Question

Let $A$ be the set of all $3\times 3$ symmetric matrices all of whose entries are either $0$ or $1$. If five of these entries are $1$ and four of them are $0$, then the number of matrices in $A$ is

• A. $12$
• B. $6$
• C. $9$
• D. $3$

Answer 1

The correct option is A $12$

Let $x$ be the number of $1^{\prime }s$ on main diagonal and $y$ be the number of $1^{\prime }s$ above main diagonal.
$x+2y=5$
$\Rightarrow x=1,y=2$ and $x=3,y=1$

Case I: If $x=1,y=2$
Here, the main diagonal can be chosen in $3$ ways and the elements above the main diagonal can be chosen in $3$ ways
So, there are $9$ such matrices.
Case II; If $x=3,y=1$
Here, the main diagonal can be chosen in $1$ way and the elements above the main diagonal can be chosen in $3$ ways
So, there are $3$ such matrices.
$\therefore$ There are in all $12$ matrices.

Hence, option A.