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)?

- sloppycanada

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- rishavraj

u know Cramer's Rule??

- sloppycanada

I don't think that was covered.

- 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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## More answers

- rishavraj

where
|dw:1435751082539:dw|

- sloppycanada

Okay... so I'm phrasing mine wrong?

- rishavraj

hmmm wht do u mean?? u tried it using matrices??

- sloppycanada

I think so, I'm thinking it's [2 3]

- rishavraj

u hav been askd to solve using matrices ...thts why u need to solve it tht way

- sloppycanada

but I don't understand how I'm supposed to do that. Just plug in the equations to what you gave me?

- rishavraj

yup....bt use with appropriate signs.....the answers will remain the same....kinda obvious :P

- sloppycanada

|dw:1435751899086:dw|

- rishavraj

like the equations are x + y = 5 and 2x - y = 1
\[a_1 = 1~~~~b_1 = 1~~~~~c_1 = 5 \]

- rishavraj

\[a_2 = 2~~~~b_2 = -1~~~~c_2 = 1\]

- sloppycanada

Oohhhhh then I add them? Or subtract?

- rishavraj

|dw:1435752333034:dw|

- rishavraj

solve D , D_1, D_2

- rishavraj

after tht
\[x = \frac{ D_1 }{ D } ~~~ and y = \frac{ D_2 }{ D }\]

- sloppycanada

|dw:1435752595951:dw|

- sloppycanada

|dw:1435752646115:dw|

- rishavraj

on solving those determinants
\[D = -3~~~~~D_1 = -6~~~~~~~~D_2 = -9\]

- sloppycanada

But it's divided... I am so confused..

- rishavraj

u need to first of all calculate D, D_1 and D_2
and then divide ..

- sloppycanada

So for the first one -
[2]
[3]

- rishavraj

u got the values of all deteminants???

- sloppycanada

-3, -6, and -9

- rishavraj

ooo cool....xD so u get x = 2
and y = 3
:))

- sloppycanada

Okay, so I was right to begin with... I just did it a different way (elimination method)

- sloppycanada

But I should phrase it [2 3]?

- rishavraj

i don't think so.....

- 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

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

we just grab the values.. you don't have to write x or y in the matrix as that is not necessary

- UsukiDoll

|dw:1435753678426:dw|
you should have a matrix that looks like this and looks like we need the Reduced row echelon form

- 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

do you know any row operations ? @sloppycanada

- sloppycanada

Yes? + or - and multiplace and division... but generally you need to [] [] of those things.

- 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

\[r_1 \leftrightarrow r_2\]
we can swap rows.. this means switch row 1 and row 2

- 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

|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

The adding thing. r1 + r2?

- 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

so since I see a 2 on the bottom of the main diagonal I have to multiply what number to the first row

- sloppycanada

2?

- UsukiDoll

yes... close... but switch signs

- UsukiDoll

what is the opposite of positive?

- UsukiDoll

instead of 2 it has to be - ?

- UsukiDoll

what's (-1)x2?

- sloppycanada

-2

- 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

so what's
-2+2
-2-1
-10+1

- sloppycanada

0
-3
-9

- 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

err I meant 0 -3 l -9

- 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

so does 0 -3 l -9 belong in the first row or second row?

- sloppycanada

Okay.. I think I'm starting to get it, kinda.

- sloppycanada

basically once you find the determinant of the first A you put it into the original matrix and then divide

- 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

determinant for 3 x 3 is possible but nasty!

- sloppycanada

I have one for practice that I've been doing as I've been listening or watching your explanation.

- UsukiDoll

|dw:1435754875634:dw|

- 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

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

|dw:1435755082603:dw|

- UsukiDoll

|dw:1435755112509:dw|

- UsukiDoll

first column is x second column is y
eventually third column is z followed by w or something similar like that

- sloppycanada

Something like this - Solve x – 3y + z = -9 using matrices.
x + y – 3z = -5
3x – y + z = 11

- UsukiDoll

oh lord... well weeee we need a 3 x 4 augmented matrix.

- UsukiDoll

|dw:1435755254310:dw|

- UsukiDoll

the main diagonal is in entries \[a_{11},a_{22},a_{33} \]

- sloppycanada

I got this - x = 4
y = 6
z = 5

- 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

3x-y+z = 11
3(4)-6+5=11
12-6+5=11
6+5=11
11=11

- UsukiDoll

x+y-3z = -5
4+6-3(5)=-5
4+6-15=-5
10-15=-5
-5=-5

- UsukiDoll

x-3y+z = -9
4-3(6)+5=-9
4-18+5=-9
-14+5=-9
-9=-9

- UsukiDoll

yup everything checks out

- sloppycanada

Thanks so much I just have five more of these stupid review problems then I can do fun things!

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