# Thread: matrix gauss-jordan method

1. ## 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

2. Hello, jayjay!

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

3. 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 $\displaystyle \begin{bmatrix{5 & 3 & 6 \\ 3 & 5 & 18\end{bmatrix}$ 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}$
(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.