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

Correct Answer True

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

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

[splitting first row]
$= \left| {\begin{array}{cccccccccccccccccccc}x&p&l\\y&q&m\\{z + c}&{r + m}&{n + h}\end{array}} \right| + \left| {\begin{array}{cccccccccccccccccccc}x&p&l\\b&v&g\\{z + c}&{r + w}&{n + h}\end{array}} \right|$

$+ \left| {\begin{array}{cccccccccccccccccccc}a&u&t\\y&q&m\\{z + c}&{r + w}&{n + h}\end{array}} \right| + \left| {\begin{array}{cccccccccccccccccccc}a&u&f\\b&v&g\\{z + c}&{r + w}&{n + h}\end{array}} \right|$

[splitting second row]
Similarly, we can split these 4 determinants in 8 determinants by splitting each one in two determinants further.

So, given statement is true.