Let A be a 2 × 2 real matrix with entries from {0,1} and |A| ≠ 0. Consider the following two stateme

Q: Wait, why do we reject the determinant being 0? The problem just says $|A| \neq 0$.

That is a great observation! The problem explicitly states $|A| \neq 0$ as a condition for the matrix $A$. In mathematics, when we are given a constraint, it acts as a filter. We are only interested in matrices that are invertible. By eliminating the case where the determinant is zero, we narrow our search space to only those matrices that are non-singular, which simplifies our analysis significantly.

Q: How did you know $ad=1$ and $bc=0$ immediately for Statement (Q)? Could it be something else?

Let's look at the equation $ad - bc = 1$. Since $a, b, c, d \in \{0, 1\}$, the products $ad$ and $bc$ can only be $0$ or $1$. If $ad=0$, then $0 - bc = 1$, which implies $bc = -1$, which is impossible. If $ad=1$ and $bc=1$, then $1 - 1 = 0$, which contradicts our premise that $|A|=1$. Therefore, the only logical combination is $ad=1$ and $bc=0$.

Q: Is there a systematic way to find counterexamples for Statement (P)? I always struggle with guessing.

Guessing is actually a valid strategy when the search space is small! Since we are dealing with a $2 \times 2$ matrix with binary entries, there are only $2^4 = 16$ possible matrices. We can systematically test them. For Statement (P), we just need one matrix that is not the identity but has a determinant of $1$. By tweaking the identity matrix slightly, we found $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$, which works perfectly.

Question

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

Options:

(A) (P) is true and (Q) is false

(B) Both (P) and (Q) are false

(C) Both (P) and (Q) are true

(D) (P) is false and (Q) is true

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are diving into a problem that might look like a simple exercise in matrix algebra, but it is actually a beautiful lesson in logical constraints.

We are dealing with a $2 	imes 2$ matrix $A$ where every entry is either $0$ or $1$. This is what we call a binary matrix.

The problem gives us a powerful constraint: the determinant $|A| 
eq 0$. This is our North Star, as it tells us that our matrix must be invertible. Let us define our matrix as:

$$A = \begin{pmatrix} a & b \ c & d \end{pmatrix}, 	ext{ where } a, b, c, d \in \{0, 1\}$$

### The Determinant Landscape

To understand the behavior of this matrix, we must look at the determinant formula: $|A| = ad - bc$. Since our entries are restricted to the set $\{0, 1\}$, the products $ad$ and $bc$ are also restricted to the same set.

There is no other outcome for these products: $0 	imes 0 = 0$, $0 	imes 1 = 0$, $1 	imes 0 = 0$, and $1 	imes 1 = 1$.

Now, let us calculate the possible values for the difference $ad - bc$. We could have $1 - 0 = 1$, $0 - 0 = 0$, $1 - 1 = 0$, or $0 - 1 = -1$.

Since the problem explicitly forbids the determinant from being zero, we eliminate the cases where the result is $0$. We are left with only two possibilities for the determinant: $|A| = 1$ or $|A| = -1$.

### The Fall of Statement (P)

Now, let us tackle Statement (P), which claims: "If $A 
eq I_2$, then $|A| = -1$." Here, $I_2$ is the identity matrix $\begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}$, which has a determinant of $1$.

Statement (P) is essentially saying that the only matrix with a determinant of $1$ is the identity matrix itself. To prove this wrong, we only need one counterexample.

Consider the matrix:

$$A = \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix}$$

This matrix is clearly not the identity matrix because of the entry in the top right corner. Let us calculate its determinant: $|A| = (1 	imes 1) - (1 	imes 0) = 1 - 0 = 1$.

We have found a matrix where $A 
eq I_2$ and yet $|A| = 1$. This directly contradicts Statement (P); therefore, **Statement (P) is false**.

### The Triumph of Statement (Q)

Finally, let us examine Statement (Q): "If $|A| = 1$, then $tr(A) = 2$." The trace of a matrix, $tr(A)$, is the sum of the diagonal elements, $a + d$.

We are given that $|A| = ad - bc = 1$. As we established earlier, since $ad$ and $bc$ can only be $0$ or $1$, the only way to get a difference of $1$ is if $ad = 1$ and $bc = 0$.

Now, focus on the product $ad = 1$. Since $a$ and $d$ are binary, the only way their product can be $1$ is if both $a = 1$ and $d = 1$.

If either were $0$, the product would be $0$. This is a rigid logical lock! If $a = 1$ and $d = 1$, then the trace is:

$$tr(A) = a + d = 1 + 1 = 2$$

**Statement (Q) is perfectly correct.** We have navigated the constraints, tested the logic, and arrived at the truth.

Select Answer:

Visualized Solution

Defining the Matrix $A$

The Determinant Formula

Possible Values of $∣ A ∣$

Analyzing Statement (P)

Finding a Counterexample for (P)

Analyzing Statement (Q)

Solving for $a d$ and $b c$

Deducing Diagonal Entries

Calculating the Trace

Final Conclusion

The Sigma Insight: Types of Matrices

Analyzing the Setup

The Determinant Landscape

The Fall of Statement (P)

The Triumph of Statement (Q)

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