# Find the solution of the following system using Gauss elimination. \begin{alignat}{3} 6x + y +...

## Question:

Find the solution of the following system using Gauss elimination.

{eq}\begin{alignat}{3} 6x + y + 5z &&=& 12, \\ 2x - y + 5z &&=& -2, \\ 5x + 3y + 4z &&=& 4. \end{alignat} {/eq}

## Pros and cons of Gaussian elimination:

When presented with a simple system of linear equations to solve, we often do so using Gaussian elimination. It is nice in that there is a simple set of steps we can follow in order to find a solution. In fact, there are three possible outcomes. We can find that there is a unique solution, infinitely many solutions, or no possible solutions. It is also possible to teach a computer how to solve a system of linear equations using Gaussian elimination fairly easily.

One potential issue is that Gaussian elimination can be slow. If you want to solve a system of hundreds of equations with hundreds of variables, the number of steps it takes to arrive at a solution grows very quickly. But for simple systems of just a few equations with a few variables, it is a great fool-proof method!

## Answer and Explanation:

We start by putting our system of linear equations into augmented matrix form:

$$\left[ \begin{array}{ccc|c} 6 & 1 & 5 & 12 \\ 2 & -1 & 5 & -2 \\ 5 & 3 & 4 & 4 \\ \end{array} \right]. $$

Now we use row operations to put this system in reduced form. We start by multiplying Row 1 by {eq}\frac16 {/eq} to get a 1 in the first entry.

$$\left[ \begin{array}{ccc|c} 1 & \frac16 & \frac56 & 2 \\ 2 & -1 & 5 & -2 \\ 5 & 3 & 4 & 4 \\ \end{array} \right]. $$

Now replace Row 2 with -2(Row 1) + Row 2, and replace Row 3 with -5(Row 1) + Row 3. This will give us zeros in the first column of Rows 2 and 3.

$$\left[ \begin{array}{ccc|c} 1 & \frac16 & \frac56 & 2 \\ 0 & -\frac43 & \frac{10}3 & -6 \\ 0 & \frac{13}6 & -\frac16 & -6 \\ \end{array} \right]. $$

Next, we multiply Row 2 by {eq}-\frac34 {/eq} to get a 1 in the second column.

$$\left[ \begin{array}{ccc|c} 1 & \frac16 & \frac56 & 2 \\ 0 & 1 & -\frac52 & \frac92 \\ 0 & \frac{13}6 & -\frac16 & -6 \\ \end{array} \right]. $$

Now replace Row 3 with {eq}-\frac{13}6 {/eq}(Row 2) + Row 3.

$$\left[ \begin{array}{ccc|c} 1 & \frac16 & \frac56 & 2 \\ 0 & 1 & -\frac52 & \frac92 \\ 0 & 0 & \frac{21}{4} & \frac{-63}{4} \\ \end{array} \right]. $$

Finally, multiply Row 3 by {eq}\frac{4}{21}. {/eq}

$$\left[ \begin{array}{ccc|c} 1 & \frac16 & \frac56 & 2 \\ 0 & 1 & -\frac52 & \frac92 \\ 0 & 0 & 1 & -3 \\ \end{array} \right]. $$

Now we can see from Row 3 that {eq}z = -3. {/eq} Using this, we solve for {eq}y {/eq} in the second row, {eq}y - \frac52(-3) = \frac92. {/eq}

That tells us that {eq}y = -3. {/eq} Finally, solve for {eq}x {/eq} in Row 1, {eq}x + \frac16(-3) + \frac56(-3) = 2. {/eq} We find that {eq}x = 5. {/eq} Thus our final solution is {eq}x = 5, y = -3, z = -3. {/eq}

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from Algebra II Textbook

Chapter 10 / Lesson 6