Help! Matrix, Gauss-Jordan elimination!

**Moon Hoplite** · December 21st 2008, 05:49 PM

Q1. Solve the following system of equations by hand. Use Gauss-Jordan elimination, on the augmented matrix, and write the row operation you used next to each new row.

3x + 2y + z = 5
x + y- z = 0
4x - y + 5z = 3

**Soroban** · December 22nd 2008, 05:05 AM

Hello, Moon Hoplite!

The problem is straight-forward.
Exactly where is your difficulty?

$1)\;\;\begin{array}{ccc}3x + 2y + z &=& 5 \\x + y- z &=& 0 \\4x - y + 5z &= &3\end{array}$

Re-order the equations: . $\begin{array}{ccc}x+y+z &=&0 \\ 3x+2y+z &=&5 \\ 4z-y+5z &=& 3\end{array}$ . . Yes, we can do that!

We have: . $\left[\begin{array}{ccc|c}1&1&1&0 \\ 3&2&1&5 \\ 4&\text{-}1&5&3 \end{array}\right]$

$\begin{array}{c}\\ R_2-3R_1 \\ R_3-4R_1\end{array} \left[\begin{array}{ccc|c}1&1&1&0 \\ 0&\text{-}1&\text{-}2&5 \\ 0&\text{-}5&1&3 \end{array}\right]$

$\begin{array}{c}R_1+R_2 \\ \text{-}1\!\cdot\!R_2 \\ R_3-5R_1 \end{array} \left[\begin{array}{ccc|c}1&0&\text{-}1&5 \\ 0&1&2&\text{-}5 \\ 0&0&11&\text{-}22 \end{array}\right]$

. . . $\begin{array}{c} \\ \\ \frac{1}{11}R_3\end{array} \left[\begin{array}{ccc|c}1&0&\text{-}1&5 \\ 0&1&2&\text{-}5 \\ 0&0&1&\text{-}2 \end{array}\right]$

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

Therefore: . $\begin{Bmatrix}x &=& 3 \\ y&=&\text{-}1 \\ z &=&\text{-}2 \end{Bmatrix}$

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