Let $$A$$ be a $$2 \times 2$$ real matrix with entries from $$\{0, 1\}$$ and $$|A| \ne 0$$. Consider the following two statements:
$$(P)$$ If $$A \ne I_2$$, then $$|A| = -1$$
$$(Q)$$ If $$|A| = 1$$, then $$tr(A) = 2$$
Where $$I_2$$ denotes $$2 \times 2$$ identity matrix and $$tr(A)$$ denotes the sum of the diagonal entries of $$A$$. Then:

We need to find all $$2 \times 2$$ matrices $$A$$ with entries from $$\{0, 1\}$$ such that $$|A| \neq 0$$, and then check statements (P) and (Q).

Let $$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ where $$a, b, c, d \in \{0, 1\}$$ and $$|A| = ad - bc \neq 0$$.

Since entries are 0 or 1, the determinant $$ad - bc$$ can only be $$-1$$, $$0$$, or $$1$$. We need $$|A| \neq 0$$, so $$|A| = 1$$ or $$|A| = -1$$.

Case 1: $$|A| = 1$$ (i.e., $$ad = 1$$ and $$bc = 0$$)

$$ad = 1$$ requires $$a = 1$$ and $$d = 1$$. $$bc = 0$$ requires at least one of $$b, c$$ to be 0.

The matrices are:

$$A_1 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I_2$$, with $$\text{tr}(A_1) = 2$$

$$A_2 = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$, with $$\text{tr}(A_2) = 2$$

$$A_3 = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$$, with $$\text{tr}(A_3) = 2$$

Case 2: $$|A| = -1$$ (i.e., $$ad = 0$$ and $$bc = 1$$)

$$bc = 1$$ requires $$b = 1$$ and $$c = 1$$. $$ad = 0$$ requires at least one of $$a, d$$ to be 0.

The matrices are:

$$A_4 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$, with $$\text{tr}(A_4) = 0$$

$$A_5 = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$$, with $$\text{tr}(A_5) = 1$$

$$A_6 = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$$, with $$\text{tr}(A_6) = 1$$

Now we check each statement.

Statement (P): If $$A \neq I_2$$, then $$|A| = -1$$.

Consider $$A_2 = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$. Here $$A_2 \neq I_2$$, but $$|A_2| = 1 \times 1 - 1 \times 0 = 1 \neq -1$$.

This is a counterexample, so statement (P) is false.

Statement (Q): If $$|A| = 1$$, then $$\text{tr}(A) = 2$$.

From Case 1, all matrices with $$|A| = 1$$ are $$A_1, A_2, A_3$$, and every one of them has $$\text{tr}(A) = 2$$.

So statement (Q) is true.

Therefore, (P) is false and (Q) is true, which corresponds to Option A.