The probability that a randomly chosen 2 x 2 matrix with all the entries from the set of first 10 pr

Q: Wait, why do we reject the case where we have three distinct primes?

That is a fantastic question! If we had three distinct primes, say $a, b, c$, then $ad = bc$ would imply that the prime $a$ must divide the product $bc$. Since $a$ is prime, by Euclid's Lemma, $a$ must divide $b$ or $c$. But since $b$ and $c$ are also primes, $a$ would have to be equal to $b$ or $c$. This contradicts the assumption that they are distinct. Thus, you can never have three distinct primes in this equation!

Q: Why is the order of $a, d$ and $b, c$ important in Case 2?

In a matrix, the position of each entry is fixed. The entry at row 1, column 1 is distinct from the entry at row 1, column 2. When we say the diagonal $\{a, d\}$ is $\{p, q\}$, we can have $(a, d) = (p, q)$ or $(a, d) = (q, p)$. Both are valid, distinct matrices. Ignoring this order would mean undercounting your favorable outcomes.

Q: Why can't we just use energy conservation or some other physics principle?

I love the enthusiasm for physics! However, this is a pure combinatorics and number theory problem. In JEE Advanced, you must switch your 'brain mode' depending on the subject. Here, the 'laws of physics' are replaced by the 'laws of arithmetic'—specifically, the Fundamental Theorem of Arithmetic. Stick to the counting principles, and you will find the path!

Question

The probability that a randomly chosen $2 	imes 2$ matrix with all the entries from the set of first 10 primes, is singular, is equal to

Options:

(A) $133/10^4$

(B) $18/10^3$

(C) $19/10^3$

(D) $271/10^4$

Sigma Percentile · Accepted Answer

### The Elegance of Primes in a Matrix

Imagine you are standing before a $2 	imes 2$ matrix, a simple grid of four numbers. These are the building blocks of all mathematics—the prime numbers.

We are choosing entries from the first ten primes: $S = \{2, 3, 5, 7, 11, 13, 17, 19, 23, 29\}$. Our goal is to find the probability that such a matrix is singular.

### The Sample Space: The Foundation

Before we dive into the mystery of singularity, we must understand the scope of our universe. We have a matrix $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$.

Each of the four positions can be filled by any of the ten primes. By the fundamental principle of counting, the total number of possible matrices is:

$$10 	imes 10 	imes 10 	imes 10 = 10^4 = 10,000$$

This is our sample space. It is a large, structured space, and we are looking for the rare, singular matrices within it.

### The Singularity Condition

A matrix is singular if and only if its determinant is zero. For our $2 	imes 2$ matrix, the determinant is $\det(A) = ad - bc$.

Setting this to zero gives us the elegant condition:

$$ad = bc$$

This is the gatekeeper of our problem. If a matrix satisfies this, it is singular; if not, it is invertible.

### The Magic of the Fundamental Theorem of Arithmetic

Here is where the beauty of prime numbers shines. The Fundamental Theorem of Arithmetic tells us that every integer has a unique prime factorization.

If $ad = bc$, then the prime factors on the left must match the prime factors on the right. This means the multiset $\{a, d\}$ must be identical to the multiset $\{b, c\}$.

This is the key that unlocks the entire problem. We don't need to test random numbers; we only need to consider how these sets can be formed.

### Case 1: The Uniform Matrix

Let's start with the simplest scenario. What if all four entries are the same prime number?

If $a = b = c = d = p$, then $p 	imes p = p 	imes p$, which is always true. Since there are ten primes in our set $S$, there are exactly $10$ such matrices.

### Case 2: The Mixed Matrix

Now, what if we use two distinct primes, $p$ and $q$? For the condition $ad = bc$ to hold, the set $\{a, d\}$ must be $\{p, q\}$ and the set $\{b, c\}$ must also be $\{p, q\}$.

First, we choose two distinct primes from our set of ten. The number of ways to do this is:

$$\binom{10}{2} = 45$$

For each pair $\{p, q\}$, we must arrange them. The main diagonal $(a, d)$ can be $(p, q)$ or $(q, p)$, which provides $2$ ways. Similarly, the off-diagonal $(b, c)$ can be $(p, q)$ or $(q, p)$, providing another $2$ ways.

Thus, for each pair, we have $2 	imes 2 = 4$ arrangements. Multiplying this by our $45$ pairs, we get $45 	imes 4 = 180$ matrices.

### The Final Synthesis

We have explored our cases. We have $10$ matrices from Case 1 and $180$ from Case 2.

Adding these together, we find $190$ singular matrices. The probability is the ratio of favorable outcomes to total outcomes:

$$P(E) = \frac{190}{10000}$$

Simplifying this, we get **$\frac{19}{1000}$**. It is a beautiful result, born from the unique properties of primes and the structured nature of matrices.

Animated Solution for Mathematics - Probability: The probability that a randomly chosen $2 \times 2$ matrix with all the entries from the set of first 10 primes, is singular, is equal to

Select Answer:

Visualized Solution

Defining the Matrix and Set

Total Possible Matrices

Condition for Singularity

Prime Factorization Logic

Case 1: All Entries Identical

Case 2: Two Distinct Primes

Arranging the Primes in Case 2

Total Favorable Outcomes

Final Probability Calculation

The Sigma Insight: Classical Definition of Probability

The Elegance of Primes in a Matrix

The Sample Space

The Singularity Condition

The Magic of the Fundamental Theorem of Arithmetic

Case 1

Case 2

The Final Synthesis

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