# matrix gauss-jordan method

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• 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
by saying R2 --> r2(1/3) - r3(1/5) which is 0, but i still get the wrong answer, please help guys, thanks
• Dec 20th 2009, 06:15 AM
Soroban
Hello, jayjay!

Quote:

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

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

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

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

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

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

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

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

• 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
by saying R2 --> r2(1/3) - r3(1/5) which is 0, but i still get the wrong answer, please help guys, thanks

It would be better if you had shown what you did! What wrong answer did you get and how do you know it is wrong?

Did you get $\begin{bmatrix{5 & 3 & 6 \\ 3 & 5 & 18\end{bmatrix}$ as your matrix or did you use $\begin{bmatrix}5 & 3 & -6 \\ 3 & 5 & -18\end{bmatrix}$? Both are okay but the thought processes are little different. Since there are only two rows, I don't know what you mean by " $r_3$". Did you mean $r_1$, the first row?

Replacing the second row with "1/3 the second row minus 1/5 the first row gives $\begin{bmatrix}5 & 3 & 6 \\ 0 5/3- 3/5 & 6- 6/5\end{bmatrix}= \begin{bmatrix}5 & 2 & 6 \\ 0 & (25- 9)/15 & (30- 6)/5\end{bmatrix}$ $= \begin{bmatrix}5 & 2 & 6 \\0 & 16/15 & 24/5\end{bmatrix}$
(If you used the second form, with -6 and -18 in the last column, just make the last column negative here.)

Is that what you got? It should be easy to finish from there.