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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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square matrix whose transpose is its negation; that is, it satisfies the condition −A = AT. If the entry in the i th row and j th column is aij, i.e. A = (aij) then the skew symmetric condition is aij = −aji

hence answer is A

0
why i ≤ n is given?? Shouldn't it be 1 ≤  i ≤  n ??
7
why we are not taking i=0 as i<=n ??
0

@ankitrazzagmail.com as array size is [n,n] by taking i=0 and i<=n it will go beyond array size...

3
how will be v[0,0] will be defined , only i will be 0 what value j will take as j domain does not have 0.
0
misprint in GO pdf
0
In question, it is given as 1 <= j <= n, means range of j is from 1 to n.

And i <= n, means what is the range of i here?  I mean i value can be ‘0’ also, i.e. range of i can be 0 to n-1 also.

Any suggestion please
0

And i <= n, means what is the range of i here?  I mean i value can be ‘0’ also, i.e. range of i can be 0 to n-1 also.

1<= j <= n( which represents column number) as given in question.

Now look at the first element of a 2-D array, it should be either at (0,0) or (1,1) according to the convention chosen, but here j should be greater than 1, so the possibility of using (0,0) is ruled out.

We have to use the convention in which first element of the 2-D array is at (1,1). Using (0,1) makes no sense here.

0

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7 Answers

47 votes
 
Best answer
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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4 Comments

this matrix is also a skew symmetric matrix, so definitely it's sum will be 0.
19
correct
0
hello here i value is greater than 1  then how can u take -1 or 0 plz let me know
1
But this is not defined for j = 0. Or what is its signifacance?
1
22 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
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

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

   

6 votes
sum of all elements =n*( $\sum_{i=1}^{n}i-\sum_{j=1}^{n}j )= 0$
4 votes
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
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@prakash What if i take, i=0,1,2.   and j=1,2,3...? then it will be skew symmetric matrix..?
0
0 votes
Answer is Option----A

Assume n=2;

then the condition is i<=2 and 1<=j<=2

now the values are i={1,2} and j={1,2}

Given, v[i] [j]= i-j;

now the values will be v ={1-1,1-2,2-1,2-2};

v={0,-1,1,0};

sum of all values present in v is 0..
Answer:

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