# Given the system of linear equations: \begin{align*} -x_1 -7x_2 + 9x_3 &= -1\\ 2x_1 + x_2 - x_3...

## Question:

Given the system of linear equations:

{eq}\begin{align*} -x_1 -7x_2 + 9x_3 &= -1\\ 2x_1 + x_2 - x_3 &= 1 \\ 3x_1 -5x_2 +7x_3 &= -1 \end{align*} {/eq}

(a) Write it in the form of Ax = b:

(b) Use Guass elimination method to find all solutions of the system of equations.

(c) Find the rank of A

(d) Find {eq}A^{-1} {/eq}

## 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.

## Answer and Explanation:

Consider the system of equations

{eq}\displaystyle -x_1 -7x_2 + 9x_3 = -1\\ \displaystyle 2x_1 + x_2 - x_3 = 1 \\ \displaystyle 3x_1 -5x_2 +7x_3 = -1 {/eq}

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

{eq}\displaystyle \begin{bmatrix} -1&-7&9 \\ 2&1&-1\\ 3&-5&7 \end{bmatrix} \begin{bmatrix} x_{1}\\x_{2} \\x_{3} \end{bmatrix}=\begin{bmatrix}-1\\1\\ -1\end{bmatrix} {/eq}

The agumented matrix is as follows

{eq}\displaystyle \left [ A:b \right ]=\begin{bmatrix} -1&-7&9&:&-1 \\ 2&1&-1&:&1 \\ 3&-5&7&:&-1 \end{bmatrix} {/eq}

Convert in row echlon form

{eq}\displaystyle R_1\:\leftrightarrow \:R_3\\ \displaystyle =\begin{bmatrix}3&-5&7&:&-1\\ 2&1&-1&:&1\\ -1&-7&9&:&-1\end{bmatrix}\\ \displaystyle R_2\:\leftarrow \:R_2-\frac{2}{3}\cdot \:R_1\\ \displaystyle R_3\:\leftarrow \:R_3+\frac{1}{3}\cdot \:R_1\\ \displaystyle =\begin{bmatrix}3&-5&7&:&-1\\ 0&\frac{13}{3}&-\frac{17}{3}&:&\frac{5}{3}\\ 0&-\frac{26}{3}&\frac{34}{3}&:&-\frac{4}{3}\end{bmatrix}\\ \displaystyle R_2\:\leftrightarrow \:R_3\\ \displaystyle =\begin{bmatrix}3&-5&7&:&-1\\ 0&-\frac{26}{3}&\frac{34}{3}&:&-\frac{4}{3}\\ 0&\frac{13}{3}&-\frac{17}{3}&:&\frac{5}{3}\end{bmatrix}\\ \displaystyle R_3\:\leftarrow \:R_3+\frac{1}{2}\cdot \:R_2\\ \displaystyle =\begin{bmatrix}3&-5&7&:&-1\\ 0&-\frac{26}{3}&\frac{34}{3}&:&-\frac{4}{3}\\ 0&0&0&:&1\end{bmatrix} {/eq}

The rank is number of non zero rows of the matrix. Therefore,

{eq}\displaystyle \text{Rank}\left [ A \right ]=2< 3 {/eq}

Hence, the inverse inverse of the matrix A does not exist.

{eq}\displaystyle \text{Rank}\left [ A:b \right ]=3\\ \displaystyle \text{If} \: \text{Rank}\left [ A:b \right ]\neq \text{Rank}\left [ A \right ] {/eq}

Then the system of equations has no solution.

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from High School Algebra II: Homework Help Resource

Chapter 8 / Lesson 9