A) \[{{a}_{ij}}\ne \]0, for \[i>j\]
B) \[{{a}_{ij}}\ne \], for \[i>j\]
C) \[{{a}_{ij}}=\]0, for \[i<j\]
D) None of these
Correct Answer: B
Solution :
\[{{[{{a}_{ij}}]}_{n\times n}}\]square matrix is a upper triangular matrix for \[{{a}_{ij}}=0,i>j\].
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