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 square matrix whose transpose is its negation; that is, it satisfies the condition −A = A^{T}. If the entry in the i th row and j th column is a_{ij}, i.e. A = (a_{ij}) then the skew symmetric condition is a_{ij} = −a_{ji}
_{hence answer is A}
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 $0$.
\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}
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
What should be the answer of question 1 in C...