Solve x + y = 5 using matrices. 2x – y = 1 I have the answer, x = 2 and y =3, how do I phrase this as a matrices? (2,3)?

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Solve x + y = 5 using matrices. 2x – y = 1 I have the answer, x = 2 and y =3, how do I phrase this as a matrices? (2,3)?

Mathematics
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u know Cramer's Rule??
I don't think that was covered.
Cramer's Rule. If the equations are \[a_1 x + b_1 y = c_1\] \[a_2 x + b_2 y = c_2\] then the solutions are are \[x = \frac{ D_1 }{ D } ~~~~ and ~~~~~ y = \frac{ D_2 }{ D }\]

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where |dw:1435751082539:dw|
Okay... so I'm phrasing mine wrong?
hmmm wht do u mean?? u tried it using matrices??
I think so, I'm thinking it's [2 3]
u hav been askd to solve using matrices ...thts why u need to solve it tht way
but I don't understand how I'm supposed to do that. Just plug in the equations to what you gave me?
yup....bt use with appropriate signs.....the answers will remain the same....kinda obvious :P
|dw:1435751899086:dw|
like the equations are x + y = 5 and 2x - y = 1 \[a_1 = 1~~~~b_1 = 1~~~~~c_1 = 5 \]
\[a_2 = 2~~~~b_2 = -1~~~~c_2 = 1\]
Oohhhhh then I add them? Or subtract?
|dw:1435752333034:dw|
solve D , D_1, D_2
after tht \[x = \frac{ D_1 }{ D } ~~~ and y = \frac{ D_2 }{ D }\]
|dw:1435752595951:dw|
|dw:1435752646115:dw|
on solving those determinants \[D = -3~~~~~D_1 = -6~~~~~~~~D_2 = -9\]
But it's divided... I am so confused..
u need to first of all calculate D, D_1 and D_2 and then divide ..
So for the first one - [2] [3]
u got the values of all deteminants???
-3, -6, and -9
ooo cool....xD so u get x = 2 and y = 3 :))
Okay, so I was right to begin with... I just did it a different way (elimination method)
But I should phrase it [2 3]?
i don't think so.....
@sloppycanada In math, if you are asked to use a specific method, you have to use the method specified to solve the problem. Otherwise the answer is considered wrong. If you have learned Gaussian Elimination (using matrices) it is considered a matrix method, so is Cramer's rule. The normal elimination method is not considered a matrix method.
Solve x + y = 5 using matrices. 2x – y = 1 So we need to have an augmented matrix in the form of a l b anything before the = sign goes in the a portion and anything after the = sign goes in the b portion
we just grab the values.. you don't have to write x or y in the matrix as that is not necessary
|dw:1435753678426:dw| you should have a matrix that looks like this and looks like we need the Reduced row echelon form
|dw:1435753725392:dw| see these triangles? this is where my main diagonal is located. Our goal for reduced row echelon form is to have all numbers below and above the main diagonal go to 0
do you know any row operations ? @sloppycanada
Yes? + or - and multiplace and division... but generally you need to [] [] of those things.
ok we have three of them either we can multiply a number throughout the row notation is like this \[2r_1 \] which means multiply the whole first row by 2
\[r_1 \leftrightarrow r_2\] we can swap rows.. this means switch row 1 and row 2
\[r_1+r_2 \rightarrow r_2 \] meaning we add row one to row two and our result goes to row 2 same thing applies with subtraction
|dw:1435753941663:dw| so back to this problem what row operations do we need to get rid of that 2 on the bottom diagonal?
The adding thing. r1 + r2?
ok good start but there's a row operation that we need to do first and it has something to do with the first row
so since I see a 2 on the bottom of the main diagonal I have to multiply what number to the first row
2?
yes... close... but switch signs
what is the opposite of positive?
instead of 2 it has to be - ?
what's (-1)x2?
-2
yes so we multiply -2 throughout the entire first row -2[ 1 1 l 5] and that row becomes -2 -2 l -10 now we add this to row 2 |dw:1435754420055:dw|
so what's -2+2 -2-1 -10+1
0 -3 -9
yes now we have 0 -3 -9 should we place this in the first row or in the second row? One of the rules for reduce row echelon form is that we need a 1 on the top left.
err I meant 0 -3 l -9
and we need a 1 on the top left ( meaning the entry of the first row and first column should always be 1)
so does 0 -3 l -9 belong in the first row or second row?
Okay.. I think I'm starting to get it, kinda.
basically once you find the determinant of the first A you put it into the original matrix and then divide
well determinants are for square matrices only like 2 x 2 (an n x n matrix) we have an m x n which is 2 rows 3 columns
determinant for 3 x 3 is possible but nasty!
I have one for practice that I've been doing as I've been listening or watching your explanation.
|dw:1435754875634:dw|
now we have to get rid of that 1 in the first row second column... let's repeat what we did the last time.. let's multiply the entire first row by ???
keep in mind that the numbers in the main diagonal can't be 0! What we are aiming for is using row operations to have a reduce row echelon form. In the end we should have something similar to an Identity Matrix
|dw:1435755082603:dw|
|dw:1435755112509:dw|
first column is x second column is y eventually third column is z followed by w or something similar like that
Something like this - Solve x – 3y + z = -9 using matrices. x + y – 3z = -5 3x – y + z = 11
oh lord... well weeee we need a 3 x 4 augmented matrix.
|dw:1435755254310:dw|
the main diagonal is in entries \[a_{11},a_{22},a_{33} \]
I got this - x = 4 y = 6 z = 5
ok so it's x =4, y =6 and z=5 now we just plug them back into the equations to see if they are true.
3x-y+z = 11 3(4)-6+5=11 12-6+5=11 6+5=11 11=11
x+y-3z = -5 4+6-3(5)=-5 4+6-15=-5 10-15=-5 -5=-5
x-3y+z = -9 4-3(6)+5=-9 4-18+5=-9 -14+5=-9 -9=-9
yup everything checks out
Thanks so much I just have five more of these stupid review problems then I can do fun things!

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