In an upper triangular matrix n x n, the minimum number of zeros are-
$A.\;\dfrac{{n\left( {n - 1} \right)}}{2}\;\;$
$B.\;\dfrac{{n\left( {n + 1} \right)}}{2}$
$C.\;\dfrac{{2n\left( {n - 1} \right)}}{2}$
$D.\;None\;of\;these$

Hint: An upper triangular matrix is the one in which aij = 0 for all i > j. That is, all the terms that are below the main diagonal are zero. For example,
\[\left[ {\begin{array}{*{20}{c}}
  1&2&3 \\
  0&1&2 \\
  0&0&1
\end{array}} \right]\]

Complete step by step answer:
We have to find the minimum number of zeros in the matrix. There are n terms in the main diagonal. The number of diagonal elements below the main diagonal are n - 1. All these n - 1 elements must be zero. Similarly, the next diagonal has n - 2 terms which are zero. This goes on until one element is left. So the total number of zeros in all these diagonals are-
(n - 1) + (n - 2) + …. + 1
1 + 2 + 3 + …. + (n - 1)
This is the sum of first n - 1 natural numbers. The sum of first n natural numbers is given by-
$\dfrac{{{\text{n}}\left( {{\text{n}} + 1} \right)}}{2}$
So the sum of n - 1 terms will be-
$\dfrac{{{\text{n}}\left( {{\text{n}} - 1} \right)}}{2}$
This is the minimum number of zeros in an upper triangular matrix. The correct option is A.
Note: We can also count the number of zeros in 2 x 2 and 3 x 3 matrices and then substitute the values and check which formula satisfies the number of zeros in each case.