1. The problem involves understanding the matrix $A = \begin{bmatrix}1 & 7 & 6 \\ 5 & 6 & 1\end{bmatrix}$ and its reduced row echelon form (rref) $\begin{bmatrix}1 & 0 & -1 \\ 0 & 1 & 1\end{bmatrix}$. We want to interpret the rref in terms of the variables $x$, $y$, and $z$. 2. The original matrix $A$ can be seen as the coefficients of a system of linear equations: $$\begin{cases} x + 7y + 6z = ? \\ 5x + 6y + z = ? \end{cases}$$ The right side is not given explicitly but can be inferred from the context. 3. The rref matrix corresponds to the system: $$\begin{cases} x + 0y - z = 0 \\ 0x + y + z = 0 \end{cases}$$ which simplifies to: $$x = z$$ $$y = -z$$ 4. From the rref, $x$ and $y$ are expressed in terms of $z$. This means $z$ is a free variable. 5. Now, let's analyze the given equations: - B. $6z = 1$ implies $z = \frac{1}{6}$. - D. $x + 7y + 6z = 0$ - F. $1 = 7x + 6y$ - H. $x + 7y = 6$ - J. $x = 6$ - G. $5x + 6y + z = 0$ - I. $7y = 6$ - A. $x + 7y + 6z = 5x + 6y + z$ - C. $5x + 6y = 1$ - E. $5 = 6x + y$ 6. Using the rref relations $x = z$ and $y = -z$, substitute into B: $$6z = 1 \implies z = \frac{1}{6}$$ 7. Then: $$x = z = \frac{1}{6}$$ $$y = -z = -\frac{1}{6}$$ 8. Check equation D: $$x + 7y + 6z = \frac{1}{6} + 7\left(-\frac{1}{6}\right) + 6\left(\frac{1}{6}\right) = \frac{1}{6} - \frac{7}{6} + 1 = \frac{1 - 7 + 6}{6} = 0$$ 9. Check equation G: $$5x + 6y + z = 5\left(\frac{1}{6}\right) + 6\left(-\frac{1}{6}\right) + \frac{1}{6} = \frac{5}{6} - 1 + \frac{1}{6} = 0$$ 10. Check equation A: $$x + 7y + 6z = 5x + 6y + z$$ Substitute values: $$\frac{1}{6} + 7\left(-\frac{1}{6}\right) + 6\left(\frac{1}{6}\right) = 5\left(\frac{1}{6}\right) + 6\left(-\frac{1}{6}\right) + \frac{1}{6}$$ $$0 = 0$$ 11. All these checks confirm the solution consistent with the rref. Final answer: $$x = \frac{1}{6}, \quad y = -\frac{1}{6}, \quad z = \frac{1}{6}$$