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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. $n^2 \frac{\left(n+1\right)}{2}$
edited | 1.8k views
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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

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

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 $0$.

 $0$ $-1$ $-2$ $-3$ $-4$ $1$ $0$ $-1$ $-2$ $-3$ $2$ $1$ $0$ $-1$ $-2$ $3$ $2$ $1$ $0$ $-1$ $4$ $3$ $2$ $1$ $0$

\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{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}

edited ago
+12
this matrix is also a skew symmetric matrix, so definitely it's sum will be 0.
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correct
+1
hello here i value is greater than 1  then how can u take -1 or 0 plz let me know
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But this is not defined for j = 0. Or what is its signifacance?
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.

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 .

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

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

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