Question

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?

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

Answer 1

Step 1: Find the sum along the respective rows.

The sum of first row$=\left(\frac{4}{11}\right)+\left(\frac{9}{11}\right)+\left(\frac{2}{11}\right)$

$$\begin{gathered} =\frac{4+9+2}{11} \\ =\frac{15}{11} \end{gathered}$$

The sum of second row$=\left(\frac{3}{11}\right)+\left(\frac{5}{11}\right)+\left(\frac{7}{11}\right)$

$$\begin{gathered} =\frac{3+5+7}{11} \\ =\frac{15}{11} \end{gathered}$$

The sum of third row$=\left(\frac{8}{11}\right)+\left(\frac{1}{11}\right)+\left(\frac{6}{11}\right)$

$$\begin{gathered} =\frac{8+1+6}{11} \\ =\frac{15}{11} \end{gathered}$$

Step 2: Find the sum along the respective columns.

The sum of first column$=\left(\frac{4}{11}\right)+\left(\frac{3}{11}\right)+\left(\frac{8}{11}\right)$

$$\begin{gathered} =\frac{4+3+8}{11} \\ =\frac{15}{11} \end{gathered}$$

The sum of second column$=\left(\frac{9}{11}\right)+\left(\frac{5}{11}\right)+\left(\frac{1}{11}\right)$

$$\begin{gathered} =\frac{9+5+1}{11} \\ =\frac{15}{11} \end{gathered}$$

The sum of third column$=\left(\frac{2}{11}\right)+\left(\frac{7}{11}\right)+\left(\frac{6}{11}\right)$

$$\begin{gathered} =\frac{2+7+6}{11} \\ =\frac{15}{11} \end{gathered}$$

Step 3: Find the sum along the respective diagonals.

The sum of first diagonal$=\left(\frac{4}{11}\right)+\left(\frac{5}{11}\right)+\left(\frac{6}{11}\right)$

$$\begin{gathered} =\frac{4+5+6}{11} \\ =\frac{15}{11} \end{gathered}$$

The sum of second diagonal$=\left(\frac{2}{11}\right)+\left(\frac{5}{11}\right)+\left(\frac{8}{11}\right)$

$$\begin{gathered} =\frac{2+5+8}{11} \\ =\frac{15}{11} \end{gathered}$$

As the sum of the numbers in each row, in each column and along the diagonals is same $\left(\mathrm{i}.\mathrm{e}.,\frac{15}{11}\right)$.

Hence, the given pattern is a magic square.