# Use Gaussian Elimination or Gauss-Jordan Elimination to solve the following system. 3y - z = -1 ...

## Question:

Use Gaussian Elimination or Gauss-Jordan Elimination to solve the following system.

3y - z = -1

x + 5y - z = -4

-3x + 6y + 2z = 11

## Gaussian Elimination:

A system of equations can be cumbersome to solve, as we have multiple equations relating to multiple variables. Constructing an augmented matrix and performing row operations on them can allow us to organize our information and steps cleanly, directing us to the solution.

## Answer and Explanation:

Let's construct the augmented matrix representing this linear system and perform row operations on it. We can construct this matrix by taking the coefficients from each equation and using them as elements of the matrix. We can then swap rows, multiply rows by constants, and add multiples of rows to one another until we've found reduced row echelon form.

{eq}\begin{align*} \begin{bmatrix} 0 & 3 & -1 & -1\\ 1 & 5 & -1 & -4\\ -3 & 6 & 2 & 11\end{bmatrix} & R_1 \leftrightarrow R_2 & \text{ Swapping the first and second rows.}\\ \begin{bmatrix}1 & 5 & -1 & -4\\ 0 & 3 & -1 & -1\\ -3 & 6 & 2 & 11\end{bmatrix} &R_3 = R_3 + 3R_1 & \text{ Replacing the third row with the sum of it and three times the first row. }\\ \begin{bmatrix}1 & 5 & -1 & -4\\ 0 & 3 & -1 & -1\\ 0 & 21 & -1 & -1 \end{bmatrix} & R_3 = R_3 - 7R_2 & \text{ Replacing the third row with the difference of it and seven times the second row.}\\ \begin{bmatrix}1 & 5 & -1 & -4\\ 0 & 3 & -1 & -1\\ 0 & 0 & 6 & 6 \end{bmatrix} &R_3 = \frac{1}{6}R_3 & \text{ Dividing the third row by 6.} \\ \begin{bmatrix}1 & 5 & -1 & -4\\ 0 & 3 & -1 & -1\\ 0 & 0 & 1 & 1 \end{bmatrix} &R_2 = R_2 + R_3 & \text{ Replacing the second row with the sum of the second and third rows.} \\ \begin{bmatrix}1 & 5 & -1 & -4\\ 0 & 3 & 0 & 0\\ 0 & 0 & 1 & 1 \end{bmatrix} &R_2 = \frac{1}{3} R_2 & \text{ Dividing the second row by three. }\\ \begin{bmatrix}1 & 5 & -1 & -4\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 1 \end{bmatrix}&R_1 = R_1 - 5R_2 & \text{ Replacing the first row with the difference of itself and five times the second row. }\\ \begin{bmatrix}1 & 0 & -1 & -4\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 1 \end{bmatrix}&R_1 = R_1 + R_3 & \text{ Replacing the first row with the sum of the first and third rows.} \\ \begin{bmatrix}1 & 0 & 0 & -3\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 1 \end{bmatrix} \end{align*} {/eq}

Therefore, the solution to this system is: {eq}x=-3 {/eq}, {eq}y=0 {/eq}, and {eq}z = 1 {/eq}.

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from Algebra II Textbook

Chapter 10 / Lesson 6