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 $a _{ ij }= a _{ ji } \forall 1 \leq i , j \leq 3$. Find the number of such determinants.

Answer 1

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 $0,0,0,0$
$\begin{vmatrix}x & a & b \\ a & x & c \\ b & c & x\end{vmatrix}$
Case-I :
If diagonal elements are $(1,1,1)$
Conjugate element are, $(0,0),(0,0),(1,1)$ as shown.
$\begin{vmatrix}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1\end{vmatrix}$
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
$\begin{vmatrix}1 & \bullet & \bullet \\ \bullet & 0 & \bullet \\ \bullet & \bullet & 0\end{vmatrix}$
and diagonal elements can be filled in $\frac{3 !}{2}=3$ ways
Hence $3 \times 3=9$ ways
$\therefore$ Total determinant $=3+9=12$