SOLUTION: Perform the row operations on the given augmented matrix. a.) R₂ = 3r₁ + r₂ b.) R₃ = -2r₁ + r₃ c.) R<sub>3</sub

Perform the row operations on the given 
augmented matrix.
a.) R2 = 3r1 + r2
b.) R3 = -2r1 + r3
c.) R3 = 4r2 + r3

The capital R's refer to the NEW rows.
The little r's refer to the OLD rows.

[ 1  2 -3 |   4]
[-3 -5  8 | -10]
[ 2  0 -1 |   4]

a.) R2 = 3r1 + r2

That means to make a NEW matrix from the OLD matrix
above by multiplying the OLD row 1 by 3 and adding it
to the OLD row 2, then putting the result as the NEW
row 2 in the new matrix. To make this easier, place
a 3 left of the OLD row 1 and a 1 left of the OLD
row 2. The NEW rows 1 and 3 will be the same
as the OLD rows 1 and 3: 

3[ 1  2 -3 |   4]
1[-3 -5  8 | -10]
 [ 2  0 -1 |   4]

You can distribute the 3 across
and the 1 and add in your head, and get

[ 1  2 -3 |   4]
[ 0  1 -1 |   2]
[ 2  0 -1 |   4]

I got that NEW row 2, which is 0 1 -1 | 2
by doing this in my head:
3 times 1 plus 1 times -3 gives 3-3 or 0
3 times 2 plus 1 times -5 gives 6-5 or 1
3 times -3 plus 1 times 8 gives -9+8 or -1
3 times 4 plus 1 times -10 gives 12-10 or 2

b.) R3 = -2r1 + r3

That means to make a NEW matrix from the OLD matrix
above by multiplying the OLD row 1 by -2 and adding it
to the OLD row 3, then putting the result as the NEW
row 3 in the new matrix. To make this easier, place
a -2 left of the OLD row 1 and a 1 left of the OLD
row 3. The NEW rows 1 and 2 will be the same
as the OLD rows 1 and 2: 

-2[ 1  2 -3 |   4]
  [ 0  1 -1 |   2]
 1[ 2  0 -1 |   4]

You can distribute the -2 across
and the 1 and add in your head, and get

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

c.) R3 = 4r2 + r3

That means to make a NEW matrix from the OLD matrix
above by multiplying the OLD row 2 by 4 and adding it
to the OLD row 3, then putting the result as the NEW
row 3 in the new matrix. To make this easier, place
a 4 left of the OLD row 2 and a 1 left of the OLD
row 3. The NEW rows 1 and 2 will be the same
as the OLD rows 1 and 2: 

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

You can distribute the 4 across
and the 1 and add in your head, and get

[ 1  2  -3 |  4]
[ 0  1  -1 |  2]
[ 0  0   1 |  4]

Notice that there are 1's on the diagonal
and 0's below the diagonal.  This is
in what is called "row echelon form".

Edwin