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Question: In an upper triangular matrix n×n, minimum number of zeros is...

In an upper triangular matrix n×n, minimum number of zeros is

A

n(n−1)2\frac{n(n - 1)}{2}

B

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

C

2n(n−1)2\frac{2n(n - 1)}{2}

D

None

Answer

n(n−1)2\frac{n(n - 1)}{2}

Explanation

Solution

As we know a square matrix A=[aij]A = \lbrack a_{ij}\rbrack is called an upper triangular matrix if aij=0a_{ij} = 0for all i>j

$A = \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} & .... & a_{1(n - 2)} & a_{1(n - 1)} & a_{1n} \ 0 & a_{22} & a_{23} & a_{24} & .... & a_{2(n - 2)} & a_{2(n - 1)} & a_{2n} \ 0 & 0 & a_{33} & a_{34} & .... & a_{3(n - 2)} & a_{3(n - 1)} & a_{3n} \ 0 & 0 & 0 & a_{44} & .... & a_{4(n - 2)} & a_{4(n - 1)} & a_{4n} \

  • & - & - & - & - & - & - & - \
  • & - & - & - & - & - & - & - \ 0 & 0 & 0 & 0 & .... & 0 & a_{(n - 1)(n - 1)} & a_{(n - 1)n} \ 0 & 0 & 0 & 0 & .... & 0 & 0 & a_{nn} \end{bmatrix}Numberofzeros= Number of zeros =(n - 1) + (n - 2) + ..... + 2 + 1 = \frac{(n - 1)n}{2}$