Question

Consider a determinant of order 3 all whose entries are either 0 or 1 . Five of these entries are 1 and four of them are ' 0 '. Also $a_{ij}=a_{ji}\forall 1\le i,j\le 3$. Find the number of such determinants.

Answer 1

Answer: 12

We have $a_{ij}=a_{ji}\Rightarrow a_{12}=a_{21}=a$ (say)
$a_{31}=a_{13}=b$ and $a_{23}=a_{32}$
$1,1,1,1,1$ and $0,0,0,0$
$\left|\begin{array}{ccc}x& a& b\\ a& x& c\\ b& c& x\end{array}\right|$
Case-I :
If diagonal elements are $(1,1,1)$
Conjugate element are, $(0,0),(0,0),(1,1)$ as shown.
$\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& 1& 1\end{array}\right|$
This can be done in $\frac{3!}{2!}=3$ ways
Case-II :
If diagonal has $(1,0,0)$ then the conjugate elements are $(1,1),(1,1),(0,0)$.
Now conjugate elements can be arranged in $\frac{3!}{2}=3$ ways
$\left|\begin{array}{ccc}1& \bullet & \bullet \\ \bullet & 0& \bullet \\ \bullet & \bullet & 0\end{array}\right|$
and diagonal elements can be filled in $\frac{3!}{2}=3$ ways
Hence $3\times 3=9$ ways
$\therefore$ Total determinant $=3+9=12$