The system of equations: \begin{cases} x+2y-z=-3, \\ x+z = -7, \\ 2x-y-z = -8 \end {cases} has a...

Question:

The system of equations:

$$\begin{cases} x+2y-z=-3, \\ x+z = -7, \\ 2x-y-z = -8 \end {cases} $$

has a unique solution. Find he solution using Gaussian elimination method or Gauss-Jordan elimination method

Solution of System of Equations:

Given a system of equations, it is possible to obtain its solution by applying the Gauss - Jordan method. This method is based on operations on the rows and columns in order to obtain an extended identity matrix with the system solution.

Answer and Explanation:

We use operations on row and columns. The operation is shown in each step.

{eq}\left(\left.\begin{matrix} 1 & 2 & -1 \\ 1 & 0 & 1 \\ 2 & -1 & -1 \end{matrix}\text{ }\right |\text{ }\begin{matrix} -3 \\ -7 \\ -8 \end{matrix}\right) {/eq}

{eq}R_2\rightarrow R_2 - R_1 \\ R_3\rightarrow R_3 - 2 R_1 {/eq}

{eq}\left(\left.\begin{matrix} 1 & 2 & -1 \\ 0 & -2 & 2 \\ 0 & -5 & 1 \end{matrix}\text{ }\right |\text{ }\begin{matrix} -3 \\ -4 \\ -2 \end{matrix}\right) {/eq}

We normalize the second row:

{eq}R_2\rightarrow \frac{R_2}{ -2 } {/eq}

{eq}\left(\left.\begin{matrix} 1 & 2 & -1 \\ 0 & 1 & -1 \\ 0 & -5 & 1 \end{matrix}\text{ }\right |\text{ }\begin{matrix} -3 \\ 2 \\ -2 \end{matrix}\right) {/eq}

{eq}R_1\rightarrow R_1 - 2 R_2 \\ R_3\rightarrow R_3 + 5 R_2 {/eq}

{eq}\left(\left.\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & -4 \end{matrix}\text{ }\right |\text{ }\begin{matrix} -7 \\ 2 \\ 8 \end{matrix}\right) {/eq}

We normalize the third row:

{eq}R_3\rightarrow \frac{R_3}{ -4 } {/eq}

{eq}\left(\left.\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{matrix}\text{ }\right |\text{ }\begin{matrix} -7 \\ 2 \\ -2 \end{matrix}\right) {/eq}

{eq}R_2\rightarrow R_1 - R_3 \\ R_2\rightarrow R_2 + R_3 {/eq}

{eq}\left(\left.\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\text{ }\right |\text{ }\begin{matrix} -5 \\ 0 \\ -2 \end{matrix}\right) {/eq}

The solution of the system is:

{eq}x=-5\\ y=0\\ z=-2 {/eq}


Learn more about this topic:

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Consistent System of Equations: Definition & Examples

from High School Algebra II: Homework Help Resource

Chapter 8 / Lesson 8
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