a = ( 8.6, 7.4, 3.9 ) 
b = ( 4.2, 2.2, -3.0 ) 
c = ( x, y, z ) 

a + b + c = 0

a + b + c = ( 12.8, 9.6, 0.9 ) + (x, y, z) = ( 12.8+x, 9.6+y, 0.9+z )  = (0, 0, 0) 

12.8 + x = 0;  x = -12.8
 9.6 + y = 0;  y =  -9.6
 0.9 + z = 0;  z =  -0.9

Algebra

The problem was: find the elements of c when

a + b + c = 0

If you didn't know they were matrices, you might have been tempted to work this using real number algebra:

  a + b  + c  = 0
  a + b  = -c + 0
 (a + b) = -c
-(a + b) =  c

In fact, this works. As long as every matrix is of the same type, and the operations are only "+" or "-", you can pretend that you are doing ordinary algebra. Notice that the last equations means "add a with b, then negate the result to get c."

To see this, look at just the first elements of the matrices:

 ( a0, ... ) + ( b0, ... ) +  ( c0, ... ) =  ( 0, ...)
 ( a0, ... ) + ( b0, ... )                = -( c0, ... ) + ( 0, ...)
 ( a0, ... ) + ( b0, ... )                = -( c0, ... )
 ( a0 + b0, ... )                         = -( c0, ... )
-( a0 + b0, ... )                         =  ( c0, ... )
-( a0 + b0)                               =    c0

If this is too ugly for you this early in the morning, mentally erase some of the junk:

 a0 + b0 + c0  =  0
 a0 + b0       = -c0 + 0 
 a0 + b0       = -c0
 a0 + b0       = -c0
-( a0 + b0)    =  c0

Of course the other elements follow the same pattern so the result is true for the matrix as a whole.

a = ( -4, 2 )T
b = ( 8, 3 )T
c = ( c0, c1 )T

a + b + c = 0