Math
posted by Lynn on .
I cannot come up with the right answers for my matrix.
I need to write as a system of equation and solve the system.
1 2 3 x 1
1 1 1 y = 12
1 1 2 z 2
The answers I come up with are
31
15
15
when I check these, they will not work, what I am doing wrong?

1 2 3 1
1 1 1 12
1 1 2 2
I want the first column to be 1 0 0 so subtract second from first and add third to first
1 2 3 1
0 1 2 11
0 3 5 3
I want the second column to be 0 1 0 so
subtract twice the second from the first and subtract 3 times the second from the third
1 0 1 23
0 1 2 11
0 0 1 36
I want the third column to be 0 0 1 so multiply the third row by 1 first
1 0 1 23
0 1 2 11
0 0 1 36
now add the third to the first and subtract twice the third from the second
1 0 0 13
0 1 0 61
0 0 1 36
so I get x = 13 , y = 61, z = 36 
when I am adding and subtracting I mean rows by the way.

The idea is that I can add anything times and row to any other row, just as I can add anything times any equation to any other equation. In fact that is exactly what I am doing.
I select what I want to multiply each by to get an identity matrix on the left, column by column, thereby separating the variables. I am really solving the equations by elimination, just in a particular fashion. 
Oh, to write as a system of equations, multiply the coefficient matrix by the variable column
1 2 3 x 1
1 1 1 y = 12
1 1 2 z 2
means
1 x + 2 y + 3 z = 1
1 x + 1 y + 1 z = 12
1 x + 1 y + 2 z = 2 
Oh, the particular fashion is often called GaussJordan reduction