Question

create two magic squares of size3x3 using rational numbers and show that the sum of row, column and diagonals are same​

Answer 1

Answer:

Sum of the rows

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

5-1-4\ =\ 05−1−4 = 0

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

-5-2+7\ \ =0−5−2+7 =0

Row 3 = 0+3+\left(-3\right)0+3+(−3)

0+3-3\ \ =0\0+3−3 =0

Sum of the columns

Column 1 = 5+\left(-5\right)+05+(−5)+0

5-5+0\ =\ 05−5+0 = 0

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

-1-2+3\ \ =\ 0−1−2+3 = 0

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

-4\ +7\ -3\ \ =\ 0−4 +7 −3 = 0

Sum of the diagonals

Diagonal 1

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

5-2-3\ \ =\ 05−2−3 = 0

Diagonal 2

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

-4-2+0\ =\ -6−4−2+0 = −6

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