3×3 magic square (square matrix) of each row, column and diagonal in part (a) add up to 15. part add up to 165 which is 11 times part (a) and part (c)add up to 0 which is 0 times part (a). we deduce this by using trial and error approach.
Part (a) solving it through trial & error method we get the 3×3 magic square (square matrix)
4 9 2
3 5 7
8 1 6
In part (a) the total of each row, column, and diagonal is 15
again taking the same approach of trial and error we get
(b)
54 59.2 52
53 55 57
58 51 56
In part (b) the total of each row, column, and diagonal is 165 which is 11 times more than the total of each row, column, and diagonal in part (a). it is because at the center of part (a) magic square ( square matrix), the number is 5 and at the center of part (b) magic square, the number is 55 which is 11 times that of part (a) magic square( square matrix). we know that the number at the center of a magic square( square matrix) is responsible to make it a magic square ( square matrix) as other numbers are placed according to the number at the center to provide balance.
(c)
-1 4.2 -3
-2 0 2
3 -4 1
In part (c) the total of each row, column, and diagonal is 0. because the digit at the center is zero. In part (c) the total of each row, column, and diagonal is 0 which is 0 times more than the total of each row, column, and diagonal in part (a). it is because at the center of part (a) magic square( square matrix) , the number is 5 and at the center of part (b) magic square( square matrix) , the number is 0 which is 0 times that of the part (a) magic square ( square matrix) when compared. we know that the number at the center of a magic square ( square matrix) is responsible to make it a magic square (square matrix) as other numbers are placed according to the number at the center to provide balance.
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