Find the general solution to the system corresponding to the following matrix. \[ A=\left[\begin{array}{llll|l}1 & 7 & 0 & 0 & 2 \\ 0 & 0 & 1 & 3 & 4 \\ 0 & 0 & 0 & 0 & 0\end{array}\right] \] \( \left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3} \\ x_{4}\end{array}\right]=\left[\begin{array}{ll}\text { Ex: 1-2x_3 } \\ \square & \\ \square\end{array}\right] \) Enter \( 1-2 x_{3} \) as: \( 1-2 \times 3 \)

To extract the general solution from the provided augmented matrix, we rewrite the system of equations represented by the matrix. The first equation gives us \( x_1 + 7x_2 = 2 \), and the second equation reveals \( x_3 + 3x_4 = 4 \). The third row, being all zeros, indicates that we have a free variable situation. Let's express \( x_2 \) and \( x_4 \) in terms of the free variable \( x_3 \). From the first equation, we isolate \( x_1 \): \[ x_1 = 2 - 7x_2 \] Then, from the second equation, solving for \( x_4 \): \[ x_4 = \frac{4 - x_3}{3} \] The complete representation of the solution is incorporated with arbitrary parameters. We can express the general solution in terms of \( x_3 \) and a free variable indicating \( x_2 \): \[ \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = \begin{pmatrix} 2 - 7x_2 \\ x_2 \\ x_3 \\ \frac{4 - x_3}{3} \end{pmatrix} \] Thus, the full solution involving substitutions can be packaged neatly. To finalize: 1. \( x_1 = 2 - 7 \times x_2 \) 2. \( x_2 = x_2 \) (free variable) 3. \( x_3 = x_3 \) (free variable) 4. \( x_4 = \frac{4 - x_3}{3} \) Expressing \( x_1 \): \[ \left[ \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array} \right] = \left[ \begin{array}{c} 2 - 7x_{2} \\ x_{2} \\ x_{3} \\ \frac{4 - x_{3}}{3} \end{array} \right] \] This gives you the framework to define \( x_2 \) and \( x_3 \) freely!