The minimum number of zero's in an upper triangular matrix of order $n \times n$ is-

\frac{n (n - 1)}{2}

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\frac{n (n + 1)}{2}

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\frac{n (n - 1) (n + 1)}{2}

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none of these

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Solution

The correct option is A $\frac{n (n - 1)}{2}$
The minimum number of zero element in upper triangular matrix is,
$= 1 + 2 + 3 + . . . . . . . . . . . . . . . + (n - 1) = \frac{n (n - 1)}{2}$

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n

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n + 1

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Find $a$

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