A square matrix $A$ will be lower triangular if and only if ($a_{mn}$ represents an element of $m^{\text{th}}$ row and $n^{\text{th}}$ column of the matrix)
Options:
(A) $a_{mn} = 0 \text{ for } n > m$
(B) $a_{mn} = 0 \text{ for } m > n$
(C) $a_{mn} \neq 0 \text{ for } m > n$
(D) $a_{mn} \neq 0 \text{ for } n > m$
Official Answer Key:
A
Step-by-Step Solution & Explanation
In a lower-triangular matrix, all entries above the main diagonal are zero, and only entries on or below the diagonal may be non-zero. An entry a_MN lies above the diagonal precisely when its column index exceeds its row index, i.e. N > M. Hence the defining condition is a_MN = 0 for N > M, which is option (A). Option (D) wrongly requires those above-diagonal entries to be non-zero (that would describe an upper triangular matrix's complement), and (B)/(C) swap the row–column inequality.
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