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$x_{1}+x_{2}+2x_{3}=-1$ $\newline$ $x_{1}-2x_{2}+x_{3}=-5$ $\newline$ $3x_{1}+x_{2}+x_{3}=3$ $\newline$ find all solutions by using the Gaussian elimination or Gauss-Jordan Reduction.

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Write an equation word problem

Full solution

Q.

x_{1}+x_{2}+2x_{3}=-1

\newline

x_{1}-2x_{2}+x_{3}=-5

\newline

3x_{1}+x_{2}+x_{3}=3

\newline

find all solutions by using the Gaussian elimination or Gauss-Jordan Reduction.

Write Augmented Matrix: Write the augmented matrix for the system of equations. $\newline$ $\begin{bmatrix} 1 & 1 & 2 & | & -1 \\ 1 & -2 & 1 & | & -5 \\ 3 & 1 & 1 & | & 3 \end{bmatrix}$
Leading $1$ in R $1$ : Perform row operations to get a leading $1$ in the first row, first column ( $R1$ ). $\newline$ No changes needed as the first element is already $1$ .
Zeros in First Column: Make zeros below the leading $1$ in the first column using R $2$ - R $1$ → R $2$ and R $3$ - $3$ R $1$ → R $3$ . $\newline$ $\begin{bmatrix} 1 & 1 & 2 & | & -1 \\ 0 & -3 & -1 & | & -4 \\ 0 & -2 & -5 & | & 6 \end{bmatrix}$
Leading $1$ in R $2$ : Make the second column's second row element into a leading $1$ by multiplying R $2$ by $-1$ / $3$ . $\newline$ $\begin{bmatrix} 1 & 1 & 2 & | & -1 \\ 0 & 1 & 1/3 & | & 4/3 \\ 0 & -2 & -5 & | & 6 \end{bmatrix}$
Eliminate First Column: Eliminate the first column's second row element by adding R $2$ to R $1$ and add $2$ R $2$ to R $3$ . $\newline$ $\begin{bmatrix} 1 & 0 & 7/3 & | & 1/3 \\ 0 & 1 & 1/3 & | & 4/3 \\ 0 & 0 & -13/3 & | & 14/3 \end{bmatrix}$
Leading $1$ in R $3$ : Convert the third row's third column element to a leading $1$ by multiplying R $3$ by $-3$ / $13$ . $\newline$ $\begin{bmatrix} 1 & 0 & 7/3 & | & 1/3 \\ 0 & 1 & 1/3 & | & 4/3 \\ 0 & 0 & 1 & | & -14/13 \end{bmatrix}$
Eliminate Third Column: Eliminate the third column's first and second row elements by subtracting ( $7$ / $3$ )R $3$ from R $1$ and subtracting ( $1$ / $3$ )R $3$ from R $2$ . $\newline$ $\begin{bmatrix} 1 & 0 & 0 & | & 29/39 \\ 0 & 1 & 0 & | & 50/39 \\ 0 & 0 & 1 & | & -14/13 \end{bmatrix}$
Check Solution: Check the solution by substituting back into the original equations. $\newline$ Substituting $x_1 = 29/39, x_2 = 50/39, x_3 = -14/13$ into the original equations confirms the solution is correct.

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