Question

prepare a magic square so that the sum of number in each row in each column and each diagonals is (-6)​

Answer 1

Answer:

solution 7: (i)

Sum of the rows

Row 1 = 5+\left(-1\right)+\left(-4\right)

5-1-4\ =\ 0

Row 2 = \left(-5\right)+\left(-2\right)\ +7

-5-2+7\ \ =0

Row 3 = 0+3+\left(-3\right)

0+3-3\ \ =0\

Sum of the columns

Column 1 = 5+\left(-5\right)+0

5-5+0\ =\ 0

Column 2 = \left(-1\right)+\left(-2\right)+\ 3\

-1-2+3\ \ =\ 0

Column 3 = \left(-4\right)+7\ +\left(-3\right)

-4\ +7\ -3\ \ =\ 0

Sum of the diagonals

Diagonal 1

\left(5\right)+\left(-2\right)+\left(-3\right)

5-2-3\ \ =\ 0

Diagonal 2

\left(-4\right)+\left(-2\right)\ +0\

-4-2+0\ =\ -6

Box (i) is not a square because all the sums not equal.

(ii)

Sum of the rows

Row 1 = 1+\left(-10\right)\ +0

1-10+0\ =-9

Row 2 = \left(-4\right)+\left(-3\right)+\left(-2\right)

-4-3-2\ \ \ =\ -9

Row 3 = \left(-6\right)+4+\left(-7\right)

-6+4-7\ \ =\ -9

Sum of the columns

Column 1 = 1+\left(-4\right)+\left(-6\right)

1-10\ =-9

Column 2 = \left(-10\right)+\left(-3\right)+4

-10\ -3+4\ \ =\ -9

Column 3 = 0+\left(-2\right)+\left(-7\right)

0-2-7\ \ =\ -9

Sum of the diagonals

Diagonal 1

1+\left(-3\right)+\left(-7\right)

1-10\ =\ -9\

Diagonal 2

0+\left(-3\right)+\left(-6\right)

-3-6\ =\ -9

Box (ii) is a magic square because all the sums are equal.