Question

Question 3
In a “magic square”. The sum of the numbers in each row, in each column and along the diagonals is the same. Is this a magic square?
$\begin{array}{ccc}\frac{4}{11}& \frac{9}{11}& \frac{2}{11}\\ \frac{3}{11}& \frac{5}{11}& \frac{7}{11}\\ \frac{8}{11}& \frac{1}{11}& \frac{6}{11}\end{array}$
$\left(\text{Along the first row}\frac{\text{4}}{\text{11}}+\frac{9}{11}+\frac{2}{11}=\frac{15}{11}\right)$

Answer 1

$\text{Sum of first row}=\frac{\text{4}}{\text{11}}+\frac{9}{11}+\frac{2}{11}=\frac{4+9+2}{11}=\frac{15}{11}[\text{Given]}\text{Sum of second row}=\frac{3}{11}+\frac{5}{11}+\frac{7}{11}=\frac{3+5+7}{11}=\frac{15}{11}\text{Sum of third row}=\frac{8}{11}+\frac{1}{11}+\frac{6}{11}=\frac{8+1+6}{11}=\frac{15}{11}\text{Sum of}\text{first column}=\frac{4}{11}+\frac{3}{11}+\frac{8}{11}=\frac{4+3+8}{11}=\frac{15}{11}\text{Sum of second column}=\frac{9}{11}+\frac{5}{11}+\frac{1}{11}=\frac{9+5+1}{11}=\frac{15}{11}$
$\text{Sum of third column}=\frac{\text{2}}{\text{11}}+\frac{7}{11}+\frac{6}{11}=\frac{2+7+6}{11}=\frac{15}{11}\text{Sum of first diagonal (left to right)}=\frac{4}{11}+\frac{5}{11}+\frac{6}{11}=\frac{4+5+6}{11}=\frac{15}{11}\text{Sum of second diagonal (left to right)}=\frac{2}{11}+\frac{5}{11}+\frac{8}{11}=\frac{2+5+8}{11}=\frac{15}{11}$
$\text{Since the sum of fractions in each row, in each column and along the diagonals are same,}\text{therefore, it's a magic square}$