# Find the solution of the following system using Gauss elimination. \begin{alignat}{3} x - 5y +...

## Question:

Find the solution of the following system using Gauss elimination.

{eq}\begin{alignat}{3} x - 5y + z &&=& 16, \\ 5y - 4z &&=& 1, \\ x + y + 5z &&=& -18. \end{alignat} {/eq}

## Gauss Elimination:

{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 an equation of 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.

The Gauss elimination method tries to reduce the coefficient matrix {eq}A {/eq} in a triangular form by performing row operations on the augmented matrix {eq}[A|B] {/eq}.

Once a triangular form of {eq}A {/eq} is obtained, the process stops and the solution to the system is obtained by back substitution.

{eq}\\ {/eq}

{eq}\begin{align*} x-5y+z &= 16 && \dots (1) \\ 5y-4z &= 1 && \dots (2) \\ x+y+5z &= -18 && \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} 1 & -5 & 1 \\ 0 & 5 & -4 \\ 1 & 1 & 5 \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} 16 \\ 1\\ -18 \end{bmatrix} \end{align*} {/eq}

The problem in matrix form is presented as:

{eq}\begin{align*} \begin{bmatrix} 1 & -5 & 1 \\ 0 & 5 & -4 \\ 1 & 1 & 5 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 16 \\ 1\\ -18 \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} 1 & -5 & 1 | & 16 \\ 0 & 5 & -4 | & 1\\ 1 & 1 & 5 | & -18 \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_3 = R_3 - R_1 {/eq}: Subtract the first row {eq}R_1 {/eq} from the third row {eq}R_3 {/eq}:

{eq}\begin{align*} [A | B] &=\begin{bmatrix} 1 & -5 & 1 | & 16 \\ 0 & 5 & -4 | & 1\\ 0 & 6 & 4 | & -34 \end{bmatrix} \\ \\ &[R_3 = 5 \times R_3 - 6 \times R_2 ] \\ \\ [A | B] &=\begin{bmatrix} 1 & -5 & 1 | & 16 \\ 0 & 5 & -4 | & 1\\ 0 & 0 & 44 | & -176 \end{bmatrix} \\ \end{align*} {/eq}

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

{eq}\begin{align*} A = \begin{bmatrix} 1 & -5 & 1 \\ 0 & 5 & -4 \\ 0 & 0 & 44 \end{bmatrix} \end{align*} {/eq}

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

{eq}\begin{align*} \begin{bmatrix} 1 & -5 & 1 \\ 0 & 5 & -4 \\ 0 & 0 & 44 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 16\\ 1 \\ -176 \end{bmatrix} \end{align*} {/eq}

By performing matrix multiplication we get back the equaitons as:

{eq}\begin{align*} x-5y+z &= 16 && \dots (1') \\ 5y-4z &=1 && \dots (2') \\ 44z &= -176 && \dots (3') \end{align*} {/eq}

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

{eq}\begin{align*} z &= -\frac{176}{4} = -4 \\ 5y-4(-4) &= 1 \implies 5y = -15 \implies y = -3\\ x+-5(-3)+(-4) &= 16 \implies x = 16-15+4 = 4\\ \\ \end{align*} {/eq}

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

{eq}\begin{align*} \textbf{x} = \begin{pmatrix} 5 \\ -3 \\ -4 \end{pmatrix} \end{align*} {/eq}