# Given the equations 2x1- 6x2-x3 =-38 -3x1- x2 + 7x3 =-34 -8x1 + x2-2x3 =-20 (a) Solve by...

## Question:

Given the equations:

{eq}\begin{align} & 2{{x}_{1}}-6{{x}_{2}}-{{x}_{3}}=-38 \\ & -3{{x}_{1}}-{{x}_{2}}+7{{x}_{3}}=-34 \\ & -8{{x}_{1}}+{{x}_{2}}-2{{x}_{3}}=-20 \\ \end{align} {/eq}

(a) Solve by Gauss elimination with partial pivoting. Show all steps of the computation.

(b) Substitute your results into the original equations to check your answer.

## Gaussian Elimination:

There are different methods to find the solution of a system of equations of three equations with three unknowns. One of them is the Gaussian elimination, which consists of reducing, through elementary operations between the rows, a matrix of order {eq}n \times n {/eq} to obtain ones in the main diagonal and then make the substitution backward.

## Answer and Explanation:

Let Given the equations:

{eq}\begin{align} & 2{{x}_{1}}-6{{x}_{2}}-{{x}_{3}}=-38 \\ & -3{{x}_{1}}-{{x}_{2}}+7{{x}_{3}}=-34 \\ & -8{{x}_{1}}+{{x}_{2}}-2{{x}_{3}}=-20 \\ \end{align} {/eq}

(a) We are going to solve the system of equations by Gauss elimination with partial pivoting.

{eq}\begin{bmatrix} 2 & -6 & -1& -38\\ -3 & -1 & 7& -34\\ -8 & 1 & -2& -20 \end{bmatrix} {/eq}

{eq}\dfrac{1}{2}F_1\rightarrow F_1 \begin{bmatrix} 1 & -3 & -\dfrac{1}{2} & -19\\ -3 & -1 & 7& -34\\ -8 & 1 & -2& -20 \end{bmatrix} {/eq}

{eq}3F_1 + F_2 \rightarrow F_2 \begin{bmatrix} 1 & -3 & -\dfrac{1}{2} & -19\\ 0 & -10 & \dfrac{11}{2} & -91\\ -8 & 1 & -2 & -20 \end{bmatrix} {/eq}

{eq}8F_1 + F_3 \rightarrow F_3 \begin{bmatrix} 1 & -3 & -\dfrac{1}{2} & -19\\ 0 & -10 & \dfrac{11}{2} & -91\\ 0 & -23 & -6 & -172 \end{bmatrix}{/eq}

{eq}-\dfrac{1}{10}F_2 \rightarrow F_2 \begin{bmatrix} 1 & -3 & -\dfrac{1}{2} & -19\\ 0 & 1 & -\dfrac{11}{20} & \dfrac{91}{10} \\ 0 & -23 & -6 & -172 \end{bmatrix}{/eq}

{eq}23F_2 + F_3 \rightarrow F_3 \begin{bmatrix} 1 & -3 & -\dfrac{1}{2} & -19\\ 0 & 1 & -\dfrac{11}{20} & \dfrac{91}{10} \\ 0 & 0 & -\dfrac{373}{20} & \dfrac{373}{10} \end{bmatrix}{/eq}

{eq}-\dfrac{20}{373}F_3 \rightarrow F_3 \begin{bmatrix} 1 & -3 & -\dfrac{1}{2} & -19\\ 0 & 1 & -\dfrac{11}{20} & \dfrac{91}{10} \\ 0 & 0 & 1 & -2 \end{bmatrix}{/eq}

The system is defined by

{eq}\left\{\begin{matrix} x_1 &-3x_2 - \dfrac{1}{2}x_3 & = -19 \\ & x_2 - \dfrac{11}{20}x_3 &= \dfrac{91}{10}\\ & x_3 &= -2 \end{matrix}\right.{/eq}

Making substitution backwards we have

{eq}x_3 = -2 \\ x_2 = \dfrac{91}{10} + \dfrac{11}{20}x_3 \to x_2 = \dfrac{91}{10} + \dfrac{11}{20}(-2) \to x_2 = 8 {/eq}

And

{eq}x_1 = 3x_2 + \dfrac{1}{2}x_3 -19 \to x_1 = 3(8) +\dfrac{1}{2}(-2) -19 \to x_1 = 4.{/eq}

Thus the values of x that satisfy the System are

{eq}x_1 = 4, x_2 = 8{/eq} and {eq}x_3= -2.{/eq}

(b) Now we are going to replace the solution found to verify that it satisfies the system.

For the first equation we have

{eq}2(4) - 6(8) - (-2)= -38 \\ -38 = -38 {/eq} equality is satisfied.

For the second equation we have

{eq}-3(4) -8 + 7(-2)= -34 \\ -34= -34 {/eq} equality is satisfied.

For the third equation we have

{eq}-8(4) + 8 -2(-2)= -20 \\ -20= -20 {/eq} equality is satisfied.

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from Algebra II Textbook

Chapter 10 / Lesson 8