# Consider the following linear system \left\{ \begin{array} x_1 + 2x_2 + 5x_4 = 1 + x_3 \\ 3x_2 -...

## Question:

Consider the following linear system

{eq}\left\{ \begin{array}{c} x_1 + 2x_2 + 5x_4 = 1 + x_3 \\ 3x_2 - 2x_3 + 6x_4 = 3 - 2x_1 \\ x_1 + x_2 - x_3 = k - x_4 \end{array} \right.{/eq} where {eq}k{/eq} is a constant.

(a) Row reduce the augmented matrix to row echelon form. Find the {eq}k{/eq} value such that the system is consistent.

(b) Find the general solution when {eq}k{/eq} is the special value in (a).

## System of Equations:

The given system of equations in a matrix form is a of order {eq}3\times 3 {/eq} . To find the solution of the system of equations 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 ] {/eq}

This implies that the system is inconsistence and it has no solution.

{eq}\text{Rank}\left [ A:b \right ]= \text{Rank}\left [ A \right ] {/eq}= Number of variables

This implies that the system is consistence and it has a unique solution.

Consider the system of equations

{eq}x_1 + 2x_2 + 5x_4 = 1 + x_3 \\ 3x_2 - 2x_3 + 6x_4 = 3 - 2x_1 \\ x_1 + x_2 - x_3 = k - x_4 \\ \text{Or}\\ x_1 + 2x_2- x_3 + 5x_4 = 1\\ 2x_1 +3x_2 - 2x_3 + 6x_4 = 3 \\ x_1 + x_2 - x_3+x_4 = k {/eq}

Rewrite the system of equations in matrix form {eq}\, \, Ax=b {/eq}

{eq}\begin{bmatrix} 1&2&-1 &5\\ 2&3&-2&6\\ 1&1&-1 &1 \end{bmatrix} \begin{bmatrix} x_1 \\x_2 \\x_3\\x_4\end{bmatrix}=\begin{bmatrix}1\\3 \\k \end{bmatrix} {/eq}

The augmented matrix is as follows

{eq}\left [ A:b \right ]=\begin{bmatrix} 1&2&-1 &5&:&1\\ 2&3&-2&6&:&3\\ 1&1&-1 &1&:&k \end{bmatrix} {/eq}

Convert in row echlon form

{eq}R_2\:\leftarrow \:R_2-\frac{1}{2}\cdot \:R_1\\ R_3\:\leftarrow \:R_3-\frac{1}{2}\cdot \:R_1\\ =\begin{bmatrix}2&3&-2&6&:&3\\ 0&\frac{1}{2}&0&2&:&-\frac{1}{2}\\ 0&-\frac{1}{2}&0&-2&:&\frac{2k-3}{2}\end{bmatrix}\\ R_3\:\leftarrow \:R_3+1\cdot \:R_2\\ =\begin{bmatrix}2&3&-2&6&:&3\\ 0&\frac{1}{2}&0&2&:&-\frac{1}{2}\\ 0&0&0&0&:&k-2\end{bmatrix} {/eq}

The system of equations will be consistent when

{eq}\text{Rank}\left [ A:b \right ]= \text{Rank}\left [ A \right ]\\ \Rightarrow k-2= 0\Rightarrow k=2 {/eq}

(b)

{eq}\text{When,}\: \: k=2,\\ =\begin{bmatrix}2&3&-2&6&:&3\\ 0&\frac{1}{2}&0&2&:&-\frac{1}{2}\\ 0&0&0&0&:&0\end{bmatrix}\\ \text{If} \: \text{Rank}\left [ A:b \right ]= \text{Rank}\left [ A \right ]=2< \text{Number of variables} {/eq}

Then the system of equations has infinite number of solution.

The last matrix represents the equivalent system

{eq}x_1 + 3x_2- 2x_3 + 6x_4 = 3\\ x_2 + 4x_4 = -1 {/eq}

Here, two equations and four variable. So, 4-2=2 variable will be a free variable. it can take any values

Let us assume that

{eq}x_3=s,\: \:x_4=t \\ x_2 + 4x_4 = -1\Rightarrow x_2 =-1-4x_4 \Rightarrow x_2 =-1-4t\\ x_1 + 3x_2- 2x_3 + 6x_4 = 3\Rightarrow x_1 = 3-3x_2+ 2x_3 - 6x_4\\ \Rightarrow x_1 = 3-3(-1-4t)+ 2s - 6t\\ \Rightarrow x_1 =6t+2s+6 {/eq}

Therefore,

{eq}x_1 =6t+2s+6,\; \; x_2=-4t-1,\; \; x_3=s,\; \; x_4=t,\quad (t\in\mathbb{R})\\ \begin{bmatrix} x_1 \\x_2 \\x_3\\x_4\end{bmatrix}=\begin{bmatrix} 2 \\0 \\1\\0\end{bmatrix}s+\begin{bmatrix} 6 \\-4\\0 \\1 \end{bmatrix}t+\begin{bmatrix} 6\\-1 \\0\\0\end{bmatrix} {/eq} 