Key Concepts and Formulas

This problem requires a thorough understanding of $2 \times 2$ matrices, their properties, and basic propositional logic. We will utilize the following definitions and formulas:

1. Matrix Representation: A $2 \times 2$ matrix $A$ is generally represented as: $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$
2. Entries from $\{0, 1\}$: In this problem, the elements $a, b, c, d$ can only be $0$ or $1$.
3. Determinant of a $2 \times 2$ Matrix: The determinant of $A$, denoted $|A|$ or $\det(A)$, is calculated as: $|A| = ad - bc$
4. Trace of a Matrix: The trace of $A$, denoted $\text{tr}(A)$, is the sum of its diagonal entries: $\text{tr}(A) = a + d$
5. Identity Matrix: The $2 \times 2$ identity matrix, $I_2$, is: $I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
6. Conditional Statements: A statement of the form "If P, then Q" is true unless P is true and Q is false. To prove it false, we need to find just one counterexample where the condition (P) is met, but the conclusion (Q) is not.

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Step 1: Identify all valid matrices based on the given conditions

We are given a $2 \times 2$ real matrix $A$ with entries from $\{0, 1\}$, such that $|A| \ne 0$. Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, where $a, b, c, d \in \{0, 1\}$.

First, let's determine the possible values for the determinant $|A| = ad - bc$. Since $a, b, c, d \in \{0, 1\}$, the products $ad$ and $bc$ can only be $0$ or $1$. Therefore, the possible values for $ad-bc$ are:

• $0 - 0 = 0$
• $1 - 0 = 1$
• $0 - 1 = -1$
• $1 - 1 = 0$

The condition $|A| \ne 0$ implies that the determinant can only be $1$ or $-1$.

Now, let's systematically list all such matrices:

Case 1: $|A| = 1$ This requires $ad=1$ and $bc=0$.

• For $ad=1$, both $a$ and $d$ must be $1$.
• For $bc=0$, at least one of $b$ or $c$ must be $0$. The possible pairs for $(b, c)$ are $(0,0)$, $(0,1)$, and $(1,0)$.

Combining these, the matrices with $|A|=1$ are:

1. $A_1 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ (where $a=1, d=1, b=0, c=0$) This is the identity matrix $I_2$. $|A_1| = (1)(1) - (0)(0) = 1$. $\text{tr}(A_1) = 1+1 = 2$.
2. $A_2 = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$ (where $a=1, d=1, b=0, c=1$) $|A_2| = (1)(1) - (0)(1) = 1$. $\text{tr}(A_2) = 1+1 = 2$.
3. $A_3 = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ (where $a=1, d=1, b=1, c=0$) $|A_3| = (1)(1) - (1)(0) = 1$. $\text{tr}(A_3) = 1+1 = 2$.

Case 2: $|A| = -1$ This requires $ad=0$ and $bc=1$.

• For $bc=1$, both $b$ and $c$ must be $1$.
• For $ad=0$, at least one of $a$ or $d$ must be $0$. The possible pairs for $(a, d)$ are $(0,0)$, $(0,1)$, and $(1,0)$.

Combining these, the matrices with $|A|=-1$ are:

1. $A_4 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ (where $b=1, c=1, a=0, d=0$) $|A_4| = (0)(0) - (1)(1) = -1$. $\text{tr}(A_4) = 0+0 = 0$.
2. $A_5 = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$ (where $b=1, c=1, a=0, d=1$) $|A_5| = (0)(1) - (1)(1) = -1$. $\text{tr}(A_5) = 0+1 = 1$.
3. $A_6 = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$ (where $b=1, c=1, a=1, d=0$) $|A_6| = (1)(0) - (1)(1) = -1$. $\text{tr}(A_6) = 1+0 = 1$.

So, there are exactly 6 matrices that satisfy the given conditions.

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Step 2: Evaluate Statement (P)

Statement (P) is: "If $A \ne I_2$, then $|A| = -1$." This is a conditional statement. To determine if it's true or false, we look for counterexamples. A counterexample would be a matrix $A$ from our list such that $A \ne I_2$ (the premise is true) AND $|A| \ne -1$ (the conclusion is false).

Let's examine the matrices that are not equal to $I_2$: $A_2, A_3, A_4, A_5, A_6$.

• Consider $A_2 = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$:
  • Is $A_2 \ne I_2$? Yes, this is true.
  • Is $|A_2| = -1$? No, $|A_2| = 1$. This is false. Since we found a matrix ($A_2$) for which the premise ($A \ne I_2$) is true, but the conclusion ($|A| = -1$) is false, statement (P) is false.

(We could also use $A_3$ as a counterexample, as $A_3 \ne I_2$ but $|A_3|=1$).

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Step 3: Evaluate Statement (Q)

Statement (Q) is: "If $|A| = 1$, then $\text{tr}(A) = 2$." This is also a conditional statement. To determine if it's true or false, we look for counterexamples. A counterexample would be a matrix $A$ from our list such that $|A|=1$ (the premise is true) AND $\text{tr}(A) \ne 2$ (the conclusion is false).

Let's examine the matrices where $|A|=1$: $A_1, A_2, A_3$.

• For $A_1 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$:
  • Is $|A_1|=1$? Yes, this is true.
  • Is $\text{tr}(A_1)=2$? Yes, $1+1=2$. This is true. (Does not contradict Q)
• For $A_2 = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$:
  • Is $|A_2|=1$? Yes, this is true.
  • Is $\text{tr}(A_2)=2$? Yes, $1+1=2$. This is true. (Does not contradict Q)
• For $A_3 = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$:
  • Is $|A_3|=1$? Yes, this is true.
  • Is $\text{tr}(A_3)=2$? Yes, $1+1=2$. This is true. (Does not contradict Q)

Since for all matrices where $|A|=1$, we found that $\text{tr}(A)=2$, there are no counterexamples to statement (Q). Therefore, statement (Q) is true.

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Step 4: Conclusion

Based on our analysis:

• Statement (P) is false.
• Statement (Q) is true.

This corresponds to option (D).

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Tips and Common Mistakes

1. Systematic Listing: When dealing with a small finite set of possibilities (like entries from $\{0, 1\}$), it's often best to systematically list all valid cases. This helps ensure no matrices are missed and avoids errors in calculation.
2. Understanding Conditional Statements: A common mistake is misinterpreting "If P, then Q". Remember it's only false if P is true AND Q is false. If P is false, the statement "If P, then Q" is considered vacuously true, but for these problems, we focus on cases where P is true.
3. Double-Check Calculations: Simple arithmetic errors in calculating determinants or traces can lead to incorrect conclusions.
4. Careful with Definitions: Ensure you correctly apply the definitions of $I_2$, $|A|$, and $\text{tr}(A)$.

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Summary / Key Takeaway

This problem emphasizes the importance of systematically exploring all possible matrices defined by the given constraints. By doing so, we can rigorously test the truth value of propositional statements about these matrices. For $2 \times 2$ matrices with entries restricted to $\{0, 1\}$ and a non-zero determinant, there are only six such matrices. Careful evaluation of each statement against these matrices reveals that (P) is false and (Q) is true.