Someone please help! Row operation problem

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October 1st 2012, 06:06 PM
Niaboc

Someone please help! Row operation problem

x + y - z = 6
2x - y + z = -6
x - 4y +3z = -36

Any help please!
October 1st 2012, 07:30 PM
mattsvip85

Re: Someone please help! Row operation problem

You can do these on your calculator fairly easily depending on your calculator. However this one isn't too hard to do by hand.
It helps to number the equations you have:
1) x + y - z = 6
2) 2x - y + z = -6
3) x - 4y +3z = -36

So starting with equation 1) we can re-arrange to...
x = 6 - y + z
Then you substitue this value of x into equation 2) which gives...
2*(6 - y + z) - y + z = -6
(expand the brackets and simplify)
12 - 3y + 3z = -6
(rearrange)
3z = -18 + 3y
(divide by 3)
z= (-6 + y)
Finally you substitue the values of x and z into equation 3) which gives...
(6 - y + z) - 4y + 3(-6 + y) = -36
(expand the brackets and simplify)
-12 - 2y + z = -36
Again we substitute the value we found for z above so...
-12 - 2y + (-6 + y) = -36
(simplify)
-18 - y = -36
-y = -18
(multiply by -1)
y = 18
Now that we have this we can easily find x and z
z = (-6 + y)
z = 12
and
x = 6 - y + z
x = 6 - 18 + 12
x = 0
October 1st 2012, 08:41 PM
Soroban

Re: Someone please help! Row operation problem

Hello, Niaboc!

You said "row operations".
Are you working with Gaussian elimination?

Quote:

$\begin{array}{ccc}x + y - z &=& 6 \\ 2x - y + z &=& \text{-}6 \\ x - 4y +3z &=& \text{-}36 \end{array}$

We have: . $\left|\begin{array}{ccc|c} 1&1&\text{-}1 & 6 \\ 2 &\text{-}1&1&\text{-}6 \\ 1&\text{-}4&3&\text{-}36 \end{array}\right|$

$\begin{array}{c}\\ R_2-2R_1 \\ R_3-R_1 \end{array}\:\left|\begin{array}{ccc|c} 1&1&\text{-}1 & 6 \\ 0&\text{-}3&3&\text{-}18 \\ 0&\text{-}5&4&\text{-}42 \end{array}\right|$

. . . $\begin{array}{c}\\ \text{-}\frac{1}{3}R_2 \\ \\ \end{array}\:\left|\begin{array}{ccc|c} 1&1&\text{-}1&6 \\ 0&1&\text{-}1&6 \\ 0&\text{-}5&4&\text{-}42 \end{array}\right|$

$\begin{array}{c}R_1-R_2 \\ \\ R_3 + 5R_2 \end{array}\:\left|\begin{array}{ccc|c}1&0&0&0 \\ 0&1&\text{-}1&6 \\ 0&0&\text{-}1&\text{-}12 \end{array}\right|$

. . . $\begin{array}{c}\\ \\ \text{-}1\!\cdot\!R_3 \end{array}\:\left|\begin{array}{ccc|c}1&0&0&0 \\ 0&1&-1&6 \\ 0&0&1&12 \end{array}\right|$

. . $\begin{array}{c} \\ R_2+R_3 \\ \\ \end{array}\:\left|\begin{array}{ccc|c}1&0&0&0 \\ 0&1&0&18 \\ 0&0&1&12 \end{array}\right|$

Therefore: . $\begin{Bmatrix}x &=& 0 \\ y &=& 18 \\ z &=& 12\end{Bmatrix}$