# Find the solution of the following system using Gauss elimination. -4x + y + 2z = -19, 4x - y +...

## Question:

Find the solution of the following system using Gauss elimination.

{eq}-4x + y + 2z = -19 {/eq}

{eq}4x - y + 4z = -17 {/eq}

{eq}3x + 3y + 4z = 15 {/eq}

## Gaussian Elimination

Gaussian elimination is a technique to solve a system of linear equations by eliminating the unknowns from the equations in such a manner that the matrix of the system is a upper triangular matrix.

To solve a system that is upper triangular, we just backsubstitute the variables in the upper equations, starting with the lowest equation.

To apply Gauss elimination, we will do row operations on the matrix of the system.

## Answer and Explanation:

To solve the system {eq}\displaystyle \begin{align} \begin{cases} -4x + y + 2z = -19\\ 4x -y + 4z = -17\\ 3x+3y + 4z = 15\\ \end{cases} \end{align} {/eq}

we will do row operations on the matrix of the system, together with the right hand side, until we obtain zero under the main diagonal entries of the matrix.

To better understand the operations we do on the rows, we will use the notation *R_i* for the *i*-th row and describe the operations on top of the equivalent symbol.

{eq}\displaystyle \begin{align} \left[\begin{array}{ccc | c} -4 & 1 & 2 & -19\\ 4 & -1&4 & -17 \\ 3 & 3& 4 & 15 \end{array}\right] \overset{ R_1+R_2}{\iff} \left[\begin{array}{ccc | c} -4 & 1 & 2 & -19\\ 0& 0&6 & 36 \\ 3 & 3& 4 & 15 \end{array}\right] \overset{ R_2 \text{ interchanges with }R_3}{\iff} \left[\begin{array}{ccc | c} -4 & 1 & 2 & -19\\ 3 & 3& 4 & 15\\ 0& 0&6 & -36 \\ \end{array}\right] \overset{\frac{3}{4}\cdot R_1+R_2}{\iff} \left[\begin{array}{ccc | c} -4 & 1 & 2 & -19\\ 0& \frac{15}{4}&\frac{11}{2} & \frac{3}{4} \\ 0 & 0& 6&-36 \end{array}\right] \overset{4\cdot R_2, \frac{1}{6}R_3}{\iff} \left[\begin{array}{ccc | c} -4 & 1 & 2 & -19\\ 0&15&22 &3 \\ 0 & 0& 1&-6 \end{array}\right] \end{align} {/eq}

We obtained an upper triangular matrix, meaning we have zeros below the diagonal terms, next, we will give the solution by writing the systems of equations corresponding to the last matrix.

{eq}\displaystyle \begin{align} &\begin{cases} \begin{array}{cccc} -4x & +y &+2z & =&-19\\ & 15y &+22z &=& 3 \\ & & z &=& -6, \text{ used in above equations:} \end{array} \end{cases} \\\\ \iff &\begin{cases} \begin{array}{cccc} -4x & +y &+2z & =&-19&\\ & &y &=&\frac{3+22\cdot 6}{15}&=9,\text{ used in above equation:} \\ & & z &=& 6& \end{array} \end{cases} \\\\ \iff &\begin{cases} \begin{array}{cccc} x & =&\frac{19}{4}+\frac{9}{4}-3&=4\\ y &=&9&\\ z &=& -6& \end{array} \end{cases} \end{align} {/eq}

Therefore, the solution is unique, {eq}\displaystyle \begin{align} \boxed{\begin{cases} \begin{array}{cccc} x & =&4\\ y &=&9\\ z&=&-6. \end{array} \end{cases}} \end{align} {/eq}

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from Algebra II Textbook

Chapter 10 / Lesson 6