{c}{x}_{2}+4\left(3-2t\right)+3\left(t\right)=2\\ {x}_{2}+12-8t+3t=2\\ {x}_{2}+12-5t=-10+5t\end{array}\right]$
Back substitute into (1): $\left[\begin{array}{c}{x}_{1}+3\left(3-2t\right)-2\left(t\right)=1\\ {x}_{1}+9-6t=1\\ {x}_{1}=-8+6t\end{array}\right]$
Result:
a. $\left\{\begin{array}{l}{x}_{1}+3{x}_{3}=1\\ {x}_{2}+4{x}_{3}+3{x}_{4}=2\\ {x}_{3}+2{x}_{4}=3\\ 0=0\end{array}$
b. Consistent, solution set: