How do I find the determinant of this matrix? I'm preparing for an exam currently, and I came across this question: 
I have noticed that A can be constructed from the matrix on the left by a series of row operations, so I had the idea maybe to express A as a product of elementary matrices as well as the matrix on the left and, maybe there was some fact about the determinants of elementary matrices? So then I could just use some properties of determinants, as well as knowing that the determinant of the matrix on the left is 2 to figure out $det(A)$
Entirely confused though because I do not know what the determinant of elementary matrices are, and if I should be multiplying all the elementary matrices together etc etc?
Any help would be hugely appreciated.
 A: Let $B = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$.
If you take the operations that transform $B$ into $A$, and apply them to the identity matrix, you get:
$$C = \begin{pmatrix} 0 & 0 & 4 \\ 0 & 1 & -1 \\ 1 & -2 & 0 \end{pmatrix}$$
So $A = CB$, which means that $\det(A) = \det(C) \det(B)$.  You should be able to calculate that $\det(C) = -4$, and you are given that $\det(B) = 2$.  Therefore, $\det(A) = -8$.
Since this is nonzero, then yes, $A$ is invertible.
A: If we are uncertain about the row / col transform we could represent it in a matrix multiplication fashion.
$$
\begin{bmatrix}
4g & 4h & 4i \\
d-g & e-h & f-i \\
a-2d & b-2e & c-2f\\
\end{bmatrix}=
\begin{bmatrix}
4 & 0 & 0\\
-1 & 1 & 0\\
0 & -2 & 1\\
\end{bmatrix}
\begin{bmatrix}
g & h & i\\
d & e & f\\
a & b & c\\
\end{bmatrix}
$$
$$
\begin{bmatrix}
g & h & i\\
d & e & f\\
a & b & c\\
\end{bmatrix}
=
\begin{bmatrix}
0 & 0 & 1\\
0 & 1 & 0\\
1 & 0 & 0\\
\end{bmatrix}
\begin{bmatrix}
a & b & c\\
d & e & f\\
g & h & i\\
\end{bmatrix}
$$
Easy to see
$$
\det\begin{bmatrix}
0 & 0 & 1\\
0 & 1 & 0\\
1 & 0 & 0\\
\end{bmatrix}=-1\\
\det\begin{bmatrix}
4 & 0 & 0\\
-1 & 1 & 0\\
0 & -2 & 1\\
\end{bmatrix}=4
$$
Since the determinant $\det AB=\det A \det B$
We get $\det A=-4\det
\begin{bmatrix}
a & b & c\\
d & e & f\\
g & h & i\\
\end{bmatrix}=-8 $
$\det A\neq 0$ so it's invertible
A: 
Entirely confused though because I do not know what the determinant of elementary matrices are...

You don't even need to know those elementary matrices. You just need to know how elementary row operations affect the determinant. In this case, we need all three types of operations, and I write the effect in the parentheses behind.

*

*Multiply a row by a non-zero number. (determinant multiplied by this number)

*Interchange two rows. (determinant multiplied by $-1$)

*Add a scalar multiple of one row to another row. (determinant unchanged)

It's obvious that we can multiply the first row of $A$ by $\frac{1}{4}$ and add to the second row (operation type 3) to get $(d\  e\  f)$. Then multiplying this row by $\frac{1}{2}$ and adding to the second row (type 3) gives us a clean row $(a\ b\ c)$. Now we have a matrix that can be related to $A$ by elementary row operations and have the same determinant as $A$:
$$
\begin{pmatrix}
4g & 4h & 4i\\
d & e & f\\
a & b & c
\end{pmatrix}
$$
Divide the first row by 4  (type 1) and interchange the first and the second last row (type 2), we get the original matrix whose determinant is known to be $2$. Since we know consequences of three types of operation, it's easy to conclude that
$$
\det (A) = -4\times 2=-8
$$

P.S.
I personally don't remember those matrices performing elementary row operations as matrix multiplications. And I don't understand the necessity of working out those operation matrces (or the overall matrix $C$ in @Dan's answer). Doing matrix multiplication is always tedious, after all. As I have said, observing that $A$ can be related to the original matrix by row operations suffice to give the answer.
