Question

Let ${\Delta }_1=\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|$ and ${\Delta }_2=\left|\begin{array}{ccc}{\alpha }_1& {\beta }_1& {\gamma }_1\\ {\alpha }_2& {\beta }_2& {\gamma }_2\\ {\alpha }_3& {\beta }_3& {\gamma }_3\end{array}\right|$
, then ${\Delta }_1\times {\Delta }_2$ can be expressed as the sum of how many determinants?

• A. $9$
• B. $3$
• C. $27$
• D. $2$

Answer 1

Answer: (C)

$27$

Given, ${\Delta }_1=\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|$ and ${\Delta }_2=\left|\begin{array}{ccc}{\alpha }_1& {\beta }_1& {\gamma }_1\\ {\alpha }_2& {\beta }_2& {\gamma }_2\\ {\alpha }_3& {\beta }_3& {\gamma }_3\end{array}\right|$
${\Delta }_1{\Delta }_2=\left|\begin{array}{ccc}{a}_1{\alpha }_1+b_1{\alpha }_2+c_1{\alpha }_3& a_1{\beta }_1+b_1{\beta }_2+c_1{\beta }_3& a_1{\gamma }_1+b_1{\gamma }_2+c_1{\gamma }_3\\ a_2{\alpha }_1+b_2{\alpha }_2+c_2{\alpha }_3& a_2{\beta }_1+b_2{\beta }_2+c_2{\beta }_3& a_2{\gamma }_1+b_2{\gamma }_2+c_2{\gamma }_3\\ a_3{\alpha }_1+b_3{\alpha }_2+c_3{\alpha }_3& a_3{\beta }_1+b_2{\beta }_2+c_3{\beta }_3& a_3{\gamma }_1+b_3{\gamma }_2+c_3{\gamma }_3\end{array}\right|$
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.
So, the number of decompositions $=3\times 3\times 3=27$