Respuesta :
Answer:
The solution is: [tex](x_{1}, x_{2}, x_{3}) = (1,0,-4)[/tex]
Step-by-step explanation:
The Gauss-Jordan elimination method is done by transforming the system's augmented matrix into reduced row-echelon form by means of row operations.
We have the following system:
[tex]2x_{1} - x_{2} + 3x_{3} = -10[/tex]
[tex]x_{1} - 2x_{2} + x_{3} = -3[/tex]
[tex]x_{1} - 5x_{2} + 2x_{3} = -7[/tex]
This system has the following augmented matrix:
[tex]\left[\begin{array}{ccc}2&-1&3|-10\\1&-2&1|-3\\1&-5&2| -7\end{array}\right][/tex]
To make the reductions easier, i am going to swap the first two lines. So
[tex]L1 <-> L2[/tex]
Now the matrix is:
[tex]\left[\begin{array}{ccc}1&-2&1|-3\\2&-1&3|-10\\1&-5&2| -7\end{array}\right][/tex]
Now we reduce the first row, doing the following operations
[tex]L2 = L2 - 2L1[/tex]
[tex]L3 = L3 - L1[/tex]
So, the matrix is:
[tex]\left[\begin{array}{ccc}1&-2&1|-3\\0&3&1|-4\\0&-3&1| -4\end{array}\right][/tex]
Now we divide L2 by 3
[tex]L2 = \frac{L2}{3}[/tex]
So we have
[tex]\left[\begin{array}{ccc}1&-2&1|-3\\0&1&\frac{1}{3}|\frac{-4}{3}\\0&-3&1| -4\end{array}\right][/tex]
Now we have:
[tex]L3 = 3L2 + L3[/tex]
So, now we have our row reduced matrix:
[tex]\left[\begin{array}{ccc}1&-2&1|-3\\0&1&\frac{1}{3}|\frac{-4}{3}\\0&0&2| -8\end{array}\right][/tex]
We start from the bottom line, where we have:
[tex]2x_{3} = -8[/tex]
[tex]x_{3} = \frac{-8}{2}[/tex]
[tex]x_{3} = -4[/tex]
At second line:
[tex]x_{2} + \frac{x_{3}}{3} = \frac{-4}{3}[/tex]
[tex]x_{2} - \frac{4}{3} = -\frac{4}{3}[/tex]
[tex]x_{2} = 0[/tex]
At the first line
[tex]x_{1} -2x_{2} + x_{3} = -3[/tex]
[tex]x_{1} - 4 = -3[/tex]
[tex]x_{1} = 1[/tex]
The solution is: [tex](x_{1}, x_{2}, x_{3}) = (1,0,-4)[/tex]
