Answer
$\{(1,-2,0)\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
x+2y+4z=-3\\
2x+7y+15z=-12\\
4x+7y+13z=-10
\end{cases}$
The augmented matrix associated with the system is:
$A=\begin{bmatrix}1&2&4&|&-3\\2&7&15&|&-12\\4&7&13&|&-10\end{bmatrix}$
Multiply $R_1$ by -2 and add it to $R_2$:
$\begin{bmatrix}1&2&4&|&-3\\0&3&7&|&-6\\4&7&13&|&-10\end{bmatrix}$
Multiply $R_1$ by -4 and add it to $R_3$:
$\begin{bmatrix}1&2&4&|&-3\\0&3&7&|&-6\\0&-1&-3&|&2\end{bmatrix}$
Multiply $R_3$ by -1 and interchange it with $R_2$:
$\begin{bmatrix}1&2&4&|&-3\\0&1&3&|&-2\\0&3&7&|&-6\end{bmatrix}$
Multiply $R_2$ by -3 and add it to $R_3$:
$\begin{bmatrix}1&2&4&|&-3\\0&1&3&|&-2\\0&0&-2&|&0\end{bmatrix}$
Multiply $R_3$ by $-\dfrac{1}{2}$:
$\begin{bmatrix}1&2&4&|&-3\\0&1&3&|&-2\\0&0&1&|&0\end{bmatrix}$
Solve the system:
$\begin{cases}
x+2y+4z=-3\\
y+3z=-2\\
z=0
\end{cases}$
$z=0$
$y+3(0)=-2$
$y=-2$
$x+2(-2)+4(0)=-3$
$x-4=-3$
$x=1$
The solution set is:
$\{(1,-2,0)\}$
