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?

$\frac{4}{11}$

$\frac{9}{11}$

$\frac{2}{11}$
(Along the first row $\frac{4}{11} + \frac{9}{11} + \frac{2}{11} + \frac{5}{11}$ )

$\frac{3}{11}$

$\frac{5}{11}$

$\frac{7}{11}$

$\frac{8}{11}$

$\frac{1}{11}$

$\frac{6}{11}$

Solution

Along the first row, sum = $\frac{4}{11} + \frac{9}{11} + \frac{2}{11} = \frac{15}{11}$

Along the second row, sum = $\frac{3}{11} + \frac{5}{11} + \frac{7}{11} = \frac{15}{11}$

Along the third row, sum = $\frac{8}{11} + \frac{1}{11} + \frac{6}{11} = \frac{15}{11}$

Along the first column, sum = $\frac{4}{11} + \frac{3}{11} + \frac{8}{11} = \frac{15}{11}$

Along the second column, sum = $\frac{9}{11} + \frac{5}{11} + \frac{1}{11} = \frac{15}{11}$

Along the third column, sum = $\frac{2}{11} + \frac{7}{11} + \frac{6}{11} = \frac{15}{11}$

Along the first diagonal, sum = $\frac{4}{11} + \frac{5}{11} + \frac{6}{11} = \frac{15}{11}$

Along the second diagonal, sum = $\frac{2}{11} + \frac{5}{11} + \frac{8}{11} = \frac{15}{11}$

Since the sum of the numbers in each row, in each column, and along the diagonals is the same, it is a magic square.

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$\frac{4}{11}$	$\frac{9}{11}$	$\frac{2}{11}$	(Along the first row $\frac{4}{11} + \frac{9}{11} + \frac{2}{11} + \frac{5}{11}$ )
$\frac{3}{11}$	$\frac{5}{11}$	$\frac{7}{11}$
$\frac{8}{11}$	$\frac{1}{11}$	$\frac{6}{11}$