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If aij=i+3ja_{ij}=i+3j, then the matrix of order 2 with elements as aija_{ij} is

Solution

✅ Correct Option: 3

Given aij=i+3ja_{ij} = i + 3j where ii is the row number and jj is the column number.

For a matrix of order 2 (a 2×2 matrix), we need to find four elements: a11a_{11}, a12a_{12}, a21a_{21}, and a22a_{22}.


For position (1,1):

a11=i+3ja_{11} = i + 3j

a11=1+3(1)a_{11} = 1 + 3(1)

a11=1+3a_{11} = 1 + 3

a11=4a_{11} = 4


For position (1,2):

a12=i+3ja_{12} = i + 3j

a12=1+3(2)a_{12} = 1 + 3(2)

a12=1+6a_{12} = 1 + 6

a12=7a_{12} = 7


For position (2,1):

a21=i+3ja_{21} = i + 3j

a21=2+3(1)a_{21} = 2 + 3(1)

a21=2+3a_{21} = 2 + 3

a21=5a_{21} = 5


For position (2,2):

a22=i+3ja_{22} = i + 3j

a22=2+3(2)a_{22} = 2 + 3(2)

a22=2+6a_{22} = 2 + 6

a22=8a_{22} = 8


Therefore, the matrix is:

[4758]\begin{bmatrix}4 & 7\\5 & 8\end{bmatrix}

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