1. **State the problem:** We are given matrix $A = \begin{bmatrix}1 & 7 & 6 \\ 5 & 6 & 1\end{bmatrix}$ which corresponds to a system of linear equations. We need to write the system of equations from $A$ and then use the reduced row echelon form $rref(A) = \begin{bmatrix}1 & 0 & -1 \\ 0 & 1 & 1\end{bmatrix}$ to find all solutions. 2. **Write the system from matrix $A$:** Matrix $A$ has 2 rows and 3 columns, so it corresponds to 2 equations with 3 variables $x, y, z$. The first row corresponds to: $$1x + 7y + 6z = 0$$ The second row corresponds to: $$5x + 6y + 1z = 0$$ So the system is: $$\begin{cases} x + 7y + 6z = 0 \\ 5x + 6y + z = 0 \end{cases}$$ 3. **Use $rref(A)$ to find solutions:** The $rref(A)$ matrix corresponds to: $$\begin{cases} x + 0y - 1z = 0 \\ 0x + y + 1z = 0 \end{cases}$$ Which simplifies to: $$\begin{cases} x - z = 0 \\ y + z = 0 \end{cases}$$ 4. **Express variables in terms of free variable $z$:** From the first equation: $$x = z$$ From the second equation: $$y = -z$$ 5. **Write the general solution:** Let $z = t$ (a free parameter), then: $$x = t, \quad y = -t, \quad z = t$$ **Final answer:** The system of equations from $A$ is: $$x + 7y + 6z = 0$$ $$5x + 6y + z = 0$$ The solution set is: $$\{(x,y,z) \mid x = t, y = -t, z = t, t \in \mathbb{R}\}$$