In this type of question, a matrix A=[aij] of a given order m×n is constructed by substituting values of i (row index) and j (column index) into a given formula.
Key Convention:
- i = row number, where 1≤i≤m
- j = column number, where 1≤j≤n
- The element aij occupies row i and column j.
Construct a matrix of order 2×3 if aij is defined by a given rule, for example:
aij=i+2j
Step-by-step solution:
For a 2×3 matrix, i∈{1,2} and j∈{1,2,3}.
Compute each element:
| j=1 | j=2 | j=3 |
|---|
| i=1 | a11=1+2(1)=3 | a12=1+2(2)=5 | a13=1+2(3)=7 |
| i=2 | a21=2+2(1)=4 | a22=2+2(2)=6 | a23=2+2(3)=8 |
So the matrix is:
A=[345678]
| Formula | Meaning |
|---|
| aij=i+j | Sum of row and column indices |
| aij=i⋅j | Product of indices |
| aij=i2−j | Row index squared minus column index |
| aij=2i−j | Difference divided by 2 |
| aij={10i=ji=j | Identity matrix pattern |
- Always identify the order of the matrix first to know the range of i and j.
- Substitute carefully — do not swap i and j.
- The resulting matrix may have real or complex entries depending on the formula.