Let A = [a_ij]_3×3 be a matrix, defined by a_ij=≤ft […] .. The number…

The correct answer is Option (3) → 4

Matrix is defined as:

$a_{ij}=\begin{cases} 2i+3j , & ij \end{cases}$

For $3\times 3$ matrix:

$A=\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$

Now, compute each element:

$a_{11}=6$

$a_{12}=2(1)+3(2)=2+6=8$

$a_{13}=2(1)+3(3)=2+9=11$

$a_{21}=3(2)-2(1)=6-2=4$

$a_{22}=6$

$a_{23}=2(2)+3(3)=4+9=13$

$a_{31}=3(3)-2(1)=9-2=7$

$a_{32}=3(3)-2(2)=9-4=5$

$a_{33}=6$

So,

$A=\begin{bmatrix} 6 & 8 & 11 \\ 4 & 6 & 13 \\ 7 & 5 & 6 \end{bmatrix}$

Elements greater than 6 are: $8, 11, 13, 7$

Number of such elements = $4$