For a $3×3 $ matrix $A= [a_{ij}],$ whose element $a_{ij}$ is given by $a_{ij}=\sqrt{i^{2}+3j}$ then what is element $a_{31}$ ?
Answer & explanation
Correct answer: option 3
The correct answer is Option 3: $2\sqrt{3}$
Given
$a_{ij}=\sqrt{i^{2}+3j}$
To find $a_{31}$
Here $i=3,\; j=1$
$a_{31}=\sqrt{3^{2}+3(1)}$
$=\sqrt{9+3}$
$=\sqrt{12}$
$=2\sqrt{3}$
The element $a_{31}$ is $2\sqrt{3}$.