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55 votes
55 votes

An $n \times n$ array $v$ is defined as follows:

$v\left[i,j\right] = i - j$ for all $i, j, i \leq n, 1 \leq j \leq n$

The sum of the elements of the array $v$ is

  1. $0$
  2. $n-1$
  3. $n^2 - 3n +2$
  4. $n^2 \frac{\left(n+1\right)}{2}$
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8 Answers

Best answer
52 votes
52 votes
The sum of the $i^{th}$ row and $i^{th}$ column is $0$ as shown below. Since, the numbers of rows equals the number of columns, the total sum will be $0$.$$\begin{array}{|l|l|l|l|l|} \hline \text{0} & \text{-1} & \text{-2} & \text{-3} & \text{-4}\\\hline \text{1} & \text{0} & \text{-1} & \text{-2} & \text{-3}  \\\hline \text{2} & \text{1} & \text{0} & \text{-1} & \text{-2} \\\hline \text{3} & \text{2} & \text{1} & \text{0} & \text{-1}\\\hline \text{4} & \text{3} & \text{2} & \text{1} & \text{0}\\\hline \end{array}$$Correct Answer: $A$
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23 votes
23 votes
If you look at code carefully it is very clear matrix getting defined is skew symmetric matrix.

Sum of all elements in skew symmetric matrix is 0.

Answer :-A 0
18 votes
18 votes
Let there are total N rows . You will find ∑ of elements of row i + ∑ of elements row (N-i+1) = 0.

So if N is even then

row 1 + row N =0

row 2 + row (N-1) =0

row 3 + row (N-2)=0

similarly row (N/2) + row (N/2+1) =0.   (So total sum is 0)

But if N is odd then row ((N+1)/2) will have no corresponding rows BUT Ithe summation of elements of this row is 0 .

So for N = even or Odd , the sum of element is 0 .
6 votes
6 votes

i think we can easily get it without drawing matrix
   as expression given v[i, j] = i - j
suppose i1-j1=k1   {for a particular index } so its opposit index shows j1-i1=-k1, for example if v[1,2]=x then v[2,1]=-x}
  so we have  cases here
1. for all  i>j v[i, j] = i - j and jut for opposit indexof v[i, j] = i - j , v[j, i] = j - i =-(i-j) :: note this is for non diagonal elements 
2. for all i=j v[i, j] = i - j=0  :: note this is for diagonal elaments
so total sum will be zero ucan easily get it by seeing above cases

   

Answer:

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