Question

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

The number of matrices $A$ in $d$ for which the system of linear equations $A\left[\begin{array}{c}X\\ Y\\ Z\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$ has a unique solution, is

• A. less than 4
• B. at least 4 but less than 7
• C. atleast 7 but less than 10
• D. at least 10

Answer 1

The correct option is B at least 4 but less than 7

$\left[\begin{array}{ccc}0& a& b\\ a& 0& c\\ b& c& 1\end{array}\right]$
either $b=0$ or $c=0\Rightarrow |A|\ne 0$
$\Rightarrow 2$ matrices
$\left[\begin{array}{ccc}0& a& b\\ a& 1& c\\ b& c& 0\end{array}\right]$
either $a=0$ or $c=0\Rightarrow |A|\ne 0$
$\Rightarrow 2$ matrices
$\left[\begin{array}{ccc}1& a& b\\ a& o& c\\ b& c& o\end{array}\right]$
either $a=0$ or $b=0\Rightarrow |A|\ne 0$
$\Rightarrow 2$ matrices.
$\left[\begin{array}{ccc}1& a& b\\ a& 1& c\\ b& c& 1\end{array}\right]$
If $a=b=0\Rightarrow |A|=0$
If $a=c=0\Rightarrow |A|=0$
If $b=c=0\Rightarrow |A|=0$
$\Rightarrow$ there will be only 6 matrices.