Animated Solution for Mathematics - Matrices and Determinants: The number of square matrices of order 5 with entries from the set $\{0,1\}$, such that the sum of all the elements in each row is 1 and the sum of all the elements in each column is also 1, is

Welcome, future engineer. Today, we aren't just solving a matrix problem; we are choreographing a dance. Imagine a blank $5\times 5$ grid.

Our goal is to fill this grid with only two characters: $0$ and $1$. We are bound by two strict, elegant rules: the sum of every row must be exactly $1$, and the sum of every column must be exactly $1$.

Let's look at the first rule: for every row $i$. Since our entries are restricted to $0$ and $1$, this means every single row must contain exactly one $1$ and four $0$s.

The second rule is its mirror image: for every column $j$. This means every column must also contain exactly one $1$ and four $0$s.

Imagine you are placing five rooks on a $5\times 5$ chessboard. A rook attacks everything in its row and column. If we want to place five rooks such that no two rooks attack each other, we must place exactly one rook in each row and exactly one in each column.

Our matrix is simply a mathematical representation of this non-attacking rook configuration!

Let's build this matrix row by row. We start with the first row. We have $5$ columns available, so we have $5$ possible positions to place our first $1$.

Let's say we choose column $k$. Now, the first row is satisfied. But wait—because of our column constraint, column $k$ is now 'consumed'. No other row can place a $1$ in column $k$.

Now, we move to the second row. We need to place another $1$. But one column is already blocked by our choice in the first row. That leaves us with $5-1=4$ available columns. We have $4$ choices for the second row.

As we move to the third row, two columns are now blocked—one by the first row and one by the second. This leaves us with $5-2=3$ choices.

The pattern is becoming clear, isn't it? For the fourth row, we have $2$ choices, and for the final fifth row, only $1$ column remains empty. We have no choice but to place the final $1$ there.

To find the total number of such matrices, we use the Fundamental Principle of Counting. We multiply the number of choices for each row:

This is the definition of $5!$ (five factorial). Calculating this, we get $120$.

These special matrices are known as Permutation Matrices. They are the building blocks of linear algebra, representing permutations of the standard basis.

For any $n\times n$ matrix of this type, the number of such matrices is simply $n!$. It is a beautiful, simple result born from a rigid set of constraints. Keep this logic in your toolkit—whenever you see constraints that restrict placement, think of the 'non-attacking' analogy. You've got this!