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 elements in $A$ 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. at least 7 but less than 10
• D. at least 10.

Answer 1

Answer: (B)

at least 4 but less than 7

$A\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$

Case$1:detA=\left|\begin{array}{ccc}1& a& a\\ a& 1& c\\ b& c& 1\end{array}\right|\ne 0$

No case not possible

Case$2$

$i:det\left(A\right)=\left|\begin{array}{ccc}1& a& a\\ a& 0& c\\ b& c& 0\end{array}\right|\ne 0$

$a,b,c\to$selected from $1,1,0$

$2$ cases

$ii.detA=\left|\begin{array}{ccc}0& a& b\\ a& 1& c\\ b& c& 0\end{array}\right|$

$2$ cases here also

$iii:det\left(A\right)=\left|\begin{array}{ccc}0& a& b\\ a& 0& c\\ b& c& 1\end{array}\right|$

$2$ cases

Total $6$

Option B