Consider the set A of all determinants of order 3 with entries 0 or 1 only. Let B be the subset of A

Q: Wait, why can't we just calculate all $2^9 = 512$ determinants and count them?

While technically possible, calculating 512 determinants is extremely tedious and prone to human error. In JEE Advanced, the goal is to find elegant, structural solutions. Using the row-interchange property allows us to solve the problem conceptually without brute force, which is the hallmark of a true mathematician.

Q: Does swapping the rows introduce any numbers other than $0$ or $1$?

No, it does not. The entries of the determinant are just rearranged. Since the original entries were only $0$ or $1$, the new matrix will also only contain $0$ or $1$. Thus, the new matrix is guaranteed to be in set $A$.

Q: Is it possible for a determinant to be $0$?

Yes, absolutely. Many of the $512$ matrices will have a determinant of $0$. However, the question specifically asks us to compare the subsets $B$ (value $1$) and $C$ (value $-1$). The matrices with determinant $0$ do not affect the relationship between $B$ and $C$.

Question

Consider the set $A$ of all determinants of order 3 with entries 0 or 1 only. Let $B$ be the subset of $A$ consisting of all determinants with value 1. Let $C$ be the subset of $A$ consisting of all determinants with value $-1$. Then

Options:

(A) (a) $C$ is empty

(B) (b) $B$ has as many elements as $C$

(C) (c) $A = B \cup C$

(D) (d) $B$ has twice as many elements as $C$

Sigma Percentile · Accepted Answer

### The Landscape of Determinants

Imagine you are standing in front of a massive wall of $512$ matrices. Each one is a $3 	imes 3$ grid, and every single cell is filled with either a $0$ or a $1$.

Your task is to categorize them based on their determinant values. Some will evaluate to $1$, some to $-1$, and many will collapse to $0$.

It feels like an impossible mountain of arithmetic, but you do not need to climb it. You only need to find the bridge that connects the two peaks we care about: the set $B$ (where $\det(D) = 1$) and the set $C$ (where $\det(D) = -1$).

### The Magic of Row Operations

In linear algebra, we have a powerful tool known as the row-interchange property. It states that if you swap any two rows of a determinant, the value of the determinant is multiplied by $-1$.

Mathematically, if $\det(D) = \Delta$, then swapping two rows results in a new determinant $D'$ such that:

$$\det(D') = -\Delta$$

This is not just a rule; it is a geometric transformation. It tells us that the sign of the determinant is intrinsically linked to the order of the rows.

### Building the Bridge

Let us define a mapping $f: B 	o C$. Take any determinant $D$ from set $B$. By definition, $\det(D) = 1$.

Now, perform a simple operation: swap the first row $R_1$ and the second row $R_2$ to create a new matrix $D'$. Because we only swapped rows, the entries of $D'$ are still just $0$s and $1$s.

This means $D'$ is still a valid member of our original set $A$. Because of the row-interchange property, the value of this new determinant is:

$$\det(D') = -\det(D) = -1$$

This confirms that $D'$ is a member of set $C$. We have successfully built a bridge from $B$ to $C$.

### The Elegance of the Bijection

To prove that the number of elements in $B$ is equal to the number of elements in $C$, we must show that our mapping $f$ is a bijection.

First, is it injective? Yes. If two different matrices $D_1$ and $D_2$ in $B$ mapped to the same matrix $D'$ in $C$, we could simply swap the rows of $D'$ back to recover the original matrices. Since the operation is reversible, $D_1$ must equal $D_2$.

Second, is it surjective? Yes. For any matrix $M$ in $C$, we can swap its first two rows to get a matrix $M'$ in $B$. Thus, every element in $C$ has a corresponding pre-image in $B$.

Because the mapping is both injective and surjective, we have a perfect one-to-one correspondence. The complexity of the $512$ matrices vanishes, replaced by the simple, beautiful symmetry of row operations.

The correct answer is that **$B$ has as many elements as $C$**.

Animated Solution for Mathematics - Matrices and Determinants: Consider the set $A$ of all determinants of order 3 with entries 0 or 1 only. Let $B$ be the subset of $A$ consisting of all determinants with value 1. Let $C$ be the subset of $A$ consisting of all determinants with value $- 1$ . Then

Select Answer:

Visualized Solution

Defining the Set $A$

Subsets $B$ and $C$

The Row-Interchange Property

Constructing the Mapping $f$

Verifying the Domain and Codomain

Proving Injectivity

Proving Surjectivity

Establishing the Bijection

Final Conclusion: $∣ B ∣ = ∣ C ∣$

The Sigma Insight: Properties of Determinants

The Landscape of Determinants

The Magic of Row Operations

Building the Bridge

The Elegance of the Bijection

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Visualized Solution

Defining the Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Subsets .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C