A square matrix A = [aᵢⱼ]ₙₓₙ is called a lower triangular matrix if aᵢⱼ = 0 for

The correct option is i < j.

To understand why this is correct, let us review the definition of a lower triangular matrix.
A square matrix $A=\left[a_{ij}\right]$ of order $n\times n$ is called a lower triangular matrix if all the entries above the main diagonal are equal to zero.

Let us represent the elements of a general $n\times n$ matrix:
$$A=\left[\begin{array}{cccc}{a}_{11}& a_{12}& \dots & a_{1n}\\ a_{21}& a_{22}& \dots & a_{2n}\\ ⋮& ⋮& \&subst;& ⋮\\ a_{n1}& a_{n2}& \dots & a_{nn}\end{array}\right]$$

The main diagonal of this matrix consists of the elements $a_{ij}$ where the row index equals the column index, i.e., $i=j$ (elements like $a_{11}$, $a_{22}$, ..., $a_{nn}$).

1. For any element lying below the main diagonal, the row index is greater than the column index, which is represented mathematically as $i>j$.
2. For any element lying above the main diagonal, the row index is less than the column index, which is represented mathematically as $i<j$.

Since a lower triangular matrix must have all entries above the main diagonal equal to zero, we require:
$a_{ij}=0$ for all $i<j$.