Question

In a magic square each row, column and diagonal have the same sum. check which of the following is a magic square​

Answer 1

Answer:

is there supposed to be a picture of the magic squares or what...???

Answer 2

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(i) Row one R1 = 5 + (-1) + (—4)

=5 – 1 – 4 = 5 – 5 = 0

Row two R2 = (-5) + (-2) + 7

= -5 – 2 + 7 = -7 + 7 = 0

Row three R3 = 0 + 3 + (-3)

= 0 + 3- 3 = 0

Column one C1t = 5 + (-5) + 0

= 5 – 5 + 0 = 0

Column two C2 = (-1) + (-2) + (3)

=-1 – 2 + 3 = -3 + 3 = 0

Column three C3 = (-4) + 7 + (-3)

= -4 + 7 – 3 = 7 – 7 = 0

Diagonal d12 = 5 + (-2) + (-3)

= 5 – 2- 3 = 5 – 5 = 0

Diagonal d2 = (-4) + (-2) + 0

= -4 – 2 + 0 = -6 + 0 = -6

Here, the sum of the integers of diagonal d2 is different from the others.

Hence, it is not a magic square.

(ii) Row one R1 = 1 + (-10) + 0

= 1 – 10 + 0 = -9

Row two R2 = (-4) + (-3) + (-2)

= -4 – 3 – 2 = -9

Row three R3 = (-6) + (4) + (-7)

= -6 + 4 – 7 = -9

Column one C3 = 1 + (-4) + (-6)

= 1 – 4 – 6 = -9

Column two C2 = (-10) + (-3) + 4

= -10 – 3 + 4 = -9

Column three C3 = 0 + (-2) + (-7)

= 0 – 2 -7 = -9

Diagonal d1 = 1 + (-3) + (-7)

= 1 – 3 – 7 = 1 – 10 = -9

Diagonal d2 = 0 + (-3) + (-6)

= 0 – 3- 6 = -9

Here, sum of the integers column wise, row wise and diagonally is same i.e. -9.

Here, sum of the integers column wise, row wise and diagonally is same i.e. -9.Hence, (ii) is a magic square.

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