If the row of a matrix is divided by a scalar, is the determinent multiplied by the scalar? Right now I'm a bit confused by my textbook. Where my textbook defines that, when a row is multiplied by a scalar, then the determinant is also multiplied by a scalar
textbook definition
Where I am getting confused, is that a youtube video that I am watching, and following gives me the correct answer is giving an opposite definition.
Picture of Conflicting definition 
Following the youtube video's definition gives me the correct answer, but seems to be the exact opposite of the website's definition.Example Problem where you can see that as I mulitply a row by a scalar (1/3), I multiply the determinant by (3) not (1/3), which gives me the correct answer.
Anyone know what am I misunderstanding here? 
 A: No, you might want to check your work.
If you multiply a row in the matrix by $k$, the determinant is ALSO multiplied by $k$. 
Your work also agrees with this. 
Let your original matrix be A and your new matrix be B.
You said that $\det(A) = 3\det(B)$, that's the same thing as saying $\det(B) = \frac{1}{3}\det(A)$. 
A: If the row of a matrix is multiplied by a scalar, the determinant is also multiplied by that scalar.
The division you're seeing happens because, when we're finding the determinant of a matrix, we don't want it to change at all.
So you might write, for example, that 
$$
   \det \begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix} =\frac1{100} \det \begin{bmatrix}100 & 200 \\ 3 & 4\end{bmatrix}.
$$
When we multiply the first row by $100$, we multiply the determinant by $100$, so we introduce a factor of $\frac1{100}$ to keep the overall expression the same.
This is equivalent to writing 
$$
   \det \begin{bmatrix}100 & 200 \\ 3 & 4\end{bmatrix} = 100\det \begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}.
$$
A: *

*If you multiply a row of a matrix by a constant $k$ then the determinant of the new matrix is $k$ times the determinant of the old matrix.  

*If when trying to calculate the determinant of the old matrix you divide a row by a constant $c \not= 0$ then the determinant of the old matrix is $c$ times the determinant of the new matrix.  
These two statements are basically saying the same thing with $c = \frac1k$ and are consistent with each other.
A: The rule is
$$
\det(a_1,\ldots,\lambda a_i,\ldots,a_n) = \lambda\det(a_1,\ldots,a_n).
$$
Hence,
$$
\det(a_1,\ldots,a_n) = c\cdot \det(a_1,\ldots,\frac 1ca_i,\ldots,a_n).
$$
