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?
On
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?
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.