# Find the solution of the following system using Gauss elimination. \left\{\begin{matrix} -4x+y...

## Question:

Find the solution of the following system using Gauss elimination.

{eq}\left\{\begin{matrix} -4x+y +2z=-19 \\ 4x- y +4z=-17 \\ 3x +3y+4z=15 \end{matrix} \right . {/eq}

## Gaussian Elimination

A system of equations contains multiple equations relating multiple variables. Solving such a system requires us to use one of many techniques, of which Gaussian elimination is one of the most popular. In this method, the system is written as an augmented matrix, and row operations simplify this system until its solutions can be expressed.

## Answer and Explanation:

The given system of equations has three variables related in three equations. We can write this as an augmented matrix by using the coefficients of the original equations as follows.

{eq}\begin{bmatrix} -4 & 1 & 2 & -19\\ 4 & -1 & 4 & -17\\ 3 & 3 & 4 & 15 \end{bmatrix} {/eq}

Let's now perform row operations to solve this system.

{eq}\begin{align*} \begin{bmatrix} -4 & 1 & 2 & -19\\ 4 & -1 & 4 & -17\\ 3 & 3 & 4 & 15 \end{bmatrix} & R_2 = R_1 +R_2\\ \begin{bmatrix} -4 & 1 & 2 & -19\\ 0 & 0 & 6 & -36\\ 3 & 3 & 4 & 15 \end{bmatrix} &R_2 \leftrightarrow R_3\\ \begin{bmatrix} -4 & 1 & 2 & -19\\ 3 & 3 & 4 & 15\\ 0 & 0 & 6 & -36\end{bmatrix} &R_1 = -R_1\\ \begin{bmatrix} 4 & -1 & -2 & 19\\ 3 & 3 & 4 & 15\\ 0 & 0 & 6 & -36\end{bmatrix} &R_1 = R_1-R_2\\ \begin{bmatrix} 1 & -4 & -6 & 4\\ 3 & 3 & 4 & 15\\ 0 & 0 & 6 & -36\end{bmatrix} &R_3 = \frac{1}{6}R_3\\ \begin{bmatrix} 1 & -4 & -6 & 4\\ 3 & 3 & 4 & 15\\ 0 & 0 & 1 & -6\end{bmatrix} &R_2 = R_2-3R_1\\ \begin{bmatrix} 1 & -4 & -6 & 4\\ 0& 15 & 22 & 3\\ 0 & 0 & 1 & -6\end{bmatrix} &R_2 = R_2 - 22R_3\\ \begin{bmatrix} 1 & -4 & -6 & 4\\ 0& 15 & 0 & 135 \\ 0 & 0 & 1 & -6\end{bmatrix} &R_2 = \frac{1}{15}R_2\\ \begin{bmatrix} 1 & -4 & -6 & 4\\ 0& 1& 0 & 9\\ 0 & 0 & 1 & -6\end{bmatrix} &R_1 = R_1 + 6R_3\\ \begin{bmatrix} 1 & -4 & 0 & -32\\ 0& 1& 0 & 9\\ 0 & 0 & 1 & -6\end{bmatrix}&R_1 = R_1 + 4R_2\\ \begin{bmatrix} 1 &0 & 0 & 4\\ 0& 1& 0 & 9\\ 0 & 0 & 1 & -6\end{bmatrix} \end{align*} {/eq}

Therefore, the solution to this system is that {eq}x = 4 {/eq}, {eq}y = 9 {/eq}, and {eq}z = -6 {/eq}.

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from Algebra II Textbook

Chapter 10 / Lesson 6