sloppycanada
  • sloppycanada
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
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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rishavraj
  • rishavraj
u know Cramer's Rule??
sloppycanada
  • sloppycanada
I don't think that was covered.
rishavraj
  • rishavraj
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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rishavraj
  • rishavraj
where |dw:1435751082539:dw|
sloppycanada
  • sloppycanada
Okay... so I'm phrasing mine wrong?
rishavraj
  • rishavraj
hmmm wht do u mean?? u tried it using matrices??
sloppycanada
  • sloppycanada
I think so, I'm thinking it's [2 3]
rishavraj
  • rishavraj
u hav been askd to solve using matrices ...thts why u need to solve it tht way
sloppycanada
  • sloppycanada
but I don't understand how I'm supposed to do that. Just plug in the equations to what you gave me?
rishavraj
  • rishavraj
yup....bt use with appropriate signs.....the answers will remain the same....kinda obvious :P
sloppycanada
  • sloppycanada
|dw:1435751899086:dw|
rishavraj
  • rishavraj
like the equations are x + y = 5 and 2x - y = 1 \[a_1 = 1~~~~b_1 = 1~~~~~c_1 = 5 \]
rishavraj
  • rishavraj
\[a_2 = 2~~~~b_2 = -1~~~~c_2 = 1\]
sloppycanada
  • sloppycanada
Oohhhhh then I add them? Or subtract?
rishavraj
  • rishavraj
|dw:1435752333034:dw|
rishavraj
  • rishavraj
solve D , D_1, D_2
rishavraj
  • rishavraj
after tht \[x = \frac{ D_1 }{ D } ~~~ and y = \frac{ D_2 }{ D }\]
sloppycanada
  • sloppycanada
|dw:1435752595951:dw|
sloppycanada
  • sloppycanada
|dw:1435752646115:dw|
rishavraj
  • rishavraj
on solving those determinants \[D = -3~~~~~D_1 = -6~~~~~~~~D_2 = -9\]
sloppycanada
  • sloppycanada
But it's divided... I am so confused..
rishavraj
  • rishavraj
u need to first of all calculate D, D_1 and D_2 and then divide ..
sloppycanada
  • sloppycanada
So for the first one - [2] [3]
rishavraj
  • rishavraj
u got the values of all deteminants???
sloppycanada
  • sloppycanada
-3, -6, and -9
rishavraj
  • rishavraj
ooo cool....xD so u get x = 2 and y = 3 :))
sloppycanada
  • sloppycanada
Okay, so I was right to begin with... I just did it a different way (elimination method)
sloppycanada
  • sloppycanada
But I should phrase it [2 3]?
rishavraj
  • rishavraj
i don't think so.....
mathmate
  • mathmate
@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.
UsukiDoll
  • UsukiDoll
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
UsukiDoll
  • UsukiDoll
we just grab the values.. you don't have to write x or y in the matrix as that is not necessary
UsukiDoll
  • UsukiDoll
|dw:1435753678426:dw| you should have a matrix that looks like this and looks like we need the Reduced row echelon form
UsukiDoll
  • UsukiDoll
|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
UsukiDoll
  • UsukiDoll
do you know any row operations ? @sloppycanada
sloppycanada
  • sloppycanada
Yes? + or - and multiplace and division... but generally you need to [] [] of those things.
UsukiDoll
  • UsukiDoll
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
UsukiDoll
  • UsukiDoll
\[r_1 \leftrightarrow r_2\] we can swap rows.. this means switch row 1 and row 2
UsukiDoll
  • UsukiDoll
\[r_1+r_2 \rightarrow r_2 \] meaning we add row one to row two and our result goes to row 2 same thing applies with subtraction
UsukiDoll
  • UsukiDoll
|dw:1435753941663:dw| so back to this problem what row operations do we need to get rid of that 2 on the bottom diagonal?
sloppycanada
  • sloppycanada
The adding thing. r1 + r2?
UsukiDoll
  • UsukiDoll
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
UsukiDoll
  • UsukiDoll
so since I see a 2 on the bottom of the main diagonal I have to multiply what number to the first row
sloppycanada
  • sloppycanada
2?
UsukiDoll
  • UsukiDoll
yes... close... but switch signs
UsukiDoll
  • UsukiDoll
what is the opposite of positive?
UsukiDoll
  • UsukiDoll
instead of 2 it has to be - ?
UsukiDoll
  • UsukiDoll
what's (-1)x2?
sloppycanada
  • sloppycanada
-2
UsukiDoll
  • UsukiDoll
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|
UsukiDoll
  • UsukiDoll
so what's -2+2 -2-1 -10+1
sloppycanada
  • sloppycanada
0 -3 -9
UsukiDoll
  • UsukiDoll
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.
UsukiDoll
  • UsukiDoll
err I meant 0 -3 l -9
UsukiDoll
  • UsukiDoll
and we need a 1 on the top left ( meaning the entry of the first row and first column should always be 1)
UsukiDoll
  • UsukiDoll
so does 0 -3 l -9 belong in the first row or second row?
sloppycanada
  • sloppycanada
Okay.. I think I'm starting to get it, kinda.
sloppycanada
  • sloppycanada
basically once you find the determinant of the first A you put it into the original matrix and then divide
UsukiDoll
  • UsukiDoll
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
UsukiDoll
  • UsukiDoll
determinant for 3 x 3 is possible but nasty!
sloppycanada
  • sloppycanada
I have one for practice that I've been doing as I've been listening or watching your explanation.
UsukiDoll
  • UsukiDoll
|dw:1435754875634:dw|
UsukiDoll
  • UsukiDoll
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 ???
UsukiDoll
  • UsukiDoll
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
UsukiDoll
  • UsukiDoll
|dw:1435755082603:dw|
UsukiDoll
  • UsukiDoll
|dw:1435755112509:dw|
UsukiDoll
  • UsukiDoll
first column is x second column is y eventually third column is z followed by w or something similar like that
sloppycanada
  • sloppycanada
Something like this - Solve x – 3y + z = -9 using matrices. x + y – 3z = -5 3x – y + z = 11
UsukiDoll
  • UsukiDoll
oh lord... well weeee we need a 3 x 4 augmented matrix.
UsukiDoll
  • UsukiDoll
|dw:1435755254310:dw|
UsukiDoll
  • UsukiDoll
the main diagonal is in entries \[a_{11},a_{22},a_{33} \]
sloppycanada
  • sloppycanada
I got this - x = 4 y = 6 z = 5
UsukiDoll
  • UsukiDoll
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.
UsukiDoll
  • UsukiDoll
3x-y+z = 11 3(4)-6+5=11 12-6+5=11 6+5=11 11=11
UsukiDoll
  • UsukiDoll
x+y-3z = -5 4+6-3(5)=-5 4+6-15=-5 10-15=-5 -5=-5
UsukiDoll
  • UsukiDoll
x-3y+z = -9 4-3(6)+5=-9 4-18+5=-9 -14+5=-9 -9=-9
UsukiDoll
  • UsukiDoll
yup everything checks out
sloppycanada
  • sloppycanada
Thanks so much I just have five more of these stupid review problems then I can do fun things!

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