Computing the determinant of a $4\times4$ matrix 
Compute the determinant of \begin{vmatrix} 2 & -1 & 1 & 3 \\ 1 & 2 & 0 & 1 \\ 2 & 1 & 1 & -1 \\ -1 & 1 & 2 & -2 \end{vmatrix}

My try:
$$\begin{vmatrix} 2 & -1 & 1 & 3 \\ 1 & 2 & 0 & 1 \\ 2 & 1 & 1 & -1 \\ -1 & 1 & 2 & -2 \end{vmatrix}_{R_2\rightarrow2R_2-R_1\\R_3\rightarrow R_3-R_1\\R_4\rightarrow2R_4+R_1}$$
$$\begin{vmatrix} 2 & -1 & 1 & 3 \\ 0 & 5 & -1 & -1 \\ 0 & 2 & 0 & -4 \\ 0 & 1 & 5 & -1 \end{vmatrix}$$
Now I took $$\begin{vmatrix} 5 & -1 & -1 \\ 2 & 0 & -4 \\ 1 & 5 & -1 \end{vmatrix}=92$$
But the answer is $46$. Where did I go wrong?
 A: When doing some operations on rows and columns of a matrix you have to be pretty careful: remember that the determinant is a multilinear map in the rows and the columns so it follows that 

Multiplying or dividing a row or a column by a scalar $\lambda$, multiplies or divides the determinant by the same scalar $\lambda$

So if $A$ is your initial matrix and $B$ is the matrix obtained by multiplying a column of $A$ by a scalar $\lambda$ you have that $$\det(B) = \lambda\det(A)$$
So in your case you've done $2$ multiplications by $2$ then the determinant of your final matrix will be $4$ times the determinant of the initial matrix. So you have that your determinant will be $$92/4\times 2 = 46$$
Obviously the times two is there for the Laplace expansion. 
A: Because you multiplied the first and the third rows by 2 while making the row operations, hence the determinant you found is 4 times bigger than the determinant of the original matrix. So the answer is $\frac{92*2}{4}$=46. 
A: I'm sure this question has been asked and answered many times, but I can't find it.
You can add or subtract a multiple of another row to a given row, without changing the value of the determinant, so $$R_1=R_1+4R_2$$
is perfectly OK.  When you multiply a row by $k$, you multiply the determinant by $k$.  When you do $$R_2=2R_2-R_1$$ you are really doing two operations.  First, you multiply row $2$ by $2$, then you subtract row $1$ from row $2$.  The effect of this is to multiply the determinant by $2$.   
It's fine to do these operations, as long as you keep track of them, so you can adjust the value of the determinant at the end.
The second mistake you made is at the end.  Because of the $2$ in the $(1,1)$ position, you should have multiplied the $3\times3$ by $2$ giving $184$, which is $4$ times the true value.  This is because you performed that "row-doubling" operation twice.
A: You multiplied two rows by two, so your answer (184) is 4 times greater than the actual determinant. This video may help.
