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)
Hint
Help VideoCorrect
Great Work!
Incorrect
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 FormFirst, 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 FormStart with transforming the first columnTransforming 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 FormRemember that each row in this matrix corresponds to a term in an equationConvert 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` -
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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)
Hint
Help VideoCorrect
Correct!
Incorrect
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 FormFirst, 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 FormStart 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 FormRemember that each row in this matrix corresponds to a term in an equationConvert 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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