Question

In a "magic square", the sum of the numbers in each row, in each column and along the diagonal is the same. Is this a magic square?

$\frac{4}{11}$	$\frac{9}{11}$	$\frac{2}{11}$
$\frac{3}{11}$	$\frac{5}{11}$	$\frac{7}{11}$
$\frac{8}{11}$	$\frac{1}{11}$	$\frac{6}{11}$

Answer 1

$$\begin{gathered} \mathrm{Sum}\mathrm{along}\mathrm{col}\mathrm{umns}\mathrm{and}\mathrm{rows}: \\ \frac{4}{11}+\frac{9}{11}+\frac{2}{11}=\frac{15}{11} \\ \frac{3}{11}+\frac{5}{11}+\frac{7}{11}=\frac{15}{11} \\ \frac{8}{11}+\frac{1}{11}+\frac{6}{11}=\frac{15}{11} \\ \frac{4}{11}+\frac{3}{11}+\frac{8}{11}=\frac{15}{11} \\ \frac{9}{11}+\frac{5}{11}+\frac{1}{11}=\frac{15}{11} \\ \frac{2}{11}+\frac{7}{11}+\frac{6}{11}=\frac{15}{11} \\ \text{S}\mathrm{um}\mathrm{along}\mathrm{diagonals}: \\ \frac{4}{11}+\frac{5}{11}+\frac{6}{11}=\frac{15}{11} \\ \frac{2}{11}+\frac{5}{11}+\frac{8}{11}=\frac{15}{11} \\ \mathrm{Since},\mathrm{all}\mathrm{the}\mathrm{sums}\text{in the square}\mathrm{are}\mathrm{equal}\mathrm{along}\mathrm{rows},\mathrm{columns}\mathrm{and}\mathrm{diagonals}, \\ \text{it}\mathrm{is}\mathrm{a}\mathrm{magic}\mathrm{square}. \end{gathered}$$