SOLUTION: Using Gauss-Jordan elimination, write the following matrix in reduced row-echelon form, unless it is an inconsistent system. {{{(matrix(3,5, 2,0,3,"|",3, 4,-3,7,"|",5, 8

We need 0's where the 4,8,0,-9,leftmost 3, and 7 are.

To get a 0 where the 4 is:
Multiply the 1st row by -2 and add it to the 2nd row,
and put the result in place of row 2

That operation is abbreviated -2R1+R2->R2



To get a 0 where the 8 is:
Multiply row 1 by -4 and add it to the row 3,
and put the result in place of row 3

That operation is abbreviated -4R1+R3->R3



We already have a 0 at the top of column 2.

To get a 0 where the -9 is:
Multiply row 2 by -3 and add it to the row 3,
and put the result in place of row 3

That operation is abbreviated -3R2+R3->R3



Now that we have our 0's, we get the 1's.

We multiply row 1 by 1/2, and row 2 by -1/3.

Those operations are abbreviated (1/2)R1->R1 and (-1/3)R2->R2



Edwin