Let A=[a_ij] be a square matrix of order 2 with entries either 0 or 1. Let E be the event that A is

Q: Wait, why can't the determinant be, say, 2 or -2? The entries are just 0 and 1.

Excellent observation! Since $a, b, c, d \in \{0, 1\}$, the products $ad$ and $bc$ can only be $0$ or $1$. Therefore, the difference $ad - bc$ is restricted to the set $\{1 - 0, 0 - 0, 0 - 1\}$, which is $\{1, 0, -1\}$. You cannot get a $2$ or $-2$ because that would require a product of $2$, which is impossible with binary entries.

Q: Is there a faster way to solve this without listing cases?

Yes! You could think about linear independence. A $2 \times 2$ matrix is invertible if its rows are linearly independent. Since the rows are vectors like $(0,0), (0,1), (1,0), (1,1)$, you can count how many pairs of rows are linearly dependent (i.e., one is a multiple of the other) and subtract that from the total. However, for a $2 \times 2$ matrix, the case-based counting we did is actually the most robust and least error-prone method.

Q: Why did we multiply the number of ways for $a,d$ and $b,c$ in the cases?

Because the choices for the pair $(a,d)$ and the pair $(b,c)$ are independent events. The Fundamental Principle of Counting states that if one event can occur in $m$ ways and another in $n$ ways, the total number of ways both can occur is $m \times n$. Here, we are essentially filling independent slots in the matrix.

Question

Let $A=[a_{ij}]$ be a square matrix of order 2 with entries either 0 or 1. Let E be the event that A is an invertible
matrix. Then the probability $P(E)$ is:

Options:

(A) $\frac{3}{16}$

(B) $\frac{5}{8}$

(C) $\frac{3}{8}$

(D) $\frac{1}{8}$

Sigma Percentile · Accepted Answer

### The Matrix Universe: A Journey into Combinatorics

Imagine you are standing before a $2 	imes 2$ grid. It is a simple structure, just four slots waiting to be filled. In the world of JEE Advanced, simplicity is often a mask for profound mathematical depth.

Today, we are going to explore the 'Matrix Universe' where every entry is restricted to a binary choice: $0$ or $1$. We aren't just filling slots; we are determining the fate of a matrix—whether it is invertible or singular.

### Phase 1: Defining the Sample Space

First, we must understand the scope of our universe. We have a matrix $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$. Each variable, $a, b, c,$ and $d$, is a binary switch that can be 'off' ($0$) or 'on' ($1$).

If you have four slots, and each slot has two independent choices, we apply the Fundamental Principle of Counting. For the first slot, we have $2$ choices, for the second $2$ choices, and so on.

The total number of matrices in our sample space is:

$$2 	imes 2 	imes 2 	imes 2 = 2^4 = 16$$

We have sixteen distinct universes to explore. Our goal is to find how many of these universes allow for an invertible matrix.

### Phase 2: The Gatekeeper of Invertibility

What makes a matrix 'invertible'? It is the determinant. The determinant is the gatekeeper.

If the determinant $|A| = ad - bc$ is non-zero, the matrix is invertible. If it is zero, the matrix is singular—it collapses, losing its inverse.

Because our entries are restricted to $\{0, 1\}$, the products $ad$ and $bc$ are also restricted to $\{0, 1\}$. We only have three potential outcomes for the determinant:

1. $1 - 0 = 1$
2. $0 - 1 = -1$
3. $0 - 0 = 0$

To be invertible, we need the determinant to be non-zero. This means we are hunting for the cases where the determinant is either $1$ or $-1$.

### Phase 3: The Case-by-Case Analysis

Let us hunt for these cases with precision.

**Case 1: The Determinant is $1$**
We need $ad - bc = 1$. Since $ad$ and $bc$ are binary, the only way to get a difference of $1$ is if $ad = 1$ and $bc = 0$.

For $ad = 1$, both $a$ and $d$ must be $1$. There is only $1$ way to do this.

For $bc = 0$, the pair $(b, c)$ can be $(0, 0), (0, 1),$ or $(1, 0)$. That gives us $3$ ways.

Multiplying these independent choices, we find $1 	imes 3 = 3$ matrices where the determinant is $1$.

**Case 2: The Determinant is $-1$**
Now, we need $ad - bc = -1$. This requires $ad = 0$ and $bc = 1$.

For $bc = 1$, both $b$ and $c$ must be $1$. Again, only $1$ way.

For $ad = 0$, the pair $(a, d)$ can be $(0, 0), (0, 1),$ or $(1, 0)$. That gives us $3$ ways.

Multiplying these, we find $3 	imes 1 = 3$ matrices where the determinant is $-1$.

### The Synthesis

We have successfully navigated the landscape. We found $3$ matrices with a determinant of $1$ and $3$ matrices with a determinant of $-1$. Adding these together, we have $3 + 3 = 6$ favorable outcomes.

Finally, the probability $P(E)$ is the ratio of favorable outcomes to the total sample space:

$$P(E) = \frac{n(E)}{n(S)} = \frac{6}{16}$$

Simplifying this fraction by dividing both numerator and denominator by $2$, we arrive at our destination:

$$\mathbf{P(E) = \frac{3}{8}}$$

It is a beautiful result, isn't it? What seemed like a daunting task of checking sixteen matrices was reduced to a logical dance of cases. Remember, in JEE Advanced, the math is rarely about brute force; it is about finding the structure within the chaos.

Animated Solution for Mathematics - Probability: Let $A = [a_{ij}]$ be a square matrix of order 2 with entries either 0 or 1. Let E be the event that A is an invertible matrix. Then the probability $P (E)$ is:

Select Answer:

Visualized Solution

Matrix Structure

Total Possible Matrices

Condition for Invertible Matrix

Possible Values of Determinant

Case 1: $∣ A ∣ = 1$

Counting Matrices for Case 1

Case 2: $∣ A ∣ = - 1$

Counting Matrices for Case 2

Total Invertible Matrices

Calculating Probability

The Sigma Insight: Classical Definition of Probability

The Matrix Universe

Phase 1

Phase 2

Phase 3

The Synthesis

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