Question

Solving 2-variable sytem by matrices
-9x+13y=3
2x-3y=-1

Answer 1

i have to use the matrices form :/

Answer 2

Should it using matrices? but we can use the simple one, like elimination for example..

Answer 3

right. but my work has to be in matrices........

Answer 4

the augmented matrix is
[-9 13 3]
[ 2 -3 -1] you can change position

Answer 5

1/-9(-3)-13(2) = [3 -13]
[-2 9]
1/27-26

1/-1 = [3 -13]
[-2 9]

right?

Answer 6

[ 2 -3 -1] new position, mult R1 by 1/2 to get
[-9 13 3]

[ 1 -3/2 -1/2] --> [ 1 -3/2 -1/2]
[-9 13 3] 9R1+R2 -> [ 0 -1/2 -3]

Answer 7

[ 2 -3 -1] =R1 row 1 or first row
[-9 13 3] =R2

Answer 8

mult R1 by 1/2 to get
[ 1 -3/2 -1/2] mult R1 by 9 to get [ 9 -27/2 -9/2] then add thus to R2
[-9 13 3] to get
--------------------
[ 0 -1/2 -3/2] so the new form now becomes
[ 1 -3/2 -1/2] =R1
[ 0 -1/2 -3/2] =R2 now R2 means 0X -1Y/2 =-3/2 NOW SOLVE FOR Y FIRST
Y=-2(-3/2)
Y=3 SUBS Y TO R1 OR R2 OF THE FIRRST EQ

Answer 9

are you getting it?

Answer 10

yes. i am thank you

Answer 11

are you in LA or SF or in SD?

Answer 12

im in LA you?

Answer 13

is this a cell phone number?

Answer 14

okay.

Answer 15

I wouldnt want to keep you up. lol

Answer 16

ok

Answer 17

this is supposed the matrix form should be:
the augmented matrix is
[-9 13 3]
[ 2 -3 -1] you can change position

[ 2 -3 -1] R1x(1/2) --> [ 1 -3/2 -1/2] --> [ 1 -3/2 -1/2]
[-9 13 3] --> [-9 13 3] 9R1+R2 --> [ 0 -1/2 -3/2]

the new matrix form are
[ 1 -3/2 -1/2]
[ 0 -1/2 -3/2] when ever you have zero here in R2 means that 0X -1Y/2=-3/2
then you can solve for y=-2(-3/2)
y=3 use this to salve for x on
either
-9x+13y=3
2x-3y=-1