GATE CSE 2000 | Question: 1.2

Question

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

• A. $0$
• B. $n-1$
• C. $n^2 - 3n +2$
• D. $n^2 \frac{\left(n+1\right)}{2}$

Comments

ditto chat gpt answer 👏

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take a small n X n size  array then run the values , it will be 0

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the range of i must be 1<= i <= n  
instead of i<=n

Answer 1

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: $A$

Comments

If you take the sum then 0 + (-1) + (-2) + 1 + 0 + (-1) + 2 + 1 + 0 = 0

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how V is an array? It is matrix right..

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@CSgirl

it is given that V is an n*n array so,you can say that V is a two dimensional array i.e v[n][n] 

Answer 2

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

Answer 3

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 .

Answer 4

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 5

sum of all elements =n*( $\sum_{i=1}^{n}i-\sum_{j=1}^{n}j )= 0$

Answer 6

ans is option (a) i.e 0

solution: suppose if n=3

then i<=3 and 1<=j<=3

i.e i=1,2,3, and j=1,2,3

=> v={i-j}=(0,-1,-2,1,0,-1,2,1,0}

sum of elements in v=0

Comments

@prakash What if i take, i=0,1,2.   and j=1,2,3...? then it will be skew symmetric matrix..?

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same here