I need help to solve this Linear Algebra's question. Can any one help me?

Question

Q.6 The reduced row echelon form of a matrix A is:

R = [1 2 0 −1

0 0 1 1

0 0 0 0]

Determine A, given that the first and third columns of A are

a1 = [1

2

1]

and a3 = [2

2

3]

Answer 1

the finished product is

1 2 0 -1

0 0 1 1

0 0 0 0

which means two rows are the same because the last row is all zer0s

the original matrix looks something like this:

1 a 2 x

2 b 2 y

1 c 3 z

in the finished matrix, row3 = row3+2*row1

1 2 0 -1

0 0 1 1

2 4 0 -2

next, row3 = 2*row2+row3

1 2 0 -1

0 0 1 1

2 4 2 0

finally swaps rows

1 2 0 -1

2 4 2 0

0 0 1 1

note that row2 matches the 2nd row of the original matrix above with b=4 and y=0

the torture continues...

row1 = row1 + 2* row3

1 2 2 1

2 4 2 0

0 0 1 1

then row3 = row3+row1 again!!!

1 2 2 1

2 4 2 0

1 2 3 2

is the alleged answer.

for your convenience,

the original matrix was

supposed to look something like this:

1 a 2 x

2 b 2 y

1 c 3 z

Comparing this to the alleged answer

above in bold...

a=2, x=1,b=4 y=0,c=2 z=2

So if Gaussian Elimination is done on

1 2 2 1

2 4 2 0

1 2 3 2

the result should be the given matrix

in it's prescribed reduced form:

begins with the alleged answer:

1 2 2 1

2 4 2 0

1 2 3 2

row2=-2*row1+row2; -row1+row3

-------------------

1 2 2 1

0 0 -2 -2

0 0 1 1

row1=row1 -2*row3;

row2 = row2 + 2*row3;

------------------------

1 2 0 -1

0 0 0 0

0 0 1 1

which is the same as

the given matrix in

reduced echelon form,

so the elleged answer is

correct