Is there a way to know if a row reduction of a matrix has been done correctly? I'm an undergrad taking the class of "Linear algebra 1". I came across a problem:
sometimes we need to apply Gaussian elimination for matrices. Very quickly this skill is not much necessary as it's not a thinking skill but purely Technic. 
Yet, often in exams there's a question that requires you to apply row reduction 
to a matrix. 
I am looking for a way to know if the Gaussian elimination has been done properly, meaning - with no Calculation mistakes, other than going over all the steps and check that the Calculation has been done correctly. as this processes will double the time I will spend on a given question, and due to the lack of time in a big course exam - which is also very stressful - such method could be much helpful for me.
note: we're allowed to use a simple scientific calculator (not a graph calculator)   
 A: The actual algorithm for Gaussian Elimination looks like this.  
I answered a similar answer showing how to perform the LU decomposition which is Gaussian Elimination without pivoting  The purpose is to zero out the row beneath is with $ \ell_{jk}$. That is why you have the operation below $ u_{j,k:m} = u_{j,k:m} - \ell_{jk} u_{k,k:m}$ It is subtracting off the ratio you just computed. 
Which yielded this.
Suppose that 
$$ A  = \begin{bmatrix} 1 &  1 & 1 \\ 3 & 5 & 6 \\ -2 & 2 & 7  \end{bmatrix} $$
$$ A = LU $$
$$ U =A, L=I$$
$$ k=1,m=3,j=2$$
$$\ell_{21} = \frac{u_{21}}{u_{11}}  = \frac{a_{21}}{a_{11}} = 3 $$
$$ u_{2,1:3} = u_{2,1:3} - 3 \cdot u_{1,1:3}  $$
Then we're going to subtract 3 times the 1st row from the 2nd row
$$ \begin{bmatrix} 3 & 5 & 6 \end{bmatrix} - 3 \cdot \begin{bmatrix} 1 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 2 & 3\end{bmatrix}  $$
Updating each of them
$$U = \begin{bmatrix} 1 &  1  & 1 \\ 0 & 2 & 3 \\ -2 & 2 & 7 \end{bmatrix} $$
$$ L =\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$
$$k=1,j=3,m=3 $$
$$\ell_{31} = \frac{u_{31}}{u_{11}} = \frac{-2}{1} = -2 $$
$$ u_{3,1:3} = u_{3,1:3} +2 \cdot u_{1,1:3} $$
Then we add two times the first row to the third row
$$ \begin{bmatrix} -2 & 2 & 7 \end{bmatrix} +  2 \cdot \begin{bmatrix} 1 & 1& 1  \end{bmatrix} = \begin{bmatrix}0 & 4 & 9  \end{bmatrix} $$
Updating 
$$ U = \begin{bmatrix} 1 &  1  & 1 \\ 0 & 2 & 3 \\ 0 & 4 & 9 \end{bmatrix}  $$
$$ L =\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 0 & 1 \end{bmatrix} $$
$$ k=2, j=3,m=3 $$
$$ \ell_{32} = \frac{u_{32}}{u_{22}} = \frac{4}{2} = 2$$
We're subtracting out little blocks
$$ u_{3,2:3} = u_{3,2:3} - 2 \cdot u_{2,2:3} $$
$$ \begin{bmatrix} 4 & 9 \end{bmatrix} - 2 \cdot\begin{bmatrix} 2& 3 \end{bmatrix} = \begin{bmatrix} 0 & 3 \end{bmatrix} $$
Updating 
$$ U = \begin{bmatrix} 1 &  1  & 1 \\ 0 & 2 & 3 \\ 0 & 0 & 3 \end{bmatrix}  $$
$$ L =\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 2 & 1 \end{bmatrix} $$
It now terminates
$$ A  = LU $$
$$ \underbrace{\begin{bmatrix} 1 &  1 & 1 \\ 3 & 5 & 6 \\ -2 & 2 & 7  \end{bmatrix}}_{A} = \underbrace{\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 2 & 1 \end{bmatrix}}_{L} \underbrace{\begin{bmatrix} 1 &  1  & 1 \\ 0 & 2 & 3 \\ 0 & 0 & 3 \end{bmatrix}}_{U}   $$
