{:\begin{align}x_{ $1$ }+x_{ $2$ }+ $2$ x_{ $3$ }&= $-1$ \x_{ $1$ } $-2$ x_{ $2$ }+x_{ $3$ }&= $-5$ (3\)x_{ $1$ }+x_{ $2$ }+x_{ $3$ }&= $3$ \end{align}:} a) find all solutions by using the Gaussian elimination & Gauss-Jordan Reduction

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Grade 8

Write an equation word problem

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Q. {:\begin{align*}x_{

1

}+x_{

2

2

x_{

3

}&=

-1

\x_{

1

}

-2

x_{

2

}+x_{

3

}&=

-5

(3\)x_{

1

}+x_{

2

}+x_{

3

}&=

3

\end{align*}:} a) find all solutions by using the Gaussian elimination & Gauss-Jordan Reduction

Write Augmented Matrix: Write the augmented matrix for the system of equations. $\newline$ $\begin{bmatrix} 1 & 1 & 2 & | & -1 \\ 1 & -2 & 1 & | & -5 \\ 3 & 1 & 1 & | & 3 \end{bmatrix}$
First Gaussian Elimination Step: 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 other elements in the first column. $\newline$ $\text{Row 2} = \text{Row 2} - \text{Row 1}$ $\newline$ $\text{Row 3} = \text{Row 3} - 3 \times \text{Row 1}$ $\newline$ $\begin{bmatrix} 1 & 1 & 2 & | & -1 \\ 0 & -3 & -1 & | & -4 \\ 0 & -2 & -5 & | & 6 \end{bmatrix}$
Second Row Normalization: Next, make the second element of the second row a $1$ by dividing the whole row by $-3$ . $\newline$ $\text{Row 2} = \frac{1}{-3} \times \text{Row 2}$ $\newline$ $\begin{bmatrix} 1 & 1 & 2 & | & -1 \\ 0 & 1 & \frac{1}{3} & | & \frac{4}{3} \\ 0 & -2 & -5 & | & 6 \end{bmatrix}$
Use Second Row for Elimination: Use the second row to zero out the second element in the first and third rows. $\newline$ $\text{Row 1} = \text{Row 1} - \text{Row 2}$ $\newline$ $\text{Row 3} = \text{Row 3} + 2 \times \text{Row 2}$ $\newline$ $\begin{bmatrix} 1 & 0 & \frac{5}{3} & | & -\frac{7}{3} \\ 0 & 1 & \frac{1}{3} & | & \frac{4}{3} \\ 0 & 0 & -\frac{7}{3} & | & \frac{14}{3} \end{bmatrix}$
Third Row Normalization: Make the third element of the third row a $1$ by dividing the whole row by - $\frac{7}{3}$ . $\newline$ $\text{Row 3} = \frac{1}{-\frac{7}{3}} \times \text{Row 3}$ $\newline$ $\begin{bmatrix} 1 & 0 & \frac{5}{3} & | & -\frac{7}{3} \\ 0 & 1 & \frac{1}{3} & | & \frac{4}{3} \\ 0 & 0 & 1 & | & -2 \end{bmatrix}$
Use Third Row for Elimination: Use the third row to zero out the third element in the first and second rows. $\newline$ $\text{Row 1} = \text{Row 1} - \frac{5}{3} \times \text{Row 3}$ $\newline$ $\text{Row 2} = \text{Row 2} - \frac{1}{3} \times \text{Row 3}$ $\newline$ $\begin{bmatrix} 1 & 0 & 0 & | & 3 \\ 0 & 1 & 0 & | & 5 \\ 0 & 0 & 1 & | & -2 \end{bmatrix}$