Question

If the determinant $\left|\begin{array}{ccc}a+p& 1+x& u+f\\ b+q& m+y& v+g\\ c+r& n+z& w+h\end{array}\right|$ splits into exactly $K$ determinants of order $3$, each elements of which contains only one term, then the value of $K$, is-

• A. $6$
• B. $8$
• C. $9$
• D. $12$

Answer 1

Answer: (B)

$8$

Now,

$\left|\begin{array}{ccc}a+p& 1+x& u+f\\ b+q& m+y& v+g\\ c+r& n+z& w+h\end{array}\right|$

$=\left|\begin{array}{ccc}a& 1+x& u+f\\ b& m+y& v+g\\ c& n+z& w+h\end{array}\right|+$$\left|\begin{array}{ccc}p& 1+x& u+f\\ q& m+y& v+g\\ r& n+z& w+h\end{array}\right|$

$=\left|\begin{array}{ccc}a& 1& u+f\\ b& m& v+g\\ c& n& w+h\end{array}\right|+$$\left|\begin{array}{ccc}a& x& u+f\\ b& y& v+g\\ c& z& w+h\end{array}\right|+$$\left|\begin{array}{ccc}p& x& u+f\\ q& y& v+g\\ r& z& w+h\end{array}\right|+$$\left|\begin{array}{ccc}p& 1& u+f\\ q& m& v+g\\ r& n& w+h\end{array}\right|$

If we proceed in this way, from above it is clear each single matrix can be further split into two matrices. Then we will get $4\times 2=8$ matrices and they can not be further split.