I have no idea how to start the question, but in the question before this were asked to prove:
$\det(A) \times \det(B) = \det(AB)$
So, I was trying to find a matrix C which would be multiplied by the first matrix to produce the second matrix, but this seems like it would be very tedious and there would be a simpler way. Any ideas?
The answer will be $2018$, because the determinant of$$\begin{pmatrix}A & b & c & D \\ e & 0 & 0 & f \\ g & 0 & 0 & h \\ I & j & k & L\end{pmatrix}$$is $c f g j - c e h j - b f g k + b e h k$. So, the choices of $A$, $D$, $I$, and $L$ are irrelevant.
The determinants are equal because of the two zeros in each of the two equal rows (or columns) which make the determinant independent of the terms $a_{11},a_{14},a_{41}$ and $a_{44}$. In fact we have both determinants are, respectively equal to
$$b\cdot\det\begin{bmatrix}0 & e & f \\ 0 & g & h \\ k & -7 & 1 \end{bmatrix}-j\cdot\det\begin{bmatrix}c & 2 & -3 \\ 0 & e & f \\ 0 & g & h \end{bmatrix}=(fg-eh)(cj-bk)$$ and
$$b\cdot\det\begin{bmatrix}0 & e & f \\ 0 & g & h \\ k & 1 & -9 \end{bmatrix}-j\cdot\det\begin{bmatrix}c & 8 & 5 \\ 0 & e & f \\ 0 & g & h \end{bmatrix}=(fg-eh)(cj-bk)$$ What I can not see clearly is how this can make checks easily the determinant of a matrix product.
The only thing I see is that for this example, the determinant of the tedious product matrix $$M=\begin{pmatrix}eb+cg+13 & 2b-3j & 2c-3k & bf+ch+37 \\ f+8e & eb+jf & ec+fk & 5e-9f\\ 8g+h & bg+jh & cg+hk & 5g-9h \\ ej+gk-55 & j-7b & k-7c & jf+hk-44\end{pmatrix}$$ should be equal to $(fg-eh)^2(cj-bk)^2$ but this is because I know the property $$\det(A) \times \det(B) = \det(AB)$$ The only way I can see is to calculate the determinant of $M$ but I renounce this, of course. Does anyone find another way?
