# If the determinant

Question:

If the determinant $\left|\begin{array}{ccc}x+a & p+u & l+f \\ y+b & q+v & m+g \\ z+c & r+w & n+h\end{array}\right|$ splits into exactly $k$ determinants of order 3 , each element of which contains only one term, then $k=$____________

Solution:

Let $\Delta=\left|\begin{array}{ccc}x+a & p+u & l+f \\ y+b & q+v & m+g \\ z+c & r+w & n+h\end{array}\right|$

$\Delta=\left|\begin{array}{ccc}x+a & p+u & l+f \\ y+b & q+v & m+g \\ z+c & r+w & n+h\end{array}\right|$

$=\left|\begin{array}{ccc}x+a & p+u & l \\ y+b & q+v & m \\ z+c & r+w & n\end{array}\right|+\left|\begin{array}{ccc}x+a & p+u & f \\ y+b & q+v & g \\ z+c & r+w & h\end{array}\right|$

$=\left|\begin{array}{ccc}x+a & p & l \\ y+b & q & m \\ z+c & r & n\end{array}\right|+\left|\begin{array}{ccc}x+a & u & l \\ y+b & v & m \\ z+c & w & n\end{array}\right|+\left|\begin{array}{ccc}x+a & p+u & f \\ y+b & q+v & g \\ z+c & r+w & h\end{array}\right|$

$=\left|\begin{array}{ccc}x & p & l \\ y & q & m \\ z & r & n\end{array}\right|+\left|\begin{array}{ccc}a & p & l \\ b & q & m \\ c & r & n\end{array}\right|+\left|\begin{array}{ccc}x+a & u & l \\ y+b & v & m \\ z+c & w & n\end{array}\right|+\left|\begin{array}{ccc}x+a & p+u & f \\ y+b & q+v & g \\ z+c & r+w & h\end{array}\right|$

$=\left|\begin{array}{ccc}x & p & l \\ y & q & m \\ z & r & n\end{array}\right|+\left|\begin{array}{ccc}a & p & l \\ b & q & m \\ c & r & n\end{array}\right|+\left|\begin{array}{ccc}x & u & l \\ y & v & m \\ z & w & n\end{array}\right|+\left|\begin{array}{ccc}a & u & l \\ b & v & m \\ c & w & n\end{array}\right|+\left|\begin{array}{ccc}x+a & p+u & f \\ y+b & q+v & g \\ z+c & r+w & h\end{array}\right|$

$=\left|\begin{array}{ccc}x & p & l \\ y & q & m \\ z & r & n\end{array}\right|+\left|\begin{array}{ccc}a & p & l \\ b & q & m \\ c & r & n\end{array}\right|+\left|\begin{array}{ccc}x & u & l \\ y & v & m \\ z & w & n\end{array}\right|+\left|\begin{array}{ccc}a & u & l \\ b & v & m \\ c & w & n\end{array}\right|+\left|\begin{array}{ccc}x+a & p & f \\ y+b & q & g \\ z+c & r & h\end{array}\right|+\left|\begin{array}{ccc}x+a & u & f \\ y+b & v & g \\ z+c & w & h\end{array}\right|$

$=\left|\begin{array}{ccc}x & p & l \\ y & q & m \\ z & r & n\end{array}\right|+\left|\begin{array}{ccc}a & p & l \\ b & q & m \\ c & r & n\end{array}\right|+\left|\begin{array}{ccc}x & u & l \\ y & v & m \\ z & w & n\end{array}\right|+\left|\begin{array}{ccc}a & u & l \\ b & v & m \\ c & w & n\end{array}\right|+\left|\begin{array}{ccc}x & p & f \\ y & q & g \\ z & r & h\end{array}\right|+\left|\begin{array}{ccc}a & p & f \\ b & q & g \\ c & r & h\end{array}\right|+\left|\begin{array}{ccc}x+a & u & f \\ y+b & v & g \\ z+c & w & h\end{array}\right|$

$=\left|\begin{array}{ccc}x & p & l \\ y & q & m \\ z & r & n\end{array}\right|+\left|\begin{array}{ccc}a & p & l \\ b & q & m \\ c & r & n\end{array}\right|+\left|\begin{array}{ccc}x & u & l \\ y & v & m \\ z & w & n\end{array}\right|+\left|\begin{array}{ccc}a & u & l \\ b & v & m \\ c & w & n\end{array}\right|+\left|\begin{array}{ccc}x & p & f \\ y & q & g \\ z & r & h\end{array}\right|+\left|\begin{array}{ccc}a & p & f \\ b & q & g \\ c & r & h\end{array}\right|+\left|\begin{array}{ccc}x & u & f \\ y & v & g \\ z & w & h\end{array}\right|+\left|\begin{array}{ccc}a & u & f \\ b & v & g \\ c & w & h\end{array}\right|$

Hence, if the determinant $\left|\begin{array}{ccc}x+a & p+u & l+f \\ y+b & q+v & m+g \\ z+c & r+w & n+h\end{array}\right|$ splits into exactly $k$ determinants of order 3 , each element of which contains only one term, then $k=\underline{8}$.