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

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Solution

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

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

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

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