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$\begin{array}{l} x_{1}+2x_{2}+3x_{3} \ x_{1}+(2)(2)+3(3) \ x_{1}+5=6 \end{array}$ An $m\times n$ matrix $A$ is \quad $x_{1}+5=6$ In reduced row echelon form if it satisfies \quad $x_{1}=6$ the following properties: $\begin{array}{l} x_{1}+x_{2}+2x_{3}=-1 \ x_{1}-2x_{2}+x_{3}=-5 \ 3x_{1}+x_{2}+x_{3}=3 \end{array}$ a) Find all solutions by using the Gaussian elimination & Gauss Jordan Reduction

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Solve linear equations with variables on both sides: word problems

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Q.

\begin{array}{l} x_{1}+2x_{2}+3x_{3} \ x_{1}+(2)(2)+3(3) \ x_{1}+5=6 \end{array}

An

m\times n

matrix

A

is \quad

x_{1}+5=6

In reduced row echelon form if it satisfies \quad

x_{1}=6

the following properties:

\begin{array}{l} x_{1}+x_{2}+2x_{3}=-1 \ x_{1}-2x_{2}+x_{3}=-5 \ 3x_{1}+x_{2}+x_{3}=3 \end{array}

a) Find all solutions by using the Gaussian elimination & Gauss Jordan Reduction

Set up augmented matrix: Set up the augmented matrix for the system of equations: $\newline$ $\begin{bmatrix} 1 & 1 & 2 & | & -1 \\ 1 & -2 & 1 & | & -5 \\ 3 & 1 & 1 & | & 3 \end{bmatrix}$
Perform Gaussian elimination: Perform the first step of Gaussian elimination: Make the first element of the first column a $1$ (already done), and use it to zero out the rest of the first column. $\newline$ - Subtract the first row from the second row: $\newline$ $\begin{bmatrix} 1 & 1 & 2 & | & -1 \\ 0 & -3 & -1 & | & -4 \\ 3 & 1 & 1 & | & 3 \end{bmatrix}$ $\newline$ - Subtract $3$ times the first row from the third row: $\newline$ $\begin{bmatrix} 1 & 1 & 2 & | & -1 \\ 0 & -3 & -1 & | & -4 \\ 0 & -2 & -5 & | & 6 \end{bmatrix}$
Normalize second row: Normalize the second row by dividing by $-3$ : $\newline$ $\begin{bmatrix} 1 & 1 & 2 & | & -1 \\ 0 & 1 & 1/3 & | & 4/3 \\ 0 & -2 & -5 & | & 6 \end{bmatrix}$ $\newline$ - Use the second row to zero out the rest of the second column: $\newline$ - Add $2$ times the second row to the third row: $\newline$ $\begin{bmatrix} 1 & 1 & 2 & | & -1 \\ 0 & 1 & 1/3 & | & 4/3 \\ 0 & 0 & -13/3 & | & 22/3 \end{bmatrix}$
Use second row: Normalize the third row by multiplying by $-3$ / $13$ : $\newline$ $\begin{bmatrix} 1 & 1 & 2 & | & -1 \\ 0 & 1 & 1/3 & | & 4/3 \\ 0 & 0 & 1 & | & -2 \end{bmatrix}$ $\newline$ - Use the third row to zero out the rest of the third column: $\newline$ - Subtract $2$ times the third row from the first row and subtract $1$ / $3$ times the third row from the second row: $\newline$ $\begin{bmatrix} 1 & 1 & 0 & | & 3 \\ 0 & 1 & 0 & | & 2 \\ 0 & 0 & 1 & | & -2 \end{bmatrix}$
Normalize third row: Use the second row to zero out the first column: $\newline$ - Subtract the second row from the first row: $\newline$ $\begin{bmatrix} 1 & 0 & 0 & | & 1 \\ 0 & 1 & 0 & | & 2 \\ 0 & 0 & 1 & | & -2 \end{bmatrix}$ $\newline$ This is the reduced row echelon form, showing that each variable has a unique solution.

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