Suppose that the augmented matrix of a system of linear
\[
\left[\begin{array}{ccc:c}
1 & 0 & -4 & \frac{23}{3} \\
3 & 1 & -10 & 27 \\
0 & 4 & 8 & 16
\end{array}\right] .
\]
Solve the system and provide the information requested.
The system has:
a unique solution
which is
\[
x=\square \quad y=\square \quad z=\square
\]
infinitely many solutions
two of which are
\[
\begin{array}{lll}
x=\square & y=\square & z=\square \\
x=\square & y=\square & z=\square
\end{array}
\]
no solution
\[ \begin{array}{lll} x=7.67 & y=4 & z=0 \ x=11.67 & y=2 & z=1 \end{array} \]
Step 1 :Suppose that the augmented matrix of a system of linear equations is given as follows: \[ \left[\begin{array}{ccc:c} 1 & 0 & -4 & \frac{23}{3} \ 3 & 1 & -10 & 27 \ 0 & 4 & 8 & 16 \end{array}\right] . \]
Step 2 :Let's define the equations based on the augmented matrix: \[ \begin{align*} x - 4z &= \frac{23}{3} \ 3x + y - 10z &= 27 \ 4y + 8z &= 16 \end{align*} \]
Step 3 :Solving the system of equations, we find that the solution is parametric in terms of \( z \). This means that there are infinitely many solutions to the system, and we can generate specific solutions by choosing specific values for \( z \).
Step 4 :For example, if we choose \( z = 0 \), we get \( x = 7.67 \) and \( y = 4 \). If we choose \( z = 1 \), we get \( x = 11.67 \) and \( y = 2 \).
Step 5 :\(\boxed{\text{Final Answer: The system has infinitely many solutions. Two of them are:}}\)
Step 6 :\[ \begin{array}{lll} x=7.67 & y=4 & z=0 \ x=11.67 & y=2 & z=1 \end{array} \]