At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
is there a typo in the second equation `x=y +z = 3` ??
there are 2 equal signs there
yes sorry its x - y + z = 3
were you able to set up the matrix?
what matrix did you get
1 1 1 1 -1 1 1 1 -1
m i right?
that's your coefficient matrix
do you know how to take the determinant of that matrix?
focus on the first two columns |dw:1443401787922:dw|
copy those columns and write it off to the right of the vertical bar |dw:1443401811957:dw|
with me so far?
ok so we have this extended matrix |dw:1443401923170:dw|
circle the elements that fall along the diagonals like this |dw:1443401944912:dw| we have 3 subgroups
ohh now i see it
multiply the numbers in the subgroups |dw:1443402013404:dw|
then add up the products: 1+1+1 = 3
next, circle the elements along the diagonals like this |dw:1443402083998:dw|
multiply out the numbers in each subgroup |dw:1443402112490:dw|
then add up the products -1+1+(-1) = -1
The first sum was 3 the second sum was -1 subtract the sums 3 - (-1) = 3+1 = 4
so in the end, the determinant of that 3x3 matrix you wrote above is 4
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
so for x we got 3
we haven't found x or y yet
whats the next step
do you see how I got the determinant to be 4?
yes you added them
so do you see how to find the determinant of any 3x3 matrix?
yes i see it
ok we'll let D = 4 be the determinant of the coefficient matrix
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)?
1 1 1 1 -1 1
we need to add another column
did you start off doing this? |dw:1443402694898:dw|
i got confused with the ones
so i got ( 21 ) ( -1) ( -1) = 21
(1) (1) (1) = 1 (1) (3) (1) = 2
good, add up those products to get ??
then you add them 21 + 1 + 3
ok keep that sum in mind now move onto the next set of diagonals
how do you make that
circling them helps me keep all the numbers straight |dw:1443402998519:dw|
1*(-1)*1 = ?? 1*1*21 = ?? -1*3*1 = ?? then add up the products
add them 17
m i right
first sum = 25 second sum = 17 subtract the sums: first - second = 25 - 17 = 8
So the determinant of this matrix (below) is 8 |dw:1443403306226:dw|
since we replaced the first column, think of it as the x column, we call this matrix Dx
to get the solution for x, we simply compute x = Dx/D
Recall earlier we found D = 4 Dx = 8 so... x = Dx/D x = 8/4 x = 2
and we do the same to find y and z
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
same with z but you replace the third column with 23,3,1 z = Dz/D
23 / 3 / 1
compute the determinant of this |dw:1443403528965:dw| to get Dy
1 * 3 *1
-3 23 1
add those products up
3 1 -23
add those values up
first sum = 21 second sum = -19 first - second = 21 - (-19) = 21+19 = 40
hold on, I think I made a typo somewhere
oh my bad
I put 23 when I should have put 21
|dw:1443404103846:dw| sorry about that
3 1 -21
-3 21 1
3 1 -21= -17
so first sum - second sum = 19 - (-17) = 19+17 = 36 making Dy = 36
y = Dy/D y = 36/4 y = 9
now we have to do the same for z
yes, replace the third column with 21,3,1 and compute the determinant
1*3*21 = 63
-21 3 1
-1 3 21
-21 3 1
first sum = 23 second sum = -17 first - second = ???
Dz = 40, z = Dz/D = ???
40 / what was d
D = 4 we got this when we computed the determinant of the first matrix we set up (the coefficient matrix)
z = 10, yes
x = 2 y = 9 z = 10
I recommend you check it with each original equation in the system
yes i will i will look at the steps again thankyou soo much for your time
btw all of this is called cramer's rule http://www.purplemath.com/modules/cramers.htm
oh ok thank you again :)