Question

Let $A$ be the set of all $3\times 3$ symmetric matrices all whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0.

The number of matrices in $A$ such that $|A|=0$, is

• A. $4$
• B. $6$
• C. $8$
• D. none of these

Answer 1

The correct option is B $6$

Let $x$ be the number of 1's on the main diagonal and $y$ be the number of 1's above the main diagonal, then
$x+2y=5$
$\Rightarrow x=1,y=2orx=3,y=1$.
When $x=1$, the main diagonal can be chosen in 3 ways, and the elements above the main diagonal in 3 ways. Therefore, there are 9 such matrices.These are
$A_1=\left[\begin{array}{ccc}1& 1& 1\\ 1& 0& 0\\ 1& 0& 0\end{array}\right]$, $A_2=\left[\begin{array}{ccc}0& 1& 1\\ 1& 1& 0\\ 1& 0& 0\end{array}\right]$
$A_3=\left[\begin{array}{ccc}0& 1& 1\\ 1& 0& 0\\ 1& 0& 1\end{array}\right]$, $A_4=\left[\begin{array}{ccc}1& 0& 1\\ 0& 0& 1\\ 1& 1& 0\end{array}\right]$
$A_5=\left[\begin{array}{ccc}0& 0& 1\\ 0& 1& 1\\ 1& 1& 0\end{array}\right]$, $A_6=\left[\begin{array}{ccc}0& 0& 1\\ 0& 0& 1\\ 1& 1& 1\end{array}\right]$
$A_7=\left[\begin{array}{ccc}1& 1& 0\\ 1& 0& 1\\ 0& 1& 0\end{array}\right]$, $A_8=\left[\begin{array}{ccc}0& 1& 0\\ 1& 1& 1\\ 0& 1& 0\end{array}\right]$
$A_9=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& 1& 1\end{array}\right]$
When $x=3$, the main diagonal can be chosen in 3 way; and the element above the main diagonal in 3 ways. Therefore, there are 3 such matrices.
$A_{10}=\left[\begin{array}{ccc}1& 1& 0\\ 1& 1& 0\\ 0& 0& 1\end{array}\right]$, $A_{11}=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 1& 0& 1\end{array}\right]$,
$A_{12}=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& 1& 1\end{array}\right]$
$|A|=0$ for six matrices.