Question

Solve the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations.

Answer 1

Answer:

x = 1 , y = - 1 , z = 2

Step-by-step explanation:

$\begin{pmatrix}1&1&-1&-2\\ 2&-1&3&9\\ 1&-1&-2&1\end{pmatrix}\\\\\\= \begin{pmatrix}2&-1&3&9\\ 1&1&-1&-2\\ 1&-1&-2&1\end{pmatrix}$                          $[ \ swap \ R_1 \ and \ R_2 \ ]$

$=\begin{pmatrix}2&-1&3&9\\ 0&\frac{3}{2}&-\frac{5}{2}&-\frac{13}{2}\\ 1&-4&-2&1\end{pmatrix}$                        $[ \ R_2 = R_2 - \frac{1}{2} R_1 \ ]$

$=\begin{pmatrix}2&-1&3&9\\ 0&\frac{3}{2}&-\frac{5}{2}&-\frac{13}{2}\\ 0&-\frac{7}{2}&-\frac{7}{2}&-\frac{7}{2}\end{pmatrix}$                        $[ \ R_3 = R_3 - \frac{1}{2} R_1 \ ]$

$=\begin{pmatrix}2&-1&3&9\\ 0&-\frac{7}{2}&-\frac{7}{2}&-\frac{7}{2}\\ 0&\frac{3}{2}&-\frac{5}{2}&-\frac{13}{2}\end{pmatrix}$                      $[ \ swap \ R_2 \ and \ R_3 \ ]$

$=\begin{pmatrix}2&-1&3&9\\ 0&-\frac{7}{2}&-\frac{7}{2}&-\frac{7}{2}\\ 0&0&-4&-8\end{pmatrix}$                        $[ \ R_3 = R_3 + \frac{3}{7}R_2 \ ]$

Therefore,

           $-4z = - 8\\\\z = 2$

        $-\frac{7}{2} y - \frac{7}{2}z = -\frac{7}{2}\\\\y + z = 1\\\\y + 2 = 1 \\\\y = 1 - 2 = - 1$

      $2x - 1y + 3z = 9\\\\2x -1(-1) + 3(2) = 9\\\\2x + 1 + 6 = 9\\\\2x = 9 - 7\\\\2x = 2 \\\\x = 1$