solving using matricies: x - y + z = 5, 2x + 3y - 3z = 0
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Check: If x = 3 and y and z can be anything, then plug in "3" for "x" and pick any numbers for y and z; say, y = z = 0:bailey07 said:
. . . . .x - y + z = 5
. . . . .[3] - [0] + [0] ?=? 5
. . . . .3 + 0 ?=? 5
. . . . .3 != 5
So the solution does not work in the first equation. (Once a solution fails, there is no need to check it in other equations.) Therefore, your solution cannot be correct (and one would have expected the solution to be parametrised, in any case).
Please reply showing your work and reasoning. Thank you!
Eliz.
Take 3 times the first: 3x-3y+3z = 15; add to second, and get 5x = 15, so x = 3.
Now we have -y+z = 2
3y - 3z = -6
the second is a multiple of the first; therefore there are infinitely many solutions, e.g., y=2, z =4, or y =1, z = 3. So while x = 3, there are not unique solutions for y, z.
Now we have -y+z = 2
3y - 3z = -6
the second is a multiple of the first; therefore there are infinitely many solutions, e.g., y=2, z =4, or y =1, z = 3. So while x = 3, there are not unique solutions for y, z.
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You have:bailey07 said:
. . . . .\(\displaystyle \left[\begin{array}{rrrr}1&-1&1&5\\2&3&-3&0\end{array}\right]\)
You multiplied the first row by 3, giving "3 -3 3 15". Then you added this to the second row, which gives the new (but equivalent) matrix:
. . . . .\(\displaystyle \left[\begin{array}{rrrr}1&-1&1&5\\5&0&0&15\end{array}\right]\)
Dividing the (new) second row by 5 gives:
. . . . .\(\displaystyle \left[\begin{array}{rrrr}1&-1&1&5\\1&0&0&3\end{array}\right]\)
At this stage, I would subtract the second row from the first to get:
. . . . .\(\displaystyle \left[\begin{array}{rrrr}0&-1&1&2\\1&0&0&3\end{array}\right]\)
...so that x = 3, and z - y = 2. Then z could be said to equal y + 2, and the full solution would have been (3, y, y + 2), with perhaps some parameter (a, q, t, etc) in place of "y".
You apparently ended up with something quite different...? Somehow the rows were the same...?
How did you arrive at your solution? (We're still missing the steps where you entirely eliminated the y and z variables, making them entirely "free".)
Please be complete, showing all of your steps, as I did above. Thank you!
Eliz.
ok but if you put the answers you got back into the equation, you get:
3-y+y+3=5
6!=5 <not equal
2(3) +3(y) -3(y+3)=0
6+3y-3y-9=0
-3!=0
did i do something wrong there?
it just seems to me that when you put them back into the equation they arent the right answers
3-y+y+3=5
6!=5 <not equal
2(3) +3(y) -3(y+3)=0
6+3y-3y-9=0
-3!=0
did i do something wrong there?
it just seems to me that when you put them back into the equation they arent the right answers