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

## Question:

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

{eq}x + y + z + w = 0 \\ 2x + 3y + z - 2w = 0 \\ 3x + 5y + z = 0 {/eq}

## Find the general solution of system of equations:

The given matrix is a of order 3x3. First we find the augmented matrix in row echlon form by apply elementary row operations. Elementary row operations on an augmented matrix never change the solution set of the associated linear system. then we find the rank of the matrix and augmented matrix if

{eq}\text{Rank}\left [ A:b \right ]\neq \text{Rank}\left [ A \right ]\\ \text{This implies that the system is inconsistence and it has no solution.}\\ \text{Rank}\left [ A:b \right ]= \text{Rank}\left [ A \right ]\\ \text{This implies that the system is consistence and it has a solution.} {/eq}

## Answer and Explanation: 1

{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}

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Chapter 8 / Lesson 8Learn what a consistent system of equations is and see how consistent and inconsistent systems differ from each other. See how consistent systems are solved.