# For the following system, find a solution using Gauss elimination: 4x+y+2z= 19 4x y+4z= 17...

## Question:

For the following system, find a solution using Gauss elimination:

{eq}4x+y+2z= 19 \\ 4x y+4z= 17 \\ 3x+3y+4z=15 {/eq}

## System of Linear Equations:

{eq}\\ {/eq}

For a linear equation, the degree of all the variables in the equation is one.

A linear equation in three variables: x, y and z, represents a plane and a system of linear equations represents a set of planes that may (or may not) intersect at a point to give a unique solution.

It is convenient to represent a system of equations in its matrix form: {eq}A \textbf{x} = B {/eq}, where {eq}A {/eq} is the coefficient matrix and {eq}B {/eq} is the force matrix.

The matrix equation can be solved by performing row-column transformations using methods like Gauss elimination or Gauss Jordan.

{eq}\\ {/eq}

{eq}\begin{align*} 4x+y+2z &= 19 && \dots (1) \\ 4x+y+4z &= 17 && \dots (2) \\ 3x+3y+4z &= 15 && \dots (3) \end{align*} {/eq}

Equations (1), (2) and (3) form a sytem of linear equations in x, y and z and they can be expressed in the form: {eq}A \textbf{x} = B {/eq}, where {eq}A {/eq} is the matrix of all the coefficients of x, y and z:

{eq}\begin{align*} A = \begin{bmatrix} 4 & 1 & 2 \\ 4 & 1 & 4 \\ 3 & 3 & 4 \end{bmatrix} \end{align*} {/eq}

{eq}\textbf{x} {/eq} is the variable matrix: {eq}[x,y,z]^T {/eq}, and {eq}B {/eq} is formed by the terms on the right hand side of the equations:

{eq}\begin{align*} B = \begin{bmatrix} 19 \\ 17\\ 15 \end{bmatrix} \end{align*} {/eq}

The problem in matrix form is presented as:

{eq}\begin{align*} \begin{bmatrix} 4 & 1 & 2 \\ 4 & 1 & 4 \\ 3 & 3 & 4 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 19 \\ 17 \\ 15 \end{bmatrix} \end{align*} {/eq}

The Gauss elimination method used to solve the system of linear equations tries to reduce the coefficient matrix into a triangular form by performing row operations on the augmented matrix {eq}[A | B] {/eq}:

{eq}\begin{align*} [A | B] &=\begin{bmatrix} 4 & 1 & 2 & | & 19 \\ 4 & 1 & 4 & | & 17\\ 3 & 3 & 4 & | & 15 \end{bmatrix} \end{align*} {/eq}

The first element of the first row of {eq}[A|B] {/eq} is called the pivot element and it is used in row operations to reduce the elements below it to zero.

{eq}R_2 = R_2 - R_1 {/eq}: Subtract the first row {eq}R_1 {/eq} from the second row {eq}R_2 {/eq}:

{eq}\begin{align*} [A | B] &=\begin{bmatrix} 4 & 1 & 2 & | & 19 \\ 0 & 0 & 2 & | & -2 \\ 3 & 3 & 4 & | & 15 \end{bmatrix} \\ \\ &[R_3 \Leftrightarrow R_2 : \text{Interchange } R_2 \ \& \ R_3 ] \\ \\ [A | B] &=\begin{bmatrix} 4 & 1 & 2 & | & 19 \\ 3 & 3 & 4 & | & 15 \\ 0 & 0 & 2 & | & -2 \end{bmatrix} \\ \\ &[R_2 = R_2 - \frac{3}{4}R_1 ] \\ \\ [A | B] &=\begin{bmatrix} 4 & 1 & 2 & | & 19 \\ 0 & 2.25 & 2.5 & | & 0.75\\ 0 & 0 & 2 & | & -2 \end{bmatrix} \\ \end{align*} {/eq}

{eq}A {/eq} has been reduced to an upper triangular form:

{eq}\begin{align*} A = \begin{bmatrix} 4 & 1 & 2 \\ 0 & 2.25 & 2.5 \\ 0 & 0 & 2 \end{bmatrix} \end{align*} {/eq}

so we can stop the process and rewrite the system in the reduced form:

{eq}\begin{align*} \begin{bmatrix} 4 & 1 & 2 \\ 0 & 2.25 & 2.5 \\ 0 & 0 & 2 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 19 \\ 0.75 \\ -2 \end{bmatrix} \end{align*} {/eq}

By performing matrix multiplication we get back the equaitons as:

{eq}\begin{align*} 4x+y+2z &= 19 && \dots (1') \\ 2.25y+2.5z &=0.75 && \dots (2') \\ 2z &= -2 && \dots (3') \end{align*} {/eq}

Solve (3') for z and then back sustitute:

{eq}\begin{align*} z &= -\frac{2}{2} = -1 \\ 2.25y + 2.5(-1) &= 0.75 \implies 2.25y = 3.25 \implies y = \frac{13}{9} \\ 4x+\frac{13}{9}-2 &= 19 \implies x = \frac{44}{9}\\ \\ \end{align*} {/eq}

Therefore, the solution to the given system of equations is:

{eq}\begin{align*} \textbf{x} = \begin{pmatrix} \frac{44}{9} \\ \frac{13}{9} \\ -1 \end{pmatrix} \end{align*} {/eq}