Animated Solution for Mathematics - Matrices and Determinants: Let $P$ be a matrix of order $3\times 3$ such that all the entries in $P$ are from the set $\{-1,0,1\}$. Then, the maximum possible value of the determinant of $P$ is ________.

Welcome, fellow explorers of the mathematical universe. Today, we are going to unravel a classic puzzle that often appears in the halls of JEE Advanced.

We are dealing with a $3\times 3$ matrix $P$, where every entry is restricted to the set $\{-1,0,1\}$. Our mission is to find the maximum possible value of $\det (P)$.

At first glance, this looks like a problem of endless combinations, but as we will see, it is a problem of elegant constraints.

The Sarrus Rule

To begin, let us visualize the determinant. We use the Sarrus Rule, a beautiful geometric way to expand a $3\times 3$ matrix.

We write the first two columns next to the third and trace the diagonals. We have three main diagonals (let us call their products $P_1,P_2,P_3$) and three anti-diagonals (let us call their products $N_1,N_2,N_3$).

The determinant is simply the sum of the positive terms minus the sum of the negative terms:

Why 6 is a Trap

Since each entry is $\{-1,0,1\}$, the product of any three entries is also $\{-1,0,1\}$. If we want to maximize the determinant, we might dream of a scenario where $P_1=P_2=P_3=1$ and $N_1=N_2=N_3=-1$.

This would give us $1+1+1-(-1-1-1)=6$. It sounds perfect, but in mathematics, if something sounds too good to be true, it usually is.

Here is where the genius of the problem lies. Let us multiply all six terms of our expansion together:

If you trace the diagonals carefully, you will realize that every single element $a_{ij}$ of the matrix appears exactly twice in this grand product. This means the product is actually the square of the product of all nine elements:

Because this is a square of a real number, the product MUST be greater than or equal to zero. Now, let us test our theoretical maximum of 6.

If $\det (P)=6$, then $P_i=1$ and $N_i=-1$. The product of these six terms would be $(1{)}^3\times (-1{)}^3=-1$.

But wait! We just proved the product must be $\ge 0$. Since $-1$ is not $\ge 0$, a determinant of 6 is mathematically impossible.

Why 5 Fails

What about 5? If we do not use any zeros, all six terms are $\pm 1$.

The sum of six odd numbers is always an even number. Since 5 is odd, it is impossible to achieve.

If we use at least one zero, the number of non-zero terms decreases, and the maximum sum we can achieve drops to 4. Thus, 5 is also out of the question.

Constructing the Maximum

Since 6 and 5 are impossible, we turn our eyes to 4. Can we achieve it? Yes!

Consider the matrix:

Expanding this along the first row, we get:

We have found our maximum! The maximum possible value of $\det (P)$ is 4.

The key takeaway here is that logical bounding and parity arguments are far more powerful than brute force. Keep this in your toolkit, and you will conquer any matrix problem that comes your way.