# matrix gauss-jordan method

• Dec 20th 2009, 04:17 AM
jayjay
matrix gauss-jordan method
Hi guys, the question is:

5x+3y+6=0
3x+5y+18=0

It should be solved by gauss-jordan method.
and the correct answers are x=1.5 and y=-4.5

when i put them in a matrix form and try to make column 1 row 2 to 0
• Dec 20th 2009, 06:15 AM
Soroban
Hello, jayjay!

Quote:

Solve by Gauss-Jordan method: .$\displaystyle \begin{array}{ccc}5x+3y+6&=& 0 \\ 3x+5y+18&=&0 \end{array}$
This is how I would solve it . . .

Switch the equations: .$\displaystyle \begin{array}{ccc} 3x + 5y &=& \text{-}18 \\ 5x + 3y &=& \text{-}6 \end{array}$

We have: .$\displaystyle \left|\begin{array}{cc|c} 3 & 5 & \text{-}18 \\ 5&3 & \text{-}6 \end{array}\right|$

$\displaystyle \begin{array}{c}2R_1-R_2 \\ \\ \end{array} \left|\begin{array}{cc|c}1 & 7 & \text{-}30 \\ 5 & 3 & \text{-}6 \end{array}\right|$

$\displaystyle \begin{array}{c} \\ R_2-5R_1 \end{array} \left|\begin{array}{cc|c}1 & 7 & \text{-}30 \\ 0 & \text{-}32 & 144 \end{array}\right|$

. . $\displaystyle \begin{array}{c}\\ \text{-}\frac{1}{32}R_2 \end{array} \left|\begin{array}{cc|c}1 & 7 & \text{-}30 \\ 0 & 1 & \text{-}\frac{9}{2}\end{array}\right|$

$\displaystyle \begin{array}{c}R_1 - 7R_2 \\ \\ \end{array} \left|\begin{array}{cc|c}1 & 0 & \frac{3}{2} \\ \\[-4mm] 0 & 1 & \text{-}\frac{9}{2}\end{array}\right|$

Therefore: .$\displaystyle \begin{Bmatrix}x &=& \dfrac{3}{2} \\ \\[-3mm] y &=& \text{-}\dfrac{9}{2} \end{Bmatrix}$

• Dec 20th 2009, 06:17 AM
HallsofIvy
Quote:

Originally Posted by jayjay
Hi guys, the question is:

5x+3y+6=0
3x+5y+18=0

It should be solved by gauss-jordan method.
and the correct answers are x=1.5 and y=-4.5

when i put them in a matrix form and try to make column 1 row 2 to 0
Did you get $\displaystyle \begin{bmatrix{5 & 3 & 6 \\ 3 & 5 & 18\end{bmatrix}$ as your matrix or did you use $\displaystyle \begin{bmatrix}5 & 3 & -6 \\ 3 & 5 & -18\end{bmatrix}$? Both are okay but the thought processes are little different. Since there are only two rows, I don't know what you mean by "$\displaystyle r_3$". Did you mean $\displaystyle r_1$, the first row?
Replacing the second row with "1/3 the second row minus 1/5 the first row gives $\displaystyle \begin{bmatrix}5 & 3 & 6 \\ 0 5/3- 3/5 & 6- 6/5\end{bmatrix}= \begin{bmatrix}5 & 2 & 6 \\ 0 & (25- 9)/15 & (30- 6)/5\end{bmatrix}$$\displaystyle = \begin{bmatrix}5 & 2 & 6 \\0 & 16/15 & 24/5\end{bmatrix}$