Maths problem. Plz help.?

Comments

• ed-i

  May 2021

  x = 1, y = 2, z = 3
• workoutgirl1235

  May 2021

  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

  May 2021

  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