Question

In a third order determinant each element of the first column consists of sum of two terms, each element of the second column consists of sum of three terms and each element of third column consists of sum of four terms, then it can be decomposed into $n$ determinants, where $n$ has the value-

• A. $1$
• B. $9$
• C. $16$
• D. $24$

Answer 1

The correct option is D $24$

$\left|\begin{array}{ccc}{a}_1+b_1& x_1+y_1+z_1& u_1+v_1+w_1+t_1\\ a_2+b_2& x_2+y_2+z_2& u_2+v_2+w_2+t_2\\ a_3+b_3& x_3+y_3+z_3& u_3+v_3+w_3+t_3\end{array}\right|$

These two small determinants can further expressed as sum of three determinants by separating the terms in second column.

Further, the six determinants each can be expressed as sum of four determinant by separating terms of third column.

Total number of determinants $=2\times 3\times 4$

$=24$