Use Gaussian Elimination of Gauss-Jordan Elimination to solve the following system. x - 3z = -2 ...

Question:

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

x - 3z = -2

2x + 2y + z = 4

3x + y - 2z = 5

Gaussian Elimination

We can solve a system of equations, which relate a number of variables using a number of equations, by using matrix row operations. We first need to use the coefficients in the equation to construct an augmented matrix, and then we can perform row operations to solve this system.

Answer and Explanation:

Let's first take the coefficients on each variable and use them to construct the augmented matrix. We can then perform any of the three row operations, which are swapping rows, multiplying rows by constants, or adding multiples of rows to one another, in order to obtain row reduced echelon form.

{eq}\begin{align*} \begin{bmatrix} 1 & 0 & -3 & -2\\ 2 & 2 & 1 & 8\\ 3 & 1 & -2 & 5 \end{bmatrix} & R_2 = R_2 - 2R_1\\ \begin{bmatrix} 1 & 0 & -3 & -2\\ 0 & 2 & 7 & 8\\ 3 & 1 & -2 & 5 \end{bmatrix} &R_3 = R_3 - 3R_1\\ \begin{bmatrix} 1 & 0 & -3 & -2\\ 0 & 2 & 7 & 8\\ 0 & 1 & 7 & 11 \end{bmatrix} & R_2 = R_2 - R_3\\ \begin{bmatrix} 1 & 0 & -3 & -2\\ 0 & 1 & 0 & -3\\ 0 & 1 & 7 & 11 \end{bmatrix} &R_3 = R_3 - R_2\\ \begin{bmatrix} 1 & 0 & -3 & -2\\ 0 & 1 & 0 & -3\\ 0 & 0 & 7 & 14 \end{bmatrix} &R_3 = \frac{1}{7}R_3\\ \begin{bmatrix} 1 & 0 & -3 & -2\\ 0 & 1 & 0 & -3\\ 0 & 0 & 1 & 2 \end{bmatrix} &R_1 = R_1 + 3R_3\\ \begin{bmatrix} 1 & 0 & 0 & 4\\ 0 & 1 & 0 & -3\\ 0 & 0 & 1 & 2 \end{bmatrix} \end{align*} {/eq}


As we've now reduced this matrix as far as we can, the final column indicates our solutions. This informs us that {eq}x = 4 {/eq}, {eq}y = -3 {/eq}, and {eq}z = 2 {/eq}.


Learn more about this topic:

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How to Solve Linear Systems Using Gaussian Elimination

from Algebra II Textbook

Chapter 10 / Lesson 6
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