Solution: Given that the system of linear equations is \begin{equation}\label{eq:23Mar2025-1} \left.\begin{aligned} x_1 - 2x_2 + x_3 + 2x_4 & = 1 \\ x_1 + x_2 - x_3 + x_4 & = 2 \\ x_1 + 7x_2 - 5x_3 - x_4 & = 3 \end{aligned}\right\} \end{equation} We will show that the above system has no solution. For that, let us consider the augmented matrix of the system \eqref{eq:23Mar2025-1}. \begin{align*} \begin{bmatrix} 1 & -2 & 1 & 2 & 1 \\ 1 & 1 & -1 & 1 & 2 \\ 1 & 7 & -5 & -1 & 3\\ \end{bmatrix}. \end{align*} We will now perform row operations to reduce the augmented matrix to row echelon form and analyze the resulting system. \begin{align*} \begin{bmatrix} 1 & -2 & 1 & 2 & 1 \\ 1 & 1 & -1 & 1 & 2 \\ 1 & 7 & -5 & -1 & 3 \end{bmatrix} & \xrightarrow{ \begin{matrix} R_2 \rightarrow R_2 - R_1 \\ R_3 \rightarrow R_3 - R_1 \end{matrix} } \begin{bmatrix} 1 & -2 & 1 & 2 & 1 \\ 0 & 3 & -2 & -1 & 1 \\ 0 & 9 & -6 & -3 & 2 \end{bmatrix} \\[2ex] & \xrightarrow{ \begin{matrix} R_3 \rightarrow R_3 - 3R_2 \end{matrix} } \begin{bmatrix} 1 & -2 & 1 & 2 & 1 \\ 0 & 3 & -2 & -1 & 1 \\ 0 & 0 & 0 & 0 & -1 \\ \end{bmatrix} \\[2ex] & \xrightarrow{ \begin{matrix} R_2 \rightarrow \frac{1}{3}R_2 \end{matrix} } \begin{bmatrix} 1 & -2 & 1 & 2 & 1 \\ 0 & 1 & -\frac{2}{3} & -\frac{1}{3} & \frac{1}{3} \\ 0 & 0 & 0 & 0 & -1 \end{bmatrix}. \end{align*} Since the first nonzero entry in the bottom row of the last matrix is in the rightmost column, the corresponding system of equations has no solution. Therefore, the system \eqref{eq:23Mar2025-1} has no solution.