Answer
Consistent
Solution set: $\left\{\left(\dfrac{1}{3},\dfrac{2}{3},1\right)\right\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
3x+y-z=\frac{2}{3}\\
2x-y+z=1\\
4x+2y=\frac{8}{3}
\end{cases}$
Write the augmented matrix:
$\begin{bmatrix}
3&1&-1&|&\frac{2}{3}\\2&-1&1&|&1\\4&2&0&|&\frac{8}{3}\end{bmatrix}$
Perform row operations to bring the matrix to the reduced row echelon form:
$R_1=-r_2+r_1$
$\begin{bmatrix}
1&2&-2&|&-\frac{1}{3}\\2&-1&1&|&1\\4&2&0&|&\frac{8}{3}\end{bmatrix}$
$R_2=-2r_1+r_2$
$\begin{bmatrix}1&2&-2&|&-\frac{1}{3}\\0&-5&5&|&\frac{5}{3}\\4&2&0&|&\frac{8}{3}\end{bmatrix}$
$R_3=-4r_1+r_3$
$\begin{bmatrix}1&2&-2&|&-\frac{1}{3}\\0&-5&5&|&\frac{5}{3}\\0&-6&8&|&4\end{bmatrix}$
$R_2=-r_3+r_2$
$\begin{bmatrix}1&2&-2&|&-\frac{1}{3}\\0&1&-3&|&-\frac{7}{3}\\0&-6&8&|&4\end{bmatrix}$
$R_3=6r_2+r_3$
$\begin{bmatrix}1&2&-2&|&-\frac{1}{3}\\0&1&-3&|&-\frac{7}{3}\\0&0&-10&|&-10\end{bmatrix}$
$R_3=-\dfrac{1}{10}r_3$
$\begin{bmatrix}1&2&-2&|&-\frac{1}{3}\\0&1&-3&|&-\frac{7}{3}\\0&0&1&|&1\end{bmatrix}$
Write the corresponding system of equations:
$\begin{cases}
x+2y-2z=-\frac{1}{3}\\
y-3z=-\frac{7}{3}\\
z=1
\end{cases}$
Solve the system by back substitution:
$z=1$
$y-3z=-\dfrac{7}{3}$
$y-3(1)=-\dfrac{7}{3}$
$y=\dfrac{2}{3}$
$x+2y-2z=-\frac{1}{3}$
$x+2\left(\dfrac{2}{3}\right)-2(1)=-\dfrac{1}{3}$
$x+\dfrac{4}{3}-2=-\dfrac{1}{3}$
$x=\dfrac{1}{3}$
The system is consistent. The solution is:
$\left\{\left(\dfrac{1}{3},\dfrac{2}{3},1\right)\right\}$
