Solve the system of linear equations using the Gauss-Jordan elimination method. 2x1 − x2 + 3x3 = −10 x1 − 2x2 + x3 = −3 x1 − 5x2 + 2x3 = −7 (x1, x2, x3) =

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]