>>987633elementary row operations mean just that - elementary row operations on the matrix of the corresponding system of linear equations.
* exchange any two rows (= equations)
* multiply a row (= equation) by a scalar
* multiply it and add it to another row (= equation)
How to do it in another way keep track of the coefficients of the original equations when solving the system (part (a)):
x + 3y = 7 (coefficients 1 0 = original values of the first equation)
x - y = -1 (coefficients 0 1 = original values of the second equation)
Subtract the second equation from the first
4y = 8 (coefficients 1-0 0-1 = 1 -1)
Divide by 4:
y = 2 (coefficients 1/4 -1/4)
Add this equation to the second
x = 1 (coefficients 1/4 1-1/4 = 3/4)
This means multiplying the original first equation by 1/4 and the second by 3/4 gives x = 1.
This concludes part (a).
Part (c): You need to compose 2x + y = 4 and you know the coefficients of x and for y, so you obtain
2(1/4 3/4) + 1(1/4 -1/4) = (3/4 5/4)
So you know that you need to multiply the original first equation by 3/4 and add it to 5/4 of the second to obtain 2x + y = 4 - but this is not an elementary row operation, and you must subdivide it into two operations.
The second equation is
-x + y = 1
which corresponds to the coefficients
-1(1/4 3/4) + 1(1/4 -1/4) = (0 -1)
which means that you only need to multiply the second equation by -1 (which you should be able to immediately see).
tldr: If you solved the first system in (a) you know the steps to obtain the solution x=1 and y=2 from the first system and you can compose these to get the steps needed to transform the first system into the second system.
