Question

Let $A$ be the set of all $3\times 3$ symmetric matrices, all of whose entries are either $0$ or $1$ and five of these entries are $1$. Consider a system of linear equations defined as $B_i\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$, where $B_i\in B\subset A$. Then which of the following is (are) CORRECT?

• A. $n\left(A\right)=12$
• B. $n\left(A\right)=9$
• C. If the system of equations has a unique solution, then $n\left(B\right)=4$
• D. If the system of equations has a unique solution, then $n\left(B\right)=6$

Answer 1

Answer: (A), (D)

The correct options are
A $n\left(A\right)=12$
D If the system of equations has a unique solution, then $n\left(B\right)=6$

Since, $A$ is symmetric, $A$ is of the form $A=\left[\begin{array}{ccc}a& b& c\\ b& d& e\\ c& e& f\end{array}\right]$

Due to symmetric nature, there arises only two cases.
Case $1:$ All three diagonal elements are $1$.
Then we have to choose only entry out of $b,c,e$ to be $1$ so that there are five $1^{\prime }s$.
Number of matrices $={}^3{C}_1=3$

Case $2:$ Two diagonal elements are $0$ and one element is $1$.
One $1$ among three diagonal entries can be done in ${}^3{C}_1$ ways.
Among off-diagonal entries, only two entries out of three can be selected to be $1$ so that there are four $1^{\prime }s$ among the off-diagonal entries. This can be done in ${}^3{C}_2$ ways.
Number of matrices $={}^3{C}_1\cdot {}^3{C}_2=9$
$\therefore$ Total number of possible matrices for $A$ is $12$.

Given, $B_i\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$
For unique solution, $det\left(B_i\right)\ne 0$
Case $1:$
$det\left(B_i\right)=\left[\begin{array}{ccc}1& b& c\\ b& 1& e\\ c& e& 1\end{array}\right]$
$=1-b^2-c^2-e^2+2bce$
$=0$ [Since, $b,c,e$ are selected from $1,0,0$]
So, this case is not possible.

Case $2:$
$\left(i\right)det\left(B_i\right)=\left[\begin{array}{ccc}1& b& c\\ b& 0& e\\ c& e& 0\end{array}\right]=2bce-e^2\ne 0$
Here, $b,c,e$ are selected from $1,1,0$.
Two cases are possible. Either $b$ and $e$ are $1$ or $c$ and $e$ are $1$.

$\left(ii\right)$ Similarly, for
$det\left(B_i\right)=\left[\begin{array}{ccc}0& b& c\\ b& 1& e\\ c& e& 0\end{array}\right]=2bce-c^2\ne 0$
two cases are possible.

and $\left(iii\right)$ $det\left(B_i\right)=\left[\begin{array}{ccc}0& b& c\\ b& 0& e\\ c& e& 1\end{array}\right]=2bce-b^2\ne 0$
two cases are possible.

Hence, there are exactly $6$ matrices for which the given system of equations has unique solution.