Answer
Consistent
Solutions set: $\left\{\left(\frac{56}{13},-\frac{7}{13},\frac{35}{13}\right)\right\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
2x+y-3z=0\\
-2x+2y+z=-7\\
3x-4y-3z=7
\end{cases}$
Write the augmented matrix:
$\begin{bmatrix}
2&1&-3&|&0\\-2&2&1&|&-7\\3&-4&-3&|&7\end{bmatrix}$
Perform row operations to bring the matrix to the reduced row echelon form:
$R_2=r_1+r_2$
$\begin{bmatrix}
2&1&-3&|&0\\0&3&-2&|&-7\\3&-4&-3&|&7\end{bmatrix}$
$R_1=r_3-r_1$
$\begin{bmatrix}1&-5&0&|&7\\0&3&-2&|&-7\\3&-4&-3&|&7\end{bmatrix}$
$R_3=-3r_1+r_3$
$\begin{bmatrix}1&-5&0&|&7\\0&3&-2&|&-7\\0&11&-3&|&-14\end{bmatrix}$
$R_2=\dfrac{1}{3}r_2$
$\begin{bmatrix}1&-5&0&|&7\\0&1&-\frac{2}{3}&|&-\frac{7}{3}\\0&11&-3&|&-14\end{bmatrix}$
$R_1=5r_2+r_1$
$\begin{bmatrix}1&0&-\frac{10}{3}&|&-\frac{14}{3}\\0&1&-\frac{2}{3}&|&-\frac{7}{3}\\0&11&-3&|&-14\end{bmatrix}$
$R_3=-11r_2+r_3$
$\begin{bmatrix}1&0&-\frac{10}{3}&|&-\frac{14}{3}\\0&1&-\frac{2}{3}&|&-\frac{7}{3}\\0&0&\frac{13}{3}&|&\frac{35}{3}\end{bmatrix}$
$R_3=\dfrac{3}{13}r_3$
$\begin{bmatrix}1&0&-\frac{10}{3}&|&-\frac{14}{3}\\0&1&-\frac{2}{3}&|&-\frac{7}{3}\\0&0&1&|&\frac{35}{13}\end{bmatrix}$
$R_1=\dfrac{10}{3}r_3+r_1$
$\begin{bmatrix}1&0&0&|&\frac{56}{13}\\0&1&-\frac{2}{3}&|&-\frac{7}{3}\\0&0&1&|&\frac{35}{13}\end{bmatrix}$
$R_2=\dfrac{2}{3}r_3+r_2$
$\begin{bmatrix}1&0&0&|&\frac{56}{13}\\0&1&0&|&-\frac{7}{13}\\0&0&1&|&\frac{35}{13}\end{bmatrix}$
Write the corresponding system of equations:
$\begin{cases}
1x+0y+0z=\frac{56}{13}\\
0x+1y+0z=-\frac{7}{13}\\
0x+0y+1z=\frac{35}{13}
\end{cases}$
$\begin{cases}
x=\frac{56}{13}\\
y=-\frac{7}{13}\\
z=\frac{35}{13}
\end{cases}$
The system is consistent. The solution set is:
$\left\{\left(\frac{56}{13},-\frac{7}{13},\frac{35}{13}\right)\right\}$
