Medal
Solve by determinants.
x + y + z = 21
x=y +z = 3
x +y - z = 1

- anonymous

Medal
Solve by determinants.
x + y + z = 21
x=y +z = 3
x +y - z = 1

- schrodinger

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- jim_thompson5910

is there a typo in the second equation `x=y +z = 3` ??

- jim_thompson5910

there are 2 equal signs there

- anonymous

yes sorry its x - y + z = 3

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## More answers

- jim_thompson5910

ok thanks

- jim_thompson5910

were you able to set up the matrix?

- anonymous

yes

- jim_thompson5910

what matrix did you get

- anonymous

1 1 1
1 -1 1
1 1 -1

- anonymous

m i right?

- jim_thompson5910

good

- jim_thompson5910

that's your coefficient matrix

- jim_thompson5910

do you know how to take the determinant of that matrix?

- anonymous

no

- jim_thompson5910

|dw:1443401707454:dw|

- jim_thompson5910

focus on the first two columns
|dw:1443401787922:dw|

- jim_thompson5910

copy those columns and write it off to the right of the vertical bar
|dw:1443401811957:dw|

- jim_thompson5910

with me so far?

- anonymous

yes

- jim_thompson5910

ok so we have this extended matrix
|dw:1443401923170:dw|

- jim_thompson5910

circle the elements that fall along the diagonals like this
|dw:1443401944912:dw|
we have 3 subgroups

- anonymous

ohh now i see it

- jim_thompson5910

multiply the numbers in the subgroups
|dw:1443402013404:dw|

- jim_thompson5910

then add up the products: 1+1+1 = 3

- jim_thompson5910

next, circle the elements along the diagonals like this
|dw:1443402083998:dw|

- jim_thompson5910

multiply out the numbers in each subgroup
|dw:1443402112490:dw|

- jim_thompson5910

then add up the products
-1+1+(-1) = -1

- jim_thompson5910

The first sum was 3
the second sum was -1
subtract the sums
3 - (-1) = 3+1 = 4

- jim_thompson5910

so in the end, the determinant of that 3x3 matrix you wrote above is 4

- jim_thompson5910

there are a lot of steps here, so go back over them if you aren't sure what's going on. And you can ask about any step

- anonymous

so for x we got 3

- jim_thompson5910

we haven't found x or y yet

- anonymous

oh ok

- anonymous

whats the next step

- jim_thompson5910

do you see how I got the determinant to be 4?

- anonymous

yes you added them

- jim_thompson5910

so do you see how to find the determinant of any 3x3 matrix?

- anonymous

yes i see it

- jim_thompson5910

ok we'll let D = 4 be the determinant of the coefficient matrix

- jim_thompson5910

if we replace the first column with the numbers 21, 3, 1 (found on the right side of each equation ) we get
|dw:1443402459893:dw|
what is the determinant of this new matrix (shown above)?

- anonymous

1 1 1
1 -1 1

- anonymous

we need to add another column

- jim_thompson5910

did you start off doing this?
|dw:1443402694898:dw|

- anonymous

i got confused with the ones

- anonymous

so i got ( 21 ) ( -1) ( -1) = 21

- anonymous

(1) (1) (1) = 1
(1) (3) (1) = 2

- anonymous

sorry 3*

- jim_thompson5910

good, add up those products to get ??

- anonymous

then you add them 21 + 1 + 3

- anonymous

25

- jim_thompson5910

ok keep that sum in mind
now move onto the next set of diagonals

- anonymous

how do you make that

- jim_thompson5910

circling them helps me keep all the numbers straight
|dw:1443402998519:dw|

- jim_thompson5910

1*(-1)*1 = ??
1*1*21 = ??
-1*3*1 = ??
then add up the products

- anonymous

-1

- anonymous

21

- anonymous

-3

- anonymous

add them 17

- anonymous

m i right

- jim_thompson5910

correct

- jim_thompson5910

first sum = 25
second sum = 17
subtract the sums: first - second = 25 - 17 = 8

- jim_thompson5910

So the determinant of this matrix (below) is 8
|dw:1443403306226:dw|

- jim_thompson5910

since we replaced the first column, think of it as the x column, we call this matrix Dx

- jim_thompson5910

to get the solution for x, we simply compute
x = Dx/D

- jim_thompson5910

Recall earlier we found D = 4
Dx = 8
so...
x = Dx/D
x = 8/4
x = 2

- anonymous

and we do the same to find y and z

- jim_thompson5910

yes you replace the second column with 23,3,1 and compute the determinant of that new matrix. That will give you Dy
to find y, compute
y = Dy/D

- jim_thompson5910

same with z but you replace the third column with 23,3,1
z = Dz/D

- anonymous

23 / 3 / 1

- jim_thompson5910

compute the determinant of this
|dw:1443403528965:dw|
to get Dy

- anonymous

1 * 3 *1

- jim_thompson5910

|dw:1443403666260:dw|

- jim_thompson5910

|dw:1443403686878:dw|

- anonymous

-3 23 1

- jim_thompson5910

add those products up

- anonymous

21

- jim_thompson5910

good

- jim_thompson5910

|dw:1443403784502:dw|

- anonymous

3 1 -23

- jim_thompson5910

add those values up

- anonymous

-19

- jim_thompson5910

first sum = 21
second sum = -19
first - second = 21 - (-19) = 21+19 = 40

- jim_thompson5910

hold on, I think I made a typo somewhere

- jim_thompson5910

oh my bad

- jim_thompson5910

I put 23 when I should have put 21

- jim_thompson5910

|dw:1443404103846:dw|
sorry about that

- anonymous

3 1 -21

- anonymous

-17

- jim_thompson5910

|dw:1443404234817:dw|

- anonymous

-3 21 1

- anonymous

19

- jim_thompson5910

|dw:1443404265866:dw|

- anonymous

3 1 -21= -17

- jim_thompson5910

so
first sum - second sum = 19 - (-17) = 19+17 = 36
making
Dy = 36

- jim_thompson5910

y = Dy/D
y = 36/4
y = 9

- anonymous

now we have to do the same for z

- jim_thompson5910

yes, replace the third column with 21,3,1 and compute the determinant

- anonymous

1*3*21 = 63

- jim_thompson5910

|dw:1443404399145:dw|

- jim_thompson5910

|dw:1443404416172:dw|

- anonymous

-21 3 1

- jim_thompson5910

|dw:1443404436727:dw|

- anonymous

-1 3 21

- anonymous

23

- jim_thompson5910

|dw:1443404481727:dw|

- anonymous

-21 3 1

- anonymous

-17

- jim_thompson5910

first sum = 23
second sum = -17
first - second = ???

- anonymous

40

- jim_thompson5910

good

- jim_thompson5910

Dz = 40,
z = Dz/D = ???

- anonymous

40 / what was d

- jim_thompson5910

D = 4
we got this when we computed the determinant of the first matrix we set up (the coefficient matrix)

- anonymous

10

- jim_thompson5910

z = 10, yes

- jim_thompson5910

x = 2
y = 9
z = 10

- jim_thompson5910

I recommend you check it with each original equation in the system

- anonymous

yes i will i will look at the steps again thankyou soo much for your time

- jim_thompson5910

btw all of this is called cramer's rule
http://www.purplemath.com/modules/cramers.htm

- anonymous

oh ok thank you again :)

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