anonymous
  • anonymous
Medal Solve by determinants. x + y + z = 21 x=y +z = 3 x +y - z = 1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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jim_thompson5910
  • jim_thompson5910
is there a typo in the second equation `x=y +z = 3` ??
jim_thompson5910
  • jim_thompson5910
there are 2 equal signs there
anonymous
  • anonymous
yes sorry its x - y + z = 3

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jim_thompson5910
  • jim_thompson5910
ok thanks
jim_thompson5910
  • jim_thompson5910
were you able to set up the matrix?
anonymous
  • anonymous
yes
jim_thompson5910
  • jim_thompson5910
what matrix did you get
anonymous
  • anonymous
1 1 1 1 -1 1 1 1 -1
anonymous
  • anonymous
m i right?
jim_thompson5910
  • jim_thompson5910
good
jim_thompson5910
  • jim_thompson5910
that's your coefficient matrix
jim_thompson5910
  • jim_thompson5910
do you know how to take the determinant of that matrix?
anonymous
  • anonymous
no
jim_thompson5910
  • jim_thompson5910
|dw:1443401707454:dw|
jim_thompson5910
  • jim_thompson5910
focus on the first two columns |dw:1443401787922:dw|
jim_thompson5910
  • jim_thompson5910
copy those columns and write it off to the right of the vertical bar |dw:1443401811957:dw|
jim_thompson5910
  • jim_thompson5910
with me so far?
anonymous
  • anonymous
yes
jim_thompson5910
  • jim_thompson5910
ok so we have this extended matrix |dw:1443401923170:dw|
jim_thompson5910
  • jim_thompson5910
circle the elements that fall along the diagonals like this |dw:1443401944912:dw| we have 3 subgroups
anonymous
  • anonymous
ohh now i see it
jim_thompson5910
  • jim_thompson5910
multiply the numbers in the subgroups |dw:1443402013404:dw|
jim_thompson5910
  • jim_thompson5910
then add up the products: 1+1+1 = 3
jim_thompson5910
  • jim_thompson5910
next, circle the elements along the diagonals like this |dw:1443402083998:dw|
jim_thompson5910
  • jim_thompson5910
multiply out the numbers in each subgroup |dw:1443402112490:dw|
jim_thompson5910
  • jim_thompson5910
then add up the products -1+1+(-1) = -1
jim_thompson5910
  • jim_thompson5910
The first sum was 3 the second sum was -1 subtract the sums 3 - (-1) = 3+1 = 4
jim_thompson5910
  • jim_thompson5910
so in the end, the determinant of that 3x3 matrix you wrote above is 4
jim_thompson5910
  • 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
  • anonymous
so for x we got 3
jim_thompson5910
  • jim_thompson5910
we haven't found x or y yet
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
whats the next step
jim_thompson5910
  • jim_thompson5910
do you see how I got the determinant to be 4?
anonymous
  • anonymous
yes you added them
jim_thompson5910
  • jim_thompson5910
so do you see how to find the determinant of any 3x3 matrix?
anonymous
  • anonymous
yes i see it
jim_thompson5910
  • jim_thompson5910
ok we'll let D = 4 be the determinant of the coefficient matrix
jim_thompson5910
  • 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
  • anonymous
1 1 1 1 -1 1
anonymous
  • anonymous
we need to add another column
jim_thompson5910
  • jim_thompson5910
did you start off doing this? |dw:1443402694898:dw|
anonymous
  • anonymous
i got confused with the ones
anonymous
  • anonymous
so i got ( 21 ) ( -1) ( -1) = 21
anonymous
  • anonymous
(1) (1) (1) = 1 (1) (3) (1) = 2
anonymous
  • anonymous
sorry 3*
jim_thompson5910
  • jim_thompson5910
good, add up those products to get ??
anonymous
  • anonymous
then you add them 21 + 1 + 3
anonymous
  • anonymous
25
jim_thompson5910
  • jim_thompson5910
ok keep that sum in mind now move onto the next set of diagonals
anonymous
  • anonymous
how do you make that
jim_thompson5910
  • jim_thompson5910
circling them helps me keep all the numbers straight |dw:1443402998519:dw|
jim_thompson5910
  • jim_thompson5910
1*(-1)*1 = ?? 1*1*21 = ?? -1*3*1 = ?? then add up the products
anonymous
  • anonymous
-1
anonymous
  • anonymous
21
anonymous
  • anonymous
-3
anonymous
  • anonymous
add them 17
anonymous
  • anonymous
m i right
jim_thompson5910
  • jim_thompson5910
correct
jim_thompson5910
  • jim_thompson5910
first sum = 25 second sum = 17 subtract the sums: first - second = 25 - 17 = 8
jim_thompson5910
  • jim_thompson5910
So the determinant of this matrix (below) is 8 |dw:1443403306226:dw|
jim_thompson5910
  • jim_thompson5910
since we replaced the first column, think of it as the x column, we call this matrix Dx
jim_thompson5910
  • jim_thompson5910
to get the solution for x, we simply compute x = Dx/D
jim_thompson5910
  • jim_thompson5910
Recall earlier we found D = 4 Dx = 8 so... x = Dx/D x = 8/4 x = 2
anonymous
  • anonymous
and we do the same to find y and z
jim_thompson5910
  • 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
  • jim_thompson5910
same with z but you replace the third column with 23,3,1 z = Dz/D
anonymous
  • anonymous
23 / 3 / 1
jim_thompson5910
  • jim_thompson5910
compute the determinant of this |dw:1443403528965:dw| to get Dy
anonymous
  • anonymous
1 * 3 *1
jim_thompson5910
  • jim_thompson5910
|dw:1443403666260:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1443403686878:dw|
anonymous
  • anonymous
-3 23 1
jim_thompson5910
  • jim_thompson5910
add those products up
anonymous
  • anonymous
21
jim_thompson5910
  • jim_thompson5910
good
jim_thompson5910
  • jim_thompson5910
|dw:1443403784502:dw|
anonymous
  • anonymous
3 1 -23
jim_thompson5910
  • jim_thompson5910
add those values up
anonymous
  • anonymous
-19
jim_thompson5910
  • jim_thompson5910
first sum = 21 second sum = -19 first - second = 21 - (-19) = 21+19 = 40
jim_thompson5910
  • jim_thompson5910
hold on, I think I made a typo somewhere
jim_thompson5910
  • jim_thompson5910
oh my bad
jim_thompson5910
  • jim_thompson5910
I put 23 when I should have put 21
jim_thompson5910
  • jim_thompson5910
|dw:1443404103846:dw| sorry about that
anonymous
  • anonymous
3 1 -21
anonymous
  • anonymous
-17
jim_thompson5910
  • jim_thompson5910
|dw:1443404234817:dw|
anonymous
  • anonymous
-3 21 1
anonymous
  • anonymous
19
jim_thompson5910
  • jim_thompson5910
|dw:1443404265866:dw|
anonymous
  • anonymous
3 1 -21= -17
jim_thompson5910
  • jim_thompson5910
so first sum - second sum = 19 - (-17) = 19+17 = 36 making Dy = 36
jim_thompson5910
  • jim_thompson5910
y = Dy/D y = 36/4 y = 9
anonymous
  • anonymous
now we have to do the same for z
jim_thompson5910
  • jim_thompson5910
yes, replace the third column with 21,3,1 and compute the determinant
anonymous
  • anonymous
1*3*21 = 63
jim_thompson5910
  • jim_thompson5910
|dw:1443404399145:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1443404416172:dw|
anonymous
  • anonymous
-21 3 1
jim_thompson5910
  • jim_thompson5910
|dw:1443404436727:dw|
anonymous
  • anonymous
-1 3 21
anonymous
  • anonymous
23
jim_thompson5910
  • jim_thompson5910
|dw:1443404481727:dw|
anonymous
  • anonymous
-21 3 1
anonymous
  • anonymous
-17
jim_thompson5910
  • jim_thompson5910
first sum = 23 second sum = -17 first - second = ???
anonymous
  • anonymous
40
jim_thompson5910
  • jim_thompson5910
good
jim_thompson5910
  • jim_thompson5910
Dz = 40, z = Dz/D = ???
anonymous
  • anonymous
40 / what was d
jim_thompson5910
  • jim_thompson5910
D = 4 we got this when we computed the determinant of the first matrix we set up (the coefficient matrix)
anonymous
  • anonymous
10
jim_thompson5910
  • jim_thompson5910
z = 10, yes
jim_thompson5910
  • jim_thompson5910
x = 2 y = 9 z = 10
jim_thompson5910
  • jim_thompson5910
I recommend you check it with each original equation in the system
anonymous
  • anonymous
yes i will i will look at the steps again thankyou soo much for your time
jim_thompson5910
  • jim_thompson5910
btw all of this is called cramer's rule http://www.purplemath.com/modules/cramers.htm
anonymous
  • anonymous
oh ok thank you again :)

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