Animated Solution for Mathematics - Probability: Let $S=\{M=[a_{ij}],a_{ij}\in \{0,1,2\},\{1\le i,j\le 2\}\}$ be a sample space and $A=\{M\in S:M\text{ is invertible}\}$ be an event. Then $P(A)$ is equal to

A Journey into Matrix Invertibility

Welcome, future engineer. Today, we are going to peel back the layers of a problem that, at first glance, looks like a tedious counting exercise.

We are dealing with a $2\times 2$ matrix $M=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$, where each entry $a,b,c,d$ is chosen from the set $\{0,1,2\}$. We want to find the probability that this matrix is invertible.

First, we must understand the size of our universe. We have four slots in our matrix, and each slot has three choices.

By the fundamental principle of counting, the total number of matrices in our sample space $S$ is:

Now, we could try to count every single invertible matrix, but that would be a chaotic endeavor. Instead, we embrace the elegance of the complement.

We know that a matrix is invertible if and only if its determinant is non-zero. Conversely, it is singular (non-invertible) if $\det (M)=0$. It is far easier to count the singular matrices and subtract that probability from $1$.

For our matrix $M$, the determinant is defined as $\det (M)=ad-bc$. For the matrix to be singular, we require $\det (M)=0$, which implies $ad=bc$.

This is the heart of the problem. We need to find how many ways we can choose $a,b,c,d$ such that the product of the main diagonal equals the product of the anti-diagonal.

Let $k$ be the value of these products. Since our entries are in $\{0,1,2\}$, the possible values for $k$ are $0,1,2,$ and $4$. Let's analyze each case systematically.

The Zero Product ($ad=bc=0$)

This is the most frequent case. For $ad=0$, at least one of $a$ or $d$ must be zero.

The possible pairs $(a,d)$ are $(0,0),(0,1),(0,2),(1,0),(2,0)$. That is $5$ pairs.

Similarly, for $bc=0$, there are $5$ pairs for $(b,c)$. Since these are independent choices, the total number of matrices where $ad=bc=0$ is:

The Unit Product ($ad=bc=1$)

This is much more restrictive. The only way to get a product of $1$ using our set is $1\times 1$.

Thus, the only pair for $(a,d)$ is $(1,1)$, and the only pair for $(b,c)$ is $(1,1)$. This gives us:

The Product of Two ($ad=bc=2$)

To get a product of $2$, we need the factors $1$ and $2$. The pairs can be $(1,2)$ or $(2,1)$.

That gives us $2$ pairs for $(a,d)$ and $2$ pairs for $(b,c)$. Multiplying these, we get:

The Product of Four ($ad=bc=4$)

Finally, the only way to get $4$ is $2\times 2$. There is only $1$ pair $(2,2)$ for $(a,d)$ and $1$ pair $(2,2)$ for $(b,c)$.

This gives us:

Now, we sum these up to find the total number of singular matrices:

The probability of picking a singular matrix is:

Finally, the probability of our event $A$ (the matrix being invertible) is:

Look at that! By breaking the problem into small, logical steps, we turned a potentially overwhelming task into a series of simple multiplications. This is the essence of JEE Advanced mathematics—not just calculation, but the strategy of simplification.