Symmetrical Matrices with entities $0$ and $1$

Question

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

Q.1 The number of matrices in A is

(A) 12

(B) 6

(C) 9

(D) 3

My approach is as follow

$\left[ {\begin{array}{*{20}{c}} a&h&g\\ h&b&f\\ g&f&c \end{array}} \right]$

CASE 1: $a,b,c\ \in{1,1,1}$ Then $f,g,h \in {0,0,1}$ Number of cases $\frac{3!}{2!}=3$

CASE 2: $a,b,c\ \in{1,0,0}$ Then $f,g,h \in {0,1,1}$ Number of cases $\frac{3!}{2!}*\frac{3!}{2!}=9$

Hence the number of matrix in set A =3+9=12 which is correct as per the official answer key.

Q.2 The number of matrices A in A for which the system of linear equations $A\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1\\ 0\\ 0 \end{array}} \right]$ 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

Q.3 The number of matrices A in A for which the system of linear equations $A\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1\\ 0\\ 0 \end{array}} \right]$ is inconsistent , is

(A) 0

(B) more than 2

(C) 2

(D) 1

From continuation from Question1 how we will proceed Question2 and Question3.

As the set A has 12 matrices

Answer 1

The equation has a unique solution if and only if $A$ is nonsingular. Suppose then, that $A$ is nonsingular.

Can $A$ have three $1$s in the same row? Then the corresponding column also has three $1$s, and the remaining four entries are $0$. We see that the other two rows are identical, and $A$ is singular. This eliminates three possibilities for $A$.

Therefore, $A$ must have two rows with two $1$s and one row with a single $1$. Can the single $1$ lie on the diagonal? Then the corresponding column also has only one $1$, so the other four entries are $1$ and again we have two identical rows. This eliminates another three possibilities for $A$.

Suppose that the first row of $A$ is $$\begin{bmatrix}0&1&0\end{bmatrix}$$ Each of the other rows has two $1$ and a $0$, and only one of them can have the $1$ on the diagonal. One possibility is $$A=\begin{bmatrix}0&1&0\\1&0&1\\0&1&1\end{bmatrix}$$ which is nonsingular.

Permuting rows and columns gives six nonsingular matrices, which accounts for all twelve possibilities for $A$.

For part $3$, I can't think of anything better than to examine the cases we looked at in part $2$. We know the equation is consistent in the six cases where $A$ is nonsingular.

The equation can be written as $$x\begin{bmatrix}{a\\h\\g}\end{bmatrix}+y\begin{bmatrix}{h\\b\\f}\end{bmatrix}+z\begin{bmatrix}{g\\f\\c}\end{bmatrix}=\begin{bmatrix}{1\\0\\0}\end{bmatrix}$$ For the case that one row has all $1$s, there are two cases to consider. Either the first row has all $1$s, or one of the the other two rows has all $1$s. This is because of the asymmetry on the right-hand side. The second and third rows are identical, but the first is different. In both cases, we easily find that the equation can be solved.

We have left the three cases where there are three $1$s on the diagonal. I leave them to you.