Symmetric Matrices Leading to Inconsistent Linear System

To find the number of matrices $A \in \mathcal{A}$ for which the system of linear equations

$$A \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$$

is inconsistent, we first determine the elements of the set $\mathcal{A}$.

A general $3 \times 3$ symmetric matrix $A$ has the form:

$$A = \begin{bmatrix} a & d & e \\ d & b & f \\ e & f & c \end{bmatrix}$$

where $a, b, c, d, e, f \in \{0, 1\}$.
The sum of all entries of $A$ is:
$\text{Sum} = a + b + c + 2(d + e + f) = 5$

Since $2(d + e + f)$ is even, the sum of the diagonal entries $a + b + c$ must be an odd integer:

1. Case 1: $a + b + c = 3 \implies a = b = c = 1$.
   Then $2(d + e + f) = 2 \implies d + e + f = 1$.
   This gives $\binom{3}{1} = 3$ matrices:
   • Matrix 1: $d=1, e=0, f=0$:

$$A_1 = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

 The system of equations is:
 

$$\begin{aligned} x + y &= 1 \\ x + y &= 0 \\ z &= 0 \end{aligned}$$

 The first two equations, $x+y=1$ and $x+y=0$, contradict each other. Thus, the system is **inconsistent**.

• Matrix 2: $d=0, e=1, f=0$:

$$A_2 = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$$

 The system of equations is:
 

$$\begin{aligned} x + z &= 1 \\ y &= 0 \\ x + z &= 0 \end{aligned}$$

 The first and third equations, $x+z=1$ and $x+z=0$, contradict each other. Thus, the system is **inconsistent**.

• Matrix 3: $d=0, e=0, f=1$:

$$A_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix}$$

 The system of equations is:
 

$$\begin{aligned} x &= 1 \\ y + z &= 0 \\ y + z &= 0 \end{aligned}$$

 This system has infinitely many solutions ($x=1, z=-y$). Thus, it is **consistent**.

2. Case 2: $a + b + c = 1$ and $d + e + f = 2$.
There are $\binom{3}{1} = 3$ choices for the diagonal entries and $\binom{3}{2} = 3$ choices for the off-diagonal entries, giving $3 \times 3 = 9$ matrices:

• For $b = 1, a = c = 0$ with $d = 1, f = 1, e = 0$:

$$A_4 = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}$$

 The system of equations is:
 

$$\begin{aligned} y &= 1 \\ x + y + z &= 0 \\ y &= 0 \end{aligned}$$

 Here, $y=1$ and $y=0$ contradict each other. Thus, the system is **inconsistent**.

• For $c = 1, a = b = 0$ with $e = 1, f = 1, d = 0$:

$$A_5 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \end{bmatrix}$$

 The system of equations is:
 

$$\begin{aligned} z &= 1 \\ z &= 0 \\ x + y + z &= 0 \end{aligned}$$

 Here, $z=1$ and $z=0$ contradict each other. Thus, the system is **inconsistent**.

We have found at least 4 matrices ($A_1, A_2, A_4, A_5$) for which the given system of linear equations is inconsistent.

Therefore, the number of such matrices is more than 2.

Correct Answer: B (more than 2)