Find the general solution to the system corresponding to the following matrix. \[ A=\left[\begin{array}{llll|l}1 & 0 & 0 & 6 & 2 \\ 0 & 1 & 0 & 7 & 0 \\ 0 & 0 & 1 & 3 & 4 \\ 0 & 0 & 0 & 0 & 0\end{array}\right] \] \( \left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3} \\ x_{4}\end{array}\right]=\left[\begin{array}{lll}\text { Ex: 1-2x_3 } \\ \square & \\ \square\end{array}\right] \) Enter \( 1-2 x_{3} \) as: \( 1-2 \times 3 \)

The system described by the given matrix represents a series of linear equations derived from the reduced row echelon form. Let's rewrite the equations based on each row: 1. \( x_1 + 6x_4 = 2 \) 2. \( x_2 + 7x_4 = 0 \) 3. \( x_3 + 3x_4 = 4 \) Now, let's express the variables in terms of the free variable \( x_4 \). From the second equation: \[ x_2 = -7x_4 \] From the third equation: \[ x_3 = 4 - 3x_4 \] Substituting \( x_3 \) into the first equation gives: \[ x_1 = 1 - 2(4 - 3x_4) - 6x_4 \] \[ x_1 = 1 - 8 + 6x_4 - 6x_4 \] \[ x_1 = 1 - 8 \] So, we summarize the general solution as follows: \[ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} -7 \\ -7x_4 \\ 4 - 3x_4 \\ x_4 \end{bmatrix} \] Thus, the complete expression in terms of \( x_4 \) becomes: \[ \begin{bmatrix} 1 - 2 \times (4 - 3x_4) \\ -7x_4 \\ 4 - 3x_4 \\ x_4 \end{bmatrix} \]