Question -
Answer -
(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.
Hence, (ii) is a magic square.