Question

lf $A=\left[\begin{array}{ccc}a+b& a+b+c& a+b+c+d\\ b+c& c+d+e& A+B+C+D\\ c+d& A+B+C& p+q+r+s\end{array}\right]$ then number of decompositions of A are

• A. $12$
• B. $6$
• C. $8$
• D. $24$

Answer 1

Answer: (D)

$24$

We know that
If the elements of row or column of a determinant of order 3 consists of m,n,p terms respectively, then determinant can be expressed in the sum of $m\times n\times p$ determinant of same order.
Here,$n=2\times 3\times 4=24$