Answer
Consistent
Solution set: $\left\{\left(3,-\dfrac{8}{3},\dfrac{1}{9}\right)\right\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
x+4y-3z=-8\\
3x-y+3z=12\\
x+y+6z=1
\end{cases}$
Write the augmented matrix:
$\begin{bmatrix}
1&4&-3&|&-8\\3&-1&3&|&12\\1&1&6&|&1\end{bmatrix}$
Perform row operations to bring the matrix to the reduced row echelon form:
$R_2=-3r_1+r_2$
$\begin{bmatrix}
1&4&-3&|&-8\\0&-13&12&|&36\\1&1&6&|&1\end{bmatrix}$
$R_3=-r_1+r_3$
$\begin{bmatrix}1&4&-3&|&-8\\0&-13&12&|&36\\0&-3&9&|&9\end{bmatrix}$
$R_2=-4r_3+r_2$
$\begin{bmatrix}1&4&-3&|&-8\\0&-1&-24&|&0\\0&-3&9&|&9\end{bmatrix}$
$R_2=-r_2$
$\begin{bmatrix}1&4&-3&|&-8\\0&1&24&|&0\\0&-3&9&|&9\end{bmatrix}$
$R_3=3r_2+r_3$
$\begin{bmatrix}1&4&-3&|&-8\\0&1&24&|&0\\0&0&81&|&9\end{bmatrix}$
$R_3=\dfrac{1}{81}r_3$
$\begin{bmatrix}1&4&-3&|&-8\\0&1&24&|&0\\0&0&1&|&\dfrac{1}{9}\end{bmatrix}$
Write the corresponding system of equations:
$\begin{cases}
x+4y-3z=-8\\
y+24z=0\\
z=\frac{1}{9}
\end{cases}$
Solve the system by back substitution:
$z=\frac{1}{9}$
$y+24z=0$
$y+24\left(\frac{1}{9}\right)=0$
$y=-\frac{8}{3}$
$x+4y-3z=-8$
$x+4\left(-\dfrac{8}{3}\right)-3\left(\dfrac{1}{9}\right)=-8$
$x-\dfrac{32}{3}-\dfrac{1}{3}=-8$
$x-11=-8$
$x=3$
The system is consistent. The solution is:
$\left\{\left(3,-\dfrac{8}{3},\dfrac{1}{9}\right)\right\}$
