Matrices

**cuteisa89** · June 26th 2007, 02:48 PM

Can someone help me with this problem? Thanks!

Solve the systems of equations. Use Gaussian elimination with back-substition or Gauss-Jordan Elimination.

2x - y + 3z = 24
2y - z = 14
7x - 5y = 6

**Soroban** · June 26th 2007, 03:30 PM

Hello, cuteisa89!

Solve the system of equations.
2x - y + 3z = 24
2y - z = 14
7x - 5y = 6

We have: . $\begin{vmatrix}2 & \text{-}1 & 3 & | & 24 \\ 0 & 2 & \text{-}1 & | & 14 \\ 7 & \text{-}5 & 0 & | & 6\end{vmatrix}$

$\begin{array}{c}\text{Move }R_3 \\ \text{to }R_1 \end{array}\; \begin{vmatrix}7 & \text{-}5 & 0 & | & 6 \\ 2 & \text{-}1 & 3 & | & 24 \\ 0 & 2 & \text{-}1 & | & 14\end{vmatrix}$

$\begin{array}{c}R_1-3R_2 \\ \\ \\ \end{array}\;\begin{vmatrix}1 & \text{-}2 & \text{-}9 & | & \text{-}66 \\ 2 & \text{-}1 & 3 & | & 24 \\ 0 & 2 & \text{-}1 & | & 14\end{vmatrix}$

$\begin{array}{c} \\ \\ R_2-2R_1 \\ \end{array}\; \begin{vmatrix}1 & \text{-}2 & \text{-}9 & | & \text{-}66 \\ 0 & 3 & 21 & | & 156 \\ 0 & 2 & \text{-}1 & | & 14\end{vmatrix}$

$\begin{array}{c} \\ \frac{1}{3}R_2 \\ \\ \end{array}\; \begin{vmatrix}1 & \text{-}2 & \text{-}9 & | & \text{-}66 \\ 0 & 1 & 7 & | & 52 \\ 0 & 2 & \text{-}1 & | & 14\end{vmatrix}$

$\begin{array}{c}R_1+2R_2 \\ \\ R_3-2R_2\end{array}\; \begin{vmatrix}1 & 0 & 5 & | & 38 \\ 0 & 1 & 7 & | & 52 \\ 0 & 0 & \text{-}15 & | & \text{-}90\end{vmatrix}$

$\begin{array}{c} \\ \\ -\frac{1}{15}R_3\end{array}\; \begin{vmatrix}1 & 0 & 5 & | & 38 \\ 0 & 1 & 7 & | & 52 \\ 0 & 0 & 1 & | & 6\end{vmatrix}$

$\begin{array}{c}R_1-5R_3 \\ R_2-7R_3\\ \\ \end{array}\; \begin{vmatrix}1 & 0 & 0 & | & 8 \\ 0 & 1 & 0 & | & 10 \\ 0 & 0 & 1 & | & 6\end{vmatrix}$

Therefore: . $(x,\,y,\,z)\;=\;(8,\,10,\,6)$

**cuteisa89** · June 26th 2007, 03:45 PM

Thank you so much!! I've been working on that problem for so long!

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