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+26 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. $n^2 \frac{\left(n+1\right)}{2}$
asked in DS by Veteran (59.8k points)
edited by | 1.7k views

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

why i ≤ n is given?? Shouldn't it be 1 ≤  i ≤  n ??

6 Answers

+31 votes
Best answer

The sum of the $i^{th}$ row and $i^{th}$ column is $0$ as shown below. Since, the numbers of rows $=$ no. of columns, the total sum will be $0$.

$0$ $-1$ $-2$ $-3$ $-4$
$1$ $0$ $-1$ $-2$ $-3$
$2$ $1$ $0$ $-1$ $-2$
$3$ $2$ $1$ $0$ $-1$
$4$ $3$ $2$ $1$ $0$
answered by Veteran (377k points)
edited by
this matrix is also a skew symmetric matrix, so definitely it's sum will be 0.
hello here i value is greater than 1  then how can u take -1 or 0 plz let me know
But this is not defined for j = 0. Or what is its signifacance?
+17 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 .
answered by Active (1.5k points)
+17 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
answered by Boss (43.4k points)
+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


answered by Boss (12.5k points)
+5 votes
sum of all elements =n*( $\sum_{i=1}^{n}i-\sum_{j=1}^{n}j )= 0$
answered by Active (3.2k points)
+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
answered by (303 points)
edited by
@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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