Gauss Jordan Elimination

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Question 1 of 2

1. Question
Solve for $x,y$ and $z$

$x+2y+z=9$

$3y+z=10$

$x+y+2z=11$
- $x=$ (1)
  
  $y=$ (2)
  
  $z=$ (4)
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Reduced Row-Echelon Form

$[[1,0,0,x],[0,1,0,x],[0,0,1,x]]$

A system of equations can be solved by using Gauss-Jordan Elimination. This process involves applying row operations to achieve the Reduced Row-Echelon Form

First, write the system of equations in matrix form by taking the constants

$x+2y+z=9$

$3y+z=10$

$x+y+2z=11$

$\begin{bmatrix} 1 & 2 & 1 & | & 9 \\ 0 & 3 & 1 & | & 10 \\ 1 & 1 & 2 & | & 11 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

Use row operations for the matrix to achieve Row-Echelon Form

Start with transforming the first column

Transforming the First Element of $R_3$ :

$\begin{bmatrix} 1 & 2 & 1 & | & 9 \\ 0 & 3 & 1 & | & 10 \\ 1 & 1 & 2 & | & 11 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$R_1-R_3→R_3$

$\begin{matrix} R_1: & 1 & 2 & 1 & | & 9 \\ -R_3: & -1 & -1 & -2 & | & -11 \\ \hline \:R_3→ & 0 & 1 & -1 & | & -2 \end{matrix}$

New matrix:

$\begin{bmatrix} 1 & 2 & 1 & | & 9 \\ 0 & 3 & 1 & | & 10 \\ 0 & 1 & -1 & | & -2 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

Next, transform $R_2$

Transforming the Third Element of $R_2$ :

$\begin{bmatrix} 1 & 2 & 1 & | & 9 \\ 0 & 3 & 1 & | & 10 \\ 0 & 1 & -1 & | & -2 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$R_3+R_2→R_2$

$\begin{matrix} R_3: & 0 & 1 & -1 & | & -2 \\ R_2: & 0 & 3 & 1 & | & 10 \\ \hline R_2→ & 0 & 4 & 0 & | & 8 \end{matrix}$

Transforming the Second Element of $R_2$ :

$\begin{bmatrix} 1 & 2 & 1 & | & 9 \\ 0 & 4 & 0 & | & 8 \\ 0 & 1 & -1 & | & -2 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$R_2\div4→R_2$

$\begin{matrix} R_2\div4→ 0 & 1 & 0 & | & 2 \end{matrix}$

New matrix:

$\begin{bmatrix} 1 & 2 & 1 & | & 9 \\ 0 & 1 & 0 & | & 2 \\ 0 & 1 & -1 & | & -2 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

Proceed with transforming $R_3$

Transforming the Second Element of $R_2$ :

$\begin{bmatrix} 1 & 2 & 1 & | & 9 \\ 0 & 1 & 0 & | & 2 \\ 0 & 1 & -1 & | & -2 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$-1R_2+R_3→R_3$

$\begin{matrix} -1R_2: & 0 & -1 & 0 & | & -2 \\ R_3: & 0 & 1 & -1 & | & -2 \\ \hline R_3→ & 0 & 0 & -1 & | & -4 \end{matrix}$

Transforming the Third Element of $R_3$ :

$\begin{bmatrix} 1 & 2 & 1 & | & 9 \\ 0 & 1 & 0 & | & 2 \\ 0 & 0 & -1 & | & -4 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$R_3\div-1→R_3$

$\begin{matrix} R_3\div-1→ 0 & 0 & 1 & | & 4 \end{matrix}$

New matrix:

$\begin{bmatrix} 1 & 2 & 1 & | & 9 \\ 0 & 1 & 0 & | & 2 \\ 0 & 0 & 1 & | & 4 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

Now, there is one row left to transform which is $R_1$

Transforming the Third Element of $R_1$ :

$\begin{bmatrix} 1 & 2 & 1 & | & 9 \\ 0 & 1 & 0 & | & 2 \\ 0 & 0 & 1 & | & 4 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$-1R_3+R_1→R_1$

$\begin{matrix} -1R_3: & 0 & 0 & -1 & | & -4 \\ R_1: & 1 & 2 & 1 & | & 9 \\ \hline R_1→ & 1 & 2 & 0 & | & 5 \end{matrix}$

Transforming the Second Element of $R_1$ :

$\begin{bmatrix} 1 & 2 & 0 & | & 5 \\ 0 & 1 & 0 & | & 2 \\ 0 & 0 & 1 & | & 4 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$-2R_2+R_1→R_1$

$\begin{matrix} -2R_2: & 0 & -2 & 0 & | & -4 \\ R_1: & 1 & 2 & 0 & | & 5 \\ \hline R_1→ & 1 & 0 & 0 & | & 1 \end{matrix}$

New matrix:

$\begin{bmatrix} 1 & 0 & 0 & | & 1 \\ 0 & 1 & 0 & | & 2 \\ 0 & 0 & 1 & | & 4 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

This new matrix is in Reduced Row-Echelon Form

Remember that each row in this matrix corresponds to a term in an equation

Convert the rows into equations

$\begin{matrix} \color{#6F6161}{\:\:\,x} & \color{#6F6161}{y} & \color{#6F6161}{z} & \, & \color{#6F6161}{C} \end{matrix}$
$\begin{bmatrix} 1 & 0 & 0 & | & 1 \\ 0 & 1 & 0 & | & 2 \\ 0 & 0 & 1 & | & 4 \end{bmatrix}$

$\begin{matrix} \color{#00880A}{\:\:\:x} & \:\: & \:\: & = & \color{#00880A}{1} \\ \:\, & \color{#00880A}{y} & \:\: & = & \color{#00880A}{2} \\ \:\: & \: & \color{#00880A}{z} & = & \color{#00880A}{4} \end{matrix}$

$x=1$

$y=2$

$z=4$
Question 2 of 2

2. Question
Solve for $x,y$ and $z$

$x-3y-z=-9$

$2y+z=2$

$-x-y=-1$
- $x=$ (-3)
  
  $y=$ (4)
  
  $z=$ (-6)
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Reduced Row-Echelon Form

$[[1,0,0,x],[0,1,0,x],[0,0,1,x]]$

A system of equations can be solved by using Gauss-Jordan Elimination. This process involves applying row operations to achieve the Reduced Row-Echelon Form

First, write the system of equations in matrix form by taking the constants

$x-3y-z=-9$

$2y+z=2$

$-x-y=-1$

$\begin{bmatrix} 1 & -3 & -1 & | & -9 \\ 0 & 2 & 1 & | & 2 \\ -1 & -1 & 0 & | & -1 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

Use row operations for the matrix to achieve Row-Echelon Form

Start with swapping $R_1$ and $R_3$ since $R_3$ already has $0$ as its third element

$\begin{bmatrix} -1 & -1 & 0 & | & -1 \\ 0 & 2 & 1 & | & 2 \\ 1 & -3 & -1 & | & -9 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

Next, transform $R_3$

Transforming the First Element of $R_3$ :

$\begin{bmatrix} -1 & -1 & 0 & | & -1 \\ 0 & 2 & 1 & | & 2 \\ 1 & -3 & -1 & | & -9 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$-R_1-R_3→R_3$

$\begin{matrix} -R_1: & 1 & 1 & 0 & | & 1 \\ -R_3: & -1 & 3 & 1 & | & 9 \\ \hline \:R_3→ & 0 & 4 & 1 & | & 10 \end{matrix}$

Transforming the Second Element of $R_3$ :

$\begin{bmatrix} -1 & -1 & 0 & | & -1 \\ 0 & 2 & 1 & | & 2 \\ 0 & 4 & 1 & | & 10 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$2R_2-R_3→R_3$

$\begin{matrix} 2R_2: & 0 & 4 & 2 & | & 4 \\ -R_3: & 0 & -4 & -1 & | & -10 \\ \hline \:R_3→ & 0 & 0 & 1 & | & -6 \end{matrix}$

New matrix:

$\begin{bmatrix} -1 & -1 & 0 & | & -1 \\ 0 & 2 & 1 & | & 2 \\ 0 & 0 & 1 & | & -6 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

Proceed with transforming $R_2$

Transforming the Third Element of $R_2$ :

$\begin{bmatrix} -1 & -1 & 0 & | & -1 \\ 0 & 2 & 1 & | & 2 \\ 0 & 0 & 1 & | & -6 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$-R_3+R_2→R_2$

$\begin{matrix} -R_3: & 0 & 0 & -1 & | & 6 \\ R_2: & 0 & 2 & 1 & | & 2 \\ \hline R_2→ & 0 & 2 & 0 & | & 8 \end{matrix}$

Transforming the Second Element of $R_2$ :

$\begin{bmatrix} -1 & -1 & 0 & | & -1 \\ 0 & 2 & 0 & | & 8 \\ 0 & 0 & 1 & | & -6 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$R_2\div2→R_2$

$\begin{matrix} R_2\div2→ 0 & 1 & 0 & | & 4 \end{matrix}$

New matrix:

$\begin{bmatrix} -1 & -1 & 0 & | & -1 \\ 0 & 1 & 0 & | & 4 \\ 0 & 0 & 1 & | & -6 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

Now, there is one row left to transform which is $R_1$

Transforming the Second Element of $R_1$ :

$\begin{bmatrix} -1 & -1 & 0 & | & -1 \\ 0 & 1 & 0 & | & 4 \\ 0 & 0 & 1 & | & -6 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$R_2+R_1→R_1$

$\begin{matrix} R_2: & 0 & 1 & 0 & | & 4 \\ R_1: & -1 & -1 & 0 & | & -1 \\ \hline R_1→ & -1 & 0 & 0 & | & 3 \end{matrix}$

Transforming the First Element of $R_1$ :

$\begin{bmatrix} -1 & 0 & 0 & | & 3 \\ 0 & 1 & 0 & | & 4 \\ 0 & 0 & 1 & | & -6 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$R_1\div-1→R_2$

$\begin{matrix} R_2\div2→ 1 & 0 & 0 & | & -3 \end{matrix}$

New matrix:

$\begin{bmatrix} 1 & 0 & 0 & | & -3 \\ 0 & 1 & 0 & | & 4 \\ 0 & 0 & 1 & | & -6 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

This new matrix is in Reduced Row-Echelon Form

Remember that each row in this matrix corresponds to a term in an equation

Convert the rows into equations

$\begin{matrix} \color{#6F6161}{\:\:\,x} & \color{#6F6161}{y} & \color{#6F6161}{z} & \, & \color{#6F6161}{\:\;C} \end{matrix}$
$\begin{bmatrix} 1 & 0 & 0 & | & -3 \\ 0 & 1 & 0 & | & 4 \\ 0 & 0 & 1 & | & -6 \end{bmatrix}$

$\begin{matrix} \color{#00880A}{\:\:\:x} & \:\: & \:\: & = & \color{#00880A}{-3} \\ \:\, & \color{#00880A}{y} & \:\: & = & \color{#00880A}{4} \\ \:\: & \: & \color{#00880A}{z} & = & \color{#00880A}{-6} \end{matrix}$

$x=-3$

$y=4$

$z=-6$

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1. Question

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Great Work!

Reduced Row-Echelon Form

2. Question

Hint

Correct!

Reduced Row-Echelon Form

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