Write the augmented matrix corresponding to the system of equations in reduced row-echelon form....

{eq}\text{Consider the system of equations }\\ x + y + z + w = 0 \\ 2x + 3y + z - 2w = 0 \\ 3x + 5y + z = 0\\ \text{This is equivalent to the matrix equation}\\ \begin{bmatrix} 1 & 1 &1&1\\ 2& 3 &1&-2 \\ 3& 5 & 1&0 \end{bmatrix}\begin{bmatrix} x\\ y\\ z\\ w \end{bmatrix}=0\\ \text{To find the solution, consider the augmented matrix.}\\ \begin{bmatrix} 1 & 1 &1&1&:&0 \\ 2& 3 &1&-2&:&0 \\ 3& 5 & 1&0&:&0 \end{bmatrix}\\ \text{Applying elementary row operations, we obtain}\\ R_2\rightarrow R_2-2R_1\\ R_3\rightarrow R_3-3R_1\\ \sim \begin{bmatrix} 1 & 1 &1&1&:&0 \\ 0& 1 &-1&-4&:&0 \\ 0&2 & -2&-3&:&0 \end{bmatrix}\\ R_3\rightarrow R_3-2R_2\\ \sim \begin{bmatrix} 1 & 1 &1&1&:&0 \\ 0& 1 &-1&-4&:&0 \\ 0&0 & 0&5&:&0 \end{bmatrix}\\ \text{Rank}\left [ A:b \right ]=3\\ \text{Rank}\left [ A \right ]=3\\ \text{Rank}\left [ A:b \right ]= \text{Rank}\left [ A \right ]= 3< 4\\ {\color{Red} {\text{This implies that the system is consistent and it has infinite number of solutions.}}}\\ \text{Corresponding system of equations are as follows}\\ x + y + z + w = 0 \\ y - z - 4w = 0 \\ w = 0\\ y - z - 4w = 0\Rightarrow y - z = 0\Rightarrow y=z\\ \text{Here, three equations and four variable}\\ \text{So, one variable will be a free variable. it can take any values}\\ \text{Let us assume that}\\ z=t\\ \Rightarrow y=t\\ x + y + z + w = 0\Rightarrow x =-2t\\ \text{Therefore,}\\ x=-2t,\; \; y=t,\; z=t,\; w=0\\ \textbf{Hence,}\\ \boxed{ {\color{Red} {\begin{bmatrix} x\\y\\z\\w \end{bmatrix}=\begin{bmatrix} -2\\ 1\\1\\ 0 \end{bmatrix}t}}} {/eq}