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The minimum n...

Question

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

n^{n + 1} - n {(n - 1)}^{n + 1} + \frac{n (n - 1)}{2!} {(n - 2)}^{n + 1} - \dots = \frac{1}{2} n (n + 1)!

n^{n} - (n + 1) {(n - 1)}^{n} + \frac{(n + 1) n}{2!} {(n - 2)}^{n} - \dots = 1

n

1^{n} - n 2^{n} + \frac{n (n - 1)}{1 \cdot 2} 3^{n} - \dots = {(- 1)}^{n} n!

{(n + p)}^{n} - n {(n + p - 1)}^{n} + \frac{n (n - 1)}{2!} {(n + p - 2)}^{n} - \dots = n!

n + 1

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

$n^{n} - n (n - 1)^{n} + \frac{n (n - 1)}{2!} (n - 2)^{n} - . . . \infty = \frac{3}{a} \times n!$ .

Find $a$

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