Q.Construct matrix $B = [b_{ij}]$ of order $2\times 2$ where $b_{ij} = \dfrac{|i-j|}{3}$.

Matrix Representation Order — The Intuition First

Imagine you're a teacher taking attendance. You have a list of 5 students and you call out their names one by one. The order matters — "Ravi, Priya, Anjali, Vikram, Sneha" is a specific sequence. If you wrote that sequence down, you'd have an ordered list.

Now imagine you're arranging those same 5 students into rows and columns for a class photo. You decide: 2 rows, 3 columns (one spot empty). The moment you fix how many rows and how many columns, you've chosen a matrix shape. But here's the key: the order in which you fill those slots — row by row, or column by column — changes who ends up where.

That's the core idea of matrix representation order: given a list of numbers (or objects), the same list can be arranged into a matrix in different ways depending on whether you fill it row-wise or column-wise.

The Precise Statement

A matrix of size $m \times n$ has $m$ rows and $n$ columns, so it holds $m \times n$ entries. If you have a sequence of $m \times n$ elements, there are two standard ways to map them into the matrix:

Let the sequence be $a_1, a_2, a_3, \dots, a_{mn}$. In row-major order, the element at position $(i, j)$ — meaning row $i$, column $j$ — is:

In column-major order, the element at $(i, j)$ is:

These formulas are not something to memorise blindly. They come from counting: in row-major, you skip $(i-1)$ full rows (each of $n$ elements) and then take the $j$-th element of that row. In column-major, you skip $(j-1)$ full columns (each of $m$ elements) and then take the $i$-th element of that column.

A Concrete Example

Take the sequence: $1, 2, 3, 4, 5, 6$. Put it into a $2 \times 3$ matrix.

Row-major (fill rows first):

Column-major (fill columns first):

Same six numbers, completely different matrices. The order of filling changes the arrangement.

Why This Matters …