Question

Let A be the set of all $3\times 3$ symmetric matrices all whose entries are either 1 or 0. Five of these entries are 1 and four of them are 0.
The number of matrices in A such that $\left|A\right|$= 0, is
$\left(A\right)$ 5
$\left(B\right)$ 6
$\left(C\right)$ 8
$\left(D\right)$ None of these.

Answer 1

Hint – In this particular type of question use the concept that in a symmetric matrices the transpose of the matrix is equal to the original matrix (i.e. $A^T=A$) and the determinant of the matrix is zero if the two rows or two columns are the same so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given data:
A be a $3\times 3$ symmetric matrices all whose entries are either 1 or 0.
As we know that in symmetric matrices the transpose of the matrix is equal to the original matrix.
$\Rightarrow A^T=A$, where T is the symbol for transpose.
Let, A = $$\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]$$, so $A^T=\left[\begin{array}{cc}{a}_{11}& a_{21}\\ a_{12}& a_{22}\end{array}\right]$
$\Rightarrow A^T=A\Rightarrow$ $$\left[\begin{array}{cc}{a}_{11}& a_{21}\\ a_{12}& a_{22}\end{array}\right]=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]$$
Now it is given that in these symmetric matrices five entries are 1 and four entries are zero.
So the possible case arise
$\left(i\right)$ When all the diagonal elements are equal to one.
When all the diagonal elements are equal to one then there is three possibilities of such symmetric matrix so that it consists of 5 one and four zero which is shown below,
$$\left[\begin{array}{ccc}1& 1& 0\\ 1& 1& 0\\ 0& 0& 1\end{array}\right],\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 1& 0& 1\end{array}\right],\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& 1& 1\end{array}\right]$$
$\left(ii\right)$ When in the main diagonal element two zero and one, one.
So in this case 9 possibilities of such symmetric matrix so that it consists of 5 one and four zero which is shown below,
$$\left[\begin{array}{ccc}1& 0& 1\\ 0& 0& 1\\ 1& 1& 0\end{array}\right],\left[\begin{array}{ccc}1& 1& 0\\ 1& 0& 1\\ 0& 1& 0\end{array}\right],\left[\begin{array}{ccc}1& 1& 1\\ 1& 0& 0\\ 1& 0& 0\end{array}\right]$$
$$\left[\begin{array}{ccc}0& 0& 1\\ 0& 1& 1\\ 1& 1& 0\end{array}\right],\left[\begin{array}{ccc}0& 1& 0\\ 1& 1& 1\\ 0& 1& 0\end{array}\right],\left[\begin{array}{ccc}0& 1& 1\\ 1& 1& 0\\ 1& 0& 0\end{array}\right]$$
$$\left[\begin{array}{ccc}0& 0& 1\\ 0& 0& 1\\ 1& 1& 1\end{array}\right],\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 1\\ 0& 1& 1\end{array}\right],\left[\begin{array}{ccc}0& 1& 1\\ 1& 0& 0\\ 1& 0& 1\end{array}\right]$$
$\left(iii\right)$ When in the main diagonal element one zero and two one.
There is no such possibility of such a symmetric matrix.
So there are total, (3+ 9) = 12 such possibilities of A.
Now we have to find out, out of these possibilities the number of matrix such that $\left|A\right|$ = 0.
Now as we know that the determinant of the matrix is zero if the two rows or two columns are the same.
So in the above shown matrix such possibilities are
$$\left[\begin{array}{ccc}1& 1& 0\\ 1& 1& 0\\ 0& 0& 1\end{array}\right],\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 1& 0& 1\end{array}\right],\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& 1& 1\end{array}\right],\left[\begin{array}{ccc}1& 1& 1\\ 1& 0& 0\\ 1& 0& 0\end{array}\right],\left[\begin{array}{ccc}0& 1& 0\\ 1& 1& 1\\ 0& 1& 0\end{array}\right],\left[\begin{array}{ccc}0& 0& 1\\ 0& 0& 1\\ 1& 1& 1\end{array}\right]$$
So as we see here are 6 cases arise, such that $\left|A\right|$ = 0.
So this is the required answer.
Hence option (B) is the correct answer.

Note – Whenever we face such types of questions the key concept we have to remember is that we have considered he only cases of such matrix which are symmetric as above, otherwise we will lead towards the wrong answer, so first find out these cases then apply the determinant rule as above we will get the required answer.