linear systems of equations/ matrices

**Shmomo89** · December 11th 2008, 08:02 PM

solve this system for x,y, and z

x + 3y + z = -2
2x + 5y + 3z = -7
x + 4y - 3z = 4

if you could help me step by step a little bit that would be much appreciated - thank you in advance

**Soroban** · December 11th 2008, 08:27 PM

Hello, Shmomo89!

$\begin{array}{ccc}x + 3y + z &=& \text{-}2 \\<br /> 2x + 5y + 3z &=& \text{-}7 \\x + 4y - 3z &=& 4 \end{array}$

We have: . $\left[\begin{array}{ccc|c}1 & 3 & 1 & \text{-}2 \\ 2 & 5 & 3 & \text{-}7 \\ 1 & 4 & \text{-}3 & 4 \end{array}\right]$

$\begin{array}{c}\\ R_2-2R_1 \\ R_3-R_1\end{array} \left[\begin{array}{ccc|c}1 & 3 & 1 & \text{-}2 \\ 0 & \text{-}1 & 1 & \text{-}3 \\ 0 & 1 & \text{-}4 & 6 \end{array}\right]$

$\begin{array}{c}R_1+3R_2 \\ \\ R_3+R_2\end{array} \left[\begin{array}{ccc|c}1 & 0 & 4 & \text{-}11 \\ 0 & \text{-}1 & 1 & \text{-}3 \\ 0 & 0 & \text{-}3 & 3 \end{array}\right]$

. . $\begin{array}{c}\\ \text{-}1\!\cdot\!R_2 \\ \text{-}\frac{1}{3}\!\cdot\!R_3 \end{array} \left[\begin{array}{ccc|c}1 & 0 & 4 & \text{-}11\\ 0 & 1 & \text{-}1 & 3 \\ 0 & 0 & 1 & \text{-}1 \end{array}\right]$

$\begin{array}{c} R_1-4R_3 \\ R_2 + R_3 \\ \\ \end{array} \left[\begin{array}{ccc|c}1&0&0 & \text{-}7 \\ 0&1&0 & 2 \\ 0&0&1 & \text{-}1 \end{array}\right]$

Therefore: . $\begin{Bmatrix}x &=& \text{-}7 \\ y &=& 2 \\ z &=& \text{-}1\end{Bmatrix}$

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