Let R = { [[a, 3, b], [c, 2, d], [0, 5, 0]] : a,b,c,d ∈ {0,3,5,7,11,13,17,19} }. Then the number of

Q: Why can't we just count the invertible matrices directly? It seems like it should be possible.

Directly counting the invertible matrices is like trying to count all the grains of sand on a beach. Because the condition $ad \neq bc$ is a negative constraint, it is much harder to count than its opposite. By using complementary counting, we turn a complex problem into a much simpler one: Total - Singular. It is a strategic retreat that leads to a faster, more accurate victory.

Q: Why do we treat the case $ad = bc = 0$ separately from the case $ad = bc \neq 0$?

Zero is the 'chaos element' in this problem. It doesn't follow the prime factorization rules that the other numbers do. If we tried to use the prime logic for zero, we would get stuck. By isolating the zero case, we keep our logic clean and ensure that our prime-based counting for the second case is perfectly valid.

Q: How did you know that $ad = bc$ implies $\{a, d\} = \{b, c\}$ for non-zero values?

This is a classic JEE trick! Since all our non-zero numbers are prime, the Fundamental Theorem of Arithmetic tells us that every number has a unique prime factorization. If the product $ad$ equals the product $bc$, then the prime factors on both sides must be identical. This means the set of numbers $\{a, d\}$ must be the same as the set $\{b, c\}$. It is a beautiful, elegant shortcut!

Question

Let $R = \left\{ \begin{pmatrix} a & 3 & b \ c & 2 & d \ 0 & 5 & 0 \end{pmatrix} : a,b,c,d \in \{0,3,5,7,11,13,17,19\} ight\}$. Then the number of invertible matrices in $R$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to tackle a problem that might look like a standard matrix question at first glance, but it is actually a beautiful, hidden puzzle of combinatorics and number theory.

We are given a matrix:

$$M = \begin{pmatrix} a & 3 & b \ c & 2 & d \ 0 & 5 & 0 \end{pmatrix}$$

The variables $a, b, c, d$ are chosen from the set $S = \{0, 3, 5, 7, 11, 13, 17, 19\}$. Our mission is to find the number of invertible matrices.

### The Determinant Shortcut

When we think of invertibility, the first thing that should flash in your mind is the determinant. A matrix is invertible if and only if its determinant is not equal to zero.

Look at the matrix $M$. The third row is a gift, as it contains two zeros and a five. Expanding along this row is the most efficient path:

$$|M| = 0 - 5(ad - bc) + 0 = -5(ad - bc)$$

For the matrix to be invertible, we require $-5(ad - bc) 
eq 0$, which simplifies to the core constraint:

$$ad 
eq bc$$

This is our golden rule.

### The Strategy of Complementary Counting

Trying to count all pairs $(a, b, c, d)$ such that $ad 
eq bc$ is overwhelming. Instead, we use the strategy of **Complementary Counting**.

We calculate the total number of possible matrices and subtract the number of singular matrices (where $ad = bc$). The total number of matrices is:

$$8^4 = 4096$$

Now, we just need to find the singular ones.

### The Prime Number Revelation

This is where the problem gets thrilling. We split the singular condition $ad = bc$ into two cases.

**Case 1:** The product is zero ($ad = 0$ and $bc = 0$). 
For $ad = 0$, either $a=0$ or $d=0$. The number of ways is $8 + 8 - 1 = 15$. Since $bc = 0$ also has 15 ways, the total for Case 1 is:

$$15 	imes 15 = 225$$

**Case 2:** The product is non-zero ($ad = bc 
eq 0$). 
Here, none of the variables can be zero. They must be chosen from the set $\{3, 5, 7, 11, 13, 17, 19\}$, which contains 7 prime numbers.

By the Fundamental Theorem of Arithmetic, if $ad = bc$, the set of prime factors must match. We break this into three subcases:

1. All four variables are the same: $7$ ways.
2. $a=b$ and $d=c$ with $a 
eq d$: $7 	imes 6 = 42$ ways.
3. $a=c$ and $d=b$ with $a 
eq d$: $7 	imes 6 = 42$ ways.

The total for Case 2 is $7 + 42 + 42 = 91$.

### The Final Tally

We have reached the end of our journey. The total number of singular matrices is:

$$225 + 91 = 316$$

Subtracting this from our total of $4096$, we get:

$$4096 - 316 = 3780$$

There you have it! By looking for the hidden structure—the sparse row, the complementary counting, and the prime number property—we turned a daunting problem into a series of logical, satisfying steps. The final answer is **3780**.

Animated Solution for Mathematics - Matrices and Determinants: Let $R = ⎩ ⎨ ⎧ a c 0 325 b d 0 : a, b, c, d \in {0, 3, 5, 7, 11, 13, 17, 19} ⎭ ⎬ ⎫$ . Then the number of invertible matrices in $R$ is

Enter Numerical Value:

Visualized Solution

Matrix and Given Set

Calculating the Determinant

The Invertibility Condition

Strategy: Complementary Counting

Total Number of Matrices

Singular Matrices: Case 1

Counting $a d = 0$

Completing Case 1

Singular Matrices: Case 2

The Prime Factorization Logic

Subcase 2a: All Four are Equal

Subcase 2b: $a = b$ and $d = c$

Subcase 2c: $a = c$ and $d = b$

Total Singular Matrices

Final Calculation

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Determinant Shortcut

The Strategy of Complementary Counting

The Prime Number Revelation

The Final Tally

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