SOLUTION: You have the following matrix, and perform the following 3 row operations. What is the resultant matrix after performing these 3 row operations? –R1 + R2  R2 2R1 +

[ 1 0 0 10] 
[ 1 1 3  5]
[-2 2 0  4]

R1 = the first (top) row = [ 1 0 0 10] 
R2 = the second (middle) row = [ 1 1 3  5]
R3 = the third (bottom) row = [-2 2 0  4]

First row operation:

      –R1 + R2 -> R2

Let's do the left side -R1 + R2 first.
Substitute [ 1 0 0 10] for R1, [ 1 1 3  5] for R2 

–R1 + R2 = -[ 1 0 0 10] + [ 1 1 3  5] = (distribute the - sign)
           [-1 0 0 -10] + [ 1 1 3  5] = (just combine corresponding elements)
           [ 0 1 3  -5]                   

The " -> R2 " tells us to replace R2 (the 2nd row) by [ 0 1 3  -5], so
the new matrix is like the old one with the middle row replaced by
what we got:

[ 1 0 0 10] 
[ 0 1 3 -5]
[-2 2 0  4]    

Now:
  
R1 = the first (top) row = [ 1 0 0 10] 
R2 = the second (middle) row = [ 0 1 3 -5]
R3 = the third (bottom) row = [-2 2 0  4]

Next row operation:

     2R1 + R3 -> R3

As before, let's do the left side 2R1 + R3 first.
Substitute [ 1 0 0 10] for R1, [-2 2 0  4] for R3 

2R1 + R3 = 2[ 1 0 0 10] + [-2 2 0  4] = (distribute the 2)
         =  [ 2 0 0 20] + [-2 2 0  4] = (combine corresponding elements)
         =  [ 0 2 0 24]

The " -> R3 " tells us to replace R3 (the 3rd row) by [ 0 2 0 24], so
the new matrix is like the previous one with the bottom row replaced by
what we got:
 
[ 1 0 0 10] 
[ 0 1 3 -5]
[ 0 2 0 24]  

Edwin