Find the general solution to the system corresponding to the following matrix. \[ A=\left[\begin{array}{llll|l}1 & 5 & 0 & 0 & 4 \\ 0 & 0 & 1 & 2 & 6 \\ 0 & 0 & 0 & 0 & 0\end{array}\right] \] \( \left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3} \\ x_{4}\end{array}\right]=\left[\begin{array}{ll}\text { Ex: 1-2x_3 } \\ \square & \\ \square\end{array}\right] \) Enter \( 1-2 x_{3} \) as: \( 1-2 \times 3 \)

To find the general solution to the system represented by the augmented matrix, we will first interpret the matrix equations. From matrix \( A \), we can break it down into the equations: 1. \( x_1 + 5x_2 + 0x_3 + 0x_4 = 4 \) or simply \( x_1 + 5x_2 = 4 \) 2. \( x_3 + 2x_4 = 6 \) The third row indicates that there are no further constraints (0 = 0). Now, let's express \( x_1 \) and \( x_3 \) in terms of the free variables \( x_2 \) and \( x_4 \): From the first equation, solving for \( x_1 \): \[ x_1 = 4 - 5x_2 \] From the second equation, solving for \( x_3 \): \[ x_3 = 6 - 2x_4 \] Thus, we can summarize our solutions with \( x_2 = t \) and \( x_4 = s \) (where \( t \) and \( s \) are free parameters): \[ \left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3} \\ x_{4}\end{array}\right]=\left[\begin{array}{l}4 - 5t \\ t \\ 6 - 2s \\ s\end{array}\right] \] Now filling in the blanks: \[ \left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3} \\ x_{4}\end{array}\right]=\left[\begin{array}{l}4 - 5 t \\ t \\ 6 - 2 s \\ s\end{array}\right] \] Did you know? The concept of solving systems of linear equations dates back to ancient civilizations, with techniques used by Babylonian mathematicians over 3,000 years ago. They tackled problems with methods surprisingly similar to modern algebra! For a real-world application, these systems find use in various fields such as economics for supply and demand modeling, engineering for circuit analysis, and computer science in algorithm complexity, where multiple conditions need to be satisfied simultaneously for successful outcomes.