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

Question

In a upper triangular matrix n × n, minimum number of zeros is (A) n(n-1) / 2 (B) n(n+1) / 2 (C) 2 n(n-1) / 2 (D) None of these

Tardigrade Admin · Accepted Answer

As we know, a square matrix A =[ai j] is called an upper triangular matrix if ai j=0 for all i>j Such as, A=[1&2&3&4 0&5&1&3 0&0&2&9 0&0&0&5]4 × 4 Number of zeros =(4(4-1)/2)=6=(n(n-1)/2).

Q. In a upper triangular matrix $n \times n$ , minimum number of zeros is

$n (n - 1) /2$

$n (n + 1) /2$

$2 n (n - 1) /2$

None of these

As we know, a square matrix $A = [a_{ij}]$ is called an upper triangular matrix if $a_{ij} = 0$ for all $i > j$
Such as, $A = ⎣ ⎡ 1000250031204395 ⎦ ⎤_{4 \times 4}$
Number of zeros $= \frac{4 ( 4 - 1 )}{2} = 6 = \frac{n ( n - 1 )}{2}$ .

Q. In a upper triangular matrix n×n, minimum number of zeros is

n(n−1)/2

n(n+1)/2

2n(n−1)/2