Maths problem. Plz help.?

anonymous

x+y+z=6, x-y+z=2, 2x+y-z=1. Solve this equation by finding the invrse of coeeficient matrix by row transformation method.

ed-i

x = 1, y = 2, z = 3

workoutgirl1235

1 ... 1 ... 1 ... 1 ... 0 ... 0

1 .. -1 ... 1 ... 0 ... 1 ... 0

2 ... 1 .. -1 ... 0 ... 0 ... 1

R1 ----------> 1 ... 1 ... 1 ... 1 ... 0 ... 0

R1 - R2 ----> 0 ... 2 ... 0 ... 1 .. -1 .. 0

R3 - 2R2 --> 0 ... 3 .. -3 ... 0 .. -2 .. 1

R1 - R3/3 -> 1 ... 0 ... 2 ... 1 ...... 2/3 . -1/3

R2/2 --------> 0 ... 1 ... 0 .. 1/2 . -1/2 .... 0

R3 ----------> 0 ... 3 .. -3 ... 0 ...... -2 ...... 1

R1 ----------> 1 ... 0 ... 2 ... 1 .... 2/3 . -1/3

R2 ----------> 0 ... 1 ... 0 .. 1/2 . -1/2 .... 0

R3 - 3R2 --> 0 ... 0 .. -3 . -3/2 . -1/2 ... 1 ===> ÷ -3 = 0 .. 0 .. 1 .. 1/2 .. 1/6 .. -1/3

R1 - 2R3 --> 1 ... 0 ... 0 ... 0 ... 1/3 ... 1/3

R2 ----------> 0 ... 1 ... 0 .. 1/2 . -1/2 .... 0

R3 ----------> 0 ... 0 ... 1 .. 1/2 .. 1/6 .. -1/3

The inverse of the original is

.0 ... 1/3 ... 1/3

1/2 . -1/2 .... 0

1/2 .. 1/6 .. -1/3

OR

(1/6) * |0 ... 2 ... 2|

-------> |3 .. -3 ... 0|

-------> |3 ... 1 .. -2|

So, the above times the matrix of constants =

(1/6) * |0 ... 2 ... 2| * [6]

-------> |3 .. -3 ... 0|->|2|

-------> |3 ... 1 .. -2|->[1]

solution = 1, 2, 3

This matrix can be solved for numerous systems where the only difference from the above system is in the constants.

I hope you have a lot of them, because other procedures would have been easier in this case!

anonymous

what's matrix thing lol

add 1st and 2nd equations

2x + 2y = 8

add 2nd and 3rd equations

3x = 3 x = 1

if x = 1, y = 3

then z = 4