Question

Consider a $n\times n$ lower triangular matrix, which is stored in row major order where each element occupies s bytes of memory space, then the address of an element A[i][j] can be given as

• A. $A\left[i\right]\left[j\right]=(i-j)+\left[\right(i-j)n-\frac{(j-1)(j-2)}{2}]\times S$
• B. $A\left[i\right]\left[j\right]=(j-1)+\frac{i(i-1)}{2}\times n$
• C. $A\left[i\right]\left[j\right]=b+\left[\right(i-1)+\frac{i(j-1)}{2}]\times S$
• D. $A\left[i\right]\left[j\right]=b+\left[\right(i-j)n-\frac{j(j-1)}{2}]\times S$

Answer 1

The correct option is C $A\left[i\right]\left[j\right]=b+\left[\right(i-1)+\frac{i(j-1)}{2}]\times S$

The matrix where the entries below the main diagonal are zero is called the upper triangular matrix.
Where as if all the entries above the main diagonal in the matrix are zero it is called as a lower triangular matrix.
The address of an element A[i][j]. can be given as :
$A\left[i\right]\left[j\right]=b+\left[\right(i-1)+\frac{i(j-1)}{2}]\times S$