# Reduce the following system of equations to the row echelon form and determine the value of x3...

## Question:

Reduce the following system of equations to the row echelon form and determine the value of x3 that is a solution to the system:

{eq}-2x1 +x2 +4.6x3 =3\\ -4.6x1 -2x2 -2x3 =5\\ 9.8x1 +8x2 +15.2x3 =-9\\ 6x1 +8x2 +9x3 =10 {/eq}

Use gauss elimination ans: -1.507

## Gaussian Elimination

When solving a system of equations with a large amount of variables, keeping this information organized is essential. We can use a matrix to both organize and solve a system, by conducting row operations in a method known as Gaussian elimination.

Let's begin by constructing the augmented matrix that corresponds to this system. We have four equations and three unknowns, so this will be a four by four matrix consisting of the coefficients.

{eq}\begin{bmatrix} -2 & 1 & 4.6 & 3\\ -4.6 & -2 & -2 & 5\\ 9.8 & 8 & 15.2 & -9\\ 6 & 8 & 9 & 10 \end{bmatrix} {/eq}

To find the value of the third variable, we need to perform operations on our rows until one row has zeros in the first two columns. We can then write an equation relating the value of the third variable to a constant, indicated by the final column's value.

{eq}\begin{align*} &\begin{bmatrix} -2 & 1 & 4.6 & 3\\ -4.6 & -2 & -2 & 5\\ 9.8 & 8 & 15.2 & -9\\ 6 & 8 & 9 & 10 \end{bmatrix} && R_4 = R_4 + 3R_1\\ &\begin{bmatrix} -2 & 1 & 4.6 & 3\\ -4.6 & -2 & -2 & 5\\ 9.8 & 8 & 15.2 & -9\\ 0 & 11 & 22.8 & 19 \end{bmatrix}& &R_3 = R_3 + 4.9R_2\\ &\begin{bmatrix} -2 & 1 & 4.6 & 3\\ -4.6 & -2 & -2 & 5\\ 0 & 12.9 & 37.74 & 0\\ 0 & 11 & 22.8 & 19 \end{bmatrix}&& R_4 = R_3 - \frac{12.9}{11} R_4\\ &\begin{bmatrix} -2 & 1 & 4.6 & 3\\ -4.6 & -2 & -2 & 5\\ 0 & 12.9 & 37.74 & 5.7\\ 0 & 0 & 11 & -16.58182 \end{bmatrix}& & R_4 = \frac{1}{11}R_4\\ &\begin{bmatrix} -2 & 1 & 4.6 & 3\\ -4.6 & -2 & -2 & 5\\ 0 & 12.9 & 37.74 & 5.7\\ 0 & 0 & 1 & -1.5074 \end{bmatrix}& \end{align*} {/eq}

Therefore, the third variable has a value of approximately -1.5074. 