College Algebra, Chapter 7, 7.1, Section 7.1, Problem 36

State whether the system of linear equations $\left\{ \begin{equation}
\begin{aligned}

3r + 2s - 3t =& 10
\\
r - s - t =& -5
\\
r + 4s - t =& 20

\end{aligned}
\end{equation} \right.$ is inconsistent or dependent. If it is dependent, find the complete solution.

We transform the system into row-echelon form.

$\left[ \begin{array}{cccc}
3 & 2 & -3 & 10 \\
1 & -1 & -1 & -5 \\
1 & 4 & -1 & 20
\end{array} \right]$

$\displaystyle \frac{1}{3} R_1$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{2}{3} & -1 & \displaystyle \frac{10}{3} \\
1 & -1 & -1 & -5 \\
1 & 4 & -1 & 20
\end{array} \right]$

$R_3 - R_1 \to R_3$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{2}{3} & -1 & \displaystyle \frac{10}{3} \\
1 & -1 & -1 & -5 \\
0 & \displaystyle \frac{10}{3} & 0 & \displaystyle \frac{50}{3}
\end{array} \right]$

$\displaystyle \frac{3}{10} R_3$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{2}{3} & -1 & \displaystyle \frac{10}{3} \\
1 & -1 & -1 & -5 \\
0 & 1 & 0 & 5
\end{array} \right]$

$\displaystyle R_2 - R_1 \to R_2$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{2}{3} & -1 & \displaystyle \frac{10}{3} \\
0 & \displaystyle \frac{-5}{3} & 0 & \displaystyle \frac{-25}{3} \\
0 & 1 & 0 & 5
\end{array} \right]$

$\displaystyle \frac{-3}{5} R_2$

$\left[ \begin{array}{cccc}
1 & \displaystyle \frac{2}{3} & -1 & \displaystyle \frac{10}{3} \\
0 & 1 & 0 & 5 \\
0 & 1 & 0 & 5
\end{array} \right]$

The matrix has infinitely many solutions to obtain the complete solution, we let $t$ represent any real number, we expresses $x$ and $z$ in terms of $t$.


$
\begin{equation}
\begin{aligned}

x + \frac{2}{3} y - t =& \frac{10}{3}
\\
\\
x =& t - \frac{2}{3} (5) + \frac{10}{3}
\\
\\
x =& t
\\
\\
y =& 5
\\
\\
z =& t

\end{aligned}
\end{equation}
$


We can also write the solution as the ordered triple $(t, 5, t)$.