The solution to the given set of linear equations using the Gauss-Jordan elimination method is [tex]x = \frac{1}{3}, y =\frac{10}{3} , z =- \frac{5}{3}[/tex]. The correct option is A. [tex]x = \frac{1}{3}, y =\frac{10}{3} , z =- \frac{5}{3}[/tex]
Gauss-Jordan elimination method
The given system of equations are
x + y + z =2
2x-3y+z= -11
-x+2y-z=8
First, convert to an augmented matrix. This becomes
[tex]\left[\begin{array}{ccc|c}1&1&1&2\\2&-3&1&-11\\-1&2&-1&8\end{array}\right][/tex]
Now, we will convert the matrix to reduced row echelon form
Subtract row 1 multiplied by 2 from row 2. That is, R₂= R₂-2R₁
[tex]\left[\begin{array}{ccc|c}1&1&1&2\\0&-5&-1&-15\\-1&2&-1&8\end{array}\right][/tex]
Add row 1 to row 3. That is, R₃ = R₃ + R₁
[tex]\left[\begin{array}{ccc|c}1&1&1&2\\0&-5&-1&-15\\0&3&0&10\end{array}\right][/tex]
Divide row 2 by -5. That is, R₂ = -R₂/5
[tex]\left[\begin{array}{ccc|c}1&1&1&2\\0&1&\frac{1}{5} &3\\0&3&0&10\end{array}\right][/tex]
Subtract row 2 from row 1. That is, R₁ = R₁ - R₂
[tex]\left[\begin{array}{ccc|c}1&0&\frac{4}{5} &-1\\0&1&\frac{1}{5} &3\\0&3&0&10\end{array}\right][/tex]
Subtract row 2 multiplied by row 3 from row 3. That is, R₃ = R₃ - 3R₂
[tex]\left[\begin{array}{ccc|c}1&0&\frac{4}{5} &-1\\0&1&\frac{1}{5} &3\\0&0&\frac{-3}{5} &1\end{array}\right][/tex]
Multiply row 3 by -5/3. That is, R₃ = -5R₃/3
[tex]\left[\begin{array}{ccc|c}1&0&\frac{4}{5} &-1\\0&1&\frac{1}{5} &3\\0&0&1&-\frac{5}{3} \end{array}\right][/tex]
Subtract row 3 multiplied by 4/5 from row 1. That is, R₁ = R₁ - 4R₃/5
[tex]\left[\begin{array}{ccc|c}1&0&0 &\frac{1}{3} \\0&1&\frac{1}{5} &3\\0&0&1&-\frac{5}{3} \end{array}\right][/tex]
Subtract row 3 multiplied by 1/5 from row 2. That is, R₂ = R₂ - R₃/5
[tex]\left[\begin{array}{ccc|c}1&0&0 &\frac{1}{3} \\0&1&0&\frac{10}{3} \\0&0&1&-\frac{5}{3} \end{array}\right][/tex]
From the result above, [tex]x = \frac{1}{3}, y =\frac{10}{3} , z =- \frac{5}{3}[/tex]
Hence, the solution to the given set of linear equations using the Gauss-Jordan elimination method is [tex]x = \frac{1}{3}, y =\frac{10}{3} , z =- \frac{5}{3}[/tex]. The correct option is A. [tex]x = \frac{1}{3}, y =\frac{10}{3} , z =- \frac{5}{3}[/tex]
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