Prove that the determinant is $0$ by expressing as a product 
I need to prove that the determinant $$\begin{vmatrix} my+nz & mq-nr &
 mb+nc \\ kz-mx & kr-mp & kb-ma \\ nx+ky & np+kq & na+kb
 \end{vmatrix}=0$$

In my book it is given as hint that the determinant can be expressed as a product of two other determinants whose value will evaluate to $0$.But I'm not being able to express the given determinant as a product of two other determinants.How should I do it?Please guide me through the procedure.
 A: Assuming there is the typo in your post pointed out by KittyL in the comment section, notice that: $$\begin{pmatrix}my-nz&mq+nr&mb+nc\\kz-mx&kr-mp&kc-ma\\nx+ky&np+kq&na+kb\end{pmatrix}=\begin{pmatrix}m&n&0\\0&k&-m\\k&0&n\end{pmatrix}\times\begin{pmatrix}y&q&b\\z&r&c\\x&p&a\end{pmatrix}.$$
A: As already pointed out there is probably some sign error in the given matrix. Otherwise one could decompose this matrix as follows:
Notice that the first column contains $x,y,z$, the second $p,q,r$ and the third $a,b,c$. So it makes sense for the second matrix to be
$$
 B =
 \begin{pmatrix}
  x & p & a \\
  y & q & b \\
  z & r & c
 \end{pmatrix},
$$
where we may have to change the signs of the entries. (We can actually assume w.l.o.g. that the signs in the first column of $B$ are already right.) Using the first column of your matrix we find that the first matrix must be
$$
 A =
 \begin{pmatrix}
  0 & m & n\\
  -m & 0 & k\\
  n & k & 0
 \end{pmatrix}.
$$
The problem is now that we still need to fix the signs in the second and third column of $B$, i.e. we need
$$
 \begin{pmatrix}
  my + nz & mq - nr & mb + nc \\
  kz - mx & kr - mp & kb - ma \\
  nx + ky & np + kq & na + kb
 \end{pmatrix}
 =
 \begin{pmatrix}
  0 & m & n\\
  -m & 0 & k\\
  n & k & 0
 \end{pmatrix}
 \begin{pmatrix}
  x & ?p & ?a \\
  y & ?q & ?b \\
  z & ?r & ?c
 \end{pmatrix}.
$$
The problem is that currently the sign of $r$ must be both $+$ and $-$.
Notice also that $\det(A) = 0$, which is what we want.
PS: Using KittyL’s suggestion we actually want to work with
$$
 \begin{pmatrix}
  my + nz & mq \color{red}{+} nr & mb + nc \\
  kz - mx & kr - mp & k\color{red}{c} - ma \\
  nx + ky & np + kq & na + kb
 \end{pmatrix}.
$$
Then the matrix $A$ stays the same, but we can now solve the sign problem with
$$
 \begin{pmatrix}
  my + nz & mq + nr & mb + nc \\
  kz - mx & kr - mp & kc - ma \\
  nx + ky & np + kq & na + kb
 \end{pmatrix}
 =
 \begin{pmatrix}
  0 & m & n\\
  -m & 0 & k\\
  n & k & 0
 \end{pmatrix}
 \begin{pmatrix}
  x & p & a \\
  y & q & b \\
  z & r & c
 \end{pmatrix}.
$$
A: Using the rule of Sarrus I get the result that
$$
\det(A)=kn(2amry + 2anrz - bkqz - bkry + 2bkrz - bmpy + bmqx - 2bmrx - b
npz - bnrx + ckqz - ckry + cmpy - cmqx + cnpz - cnrx)
$$
which is not identically zero. So, if I am not mistaken, something has to be changed in the question.
A: As Dietrich Burde mentions (I didn't know this had a name, thank you!), use the Rule of Sarrus (which works ONLY for $3 \times 3$ matrices). 
Here's the method:


*

*Replicate the first two columns. 


$$\begin{bmatrix}
my + nz & mq - nr & mb + nc & my + nz & mq - nr \\
kz - mx & kr - mp & kb - ma & kz - mx & kr - mp\\ 
nx + ky & np + kq & na + kb & nx + ky & np + kq 
\end{bmatrix}$$


*Find the product of each individual cell in all length-3 diagonals starting from the top row to the bottom row, and add them.


These length 3-diagonals going from the top row to the bottom row are given by the following:
$$\begin{bmatrix}
\color{blue}{my + nz} & mq - nr & mb + nc & my + nz & mq - nr \\
kz - mx & \color{blue}{kr - mp} & kb - ma & kz - mx & kr - mp\\ 
nx + ky & np + kq & \color{blue}{na + kb} & nx + ky & np + kq 
\end{bmatrix}$$
$$\begin{bmatrix}
my + nz & \color{blue}{mq - nr} & mb + nc & my + nz & mq - nr \\
kz - mx & kr - mp & \color{blue}{kb - ma} & kz - mx & kr - mp\\ 
nx + ky & np + kq & na + kb & \color{blue}{nx + ky} & np + kq 
\end{bmatrix}$$
$$\begin{bmatrix}
my + nz & mq - nr & \color{blue}{mb + nc} & my + nz & mq - nr \\
kz - mx & kr - mp & kb - ma & \color{blue}{kz - mx} & kr - mp\\ 
nx + ky & np + kq & na + kb & nx + ky & \color{blue}{np + kq} 
\end{bmatrix}$$
which gives $(my+nz)(kr-mp)(na+kb)+(mq-nr)(kb-ma)(nx+ky)+(mb+nc)(kz-mx)(np+kq)$.


*Find the product of each individual cell in all length-3 diagonals starting from the bottom row to the top row, and add them.


$$\begin{bmatrix}
my + nz & mq - nr & \color{red}{mb + nc} & my + nz & mq - nr \\
kz - mx & \color{red}{kr - mp} & kb - ma & kz - mx & kr - mp\\ 
\color{red}{nx + ky} & np + kq & na + kb & nx + ky & np + kq 
\end{bmatrix}$$
$$\begin{bmatrix}
my + nz & mq - nr & mb + nc & \color{red}{my + nz} & mq - nr \\
kz - mx & kr - mp & \color{red}{kb - ma} & kz - mx & kr - mp\\ 
nx + ky & \color{red}{np + kq} & na + kb & nx + ky & np + kq 
\end{bmatrix}$$
$$\begin{bmatrix}
my + nz & mq - nr & mb + nc & my + nz & \color{red}{mq - nr} \\
kz - mx & kr - mp & kb - ma & \color{red}{kz - mx} & kr - mp\\ 
nx + ky & np + kq & \color{red}{na + kb} & nx + ky & np + kq
\end{bmatrix}$$
which gives $(nx+ky)(kr-mp)(mb+nc)+(np+kq)(kb-ma)(my+nz)+(na+kb)(kz-mx)(mq-nr)$.


*Take what you get in step 2, and subtract what you get in step 3.


So, the determinant is given by
$$(my+nz)(kr-mp)(na+kb)+(mq-nr)(kb-ma)(nx+ky)+(mb+nc)(kz-mx)(np+kq)-\left[(nx+ky)(kr-mp)(mb+nc)+(np+kq)(kb-ma)(my+nz)+(na+kb)(kz-mx)(mq-nr)\right]\text{.}$$
