Magic Matrix

Algebra Level 5

[ a b c d e f g h i ] \begin{bmatrix}{a} && {b} && {c} \\ {d} && {e} && {f} \\ {g} && {h} && {i}\end{bmatrix}

You are given a matrix S S as shown above whose elements are denoted as 9 letters from a a to i i , and each letter is a distinct positive integer between 1 to 9. If this matrix is not invertible and it satisfies the constraints a + d = b + e = c + f = g + h = i a+d=b+e=c+f=g+h=i and a b = d c = 1 a-b=d-c=1 , what is the value of 9-digit integer a b c d e f g h i \overline{ abcdefghi} ?


The answer is 216783549.

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1 solution

From the first constraint, we can deduce that i is 9 as it is the only digit that can be the sum of the other four additions.

Then we can conclude that d = 9-a; e = 9-b; f = 9-c; g = 9-h. So now we have 4 variables in terms of a, b, c, h.

From the second constraint, b = a-1; d = c+1 = 9-a; c = 8-a; f = 9-c = a+1. So now the variables b, c, d are rewritten in terms of a. We have put the first two rows of matrix in terms of a.

With the a and h variables, when we calculate the determinant, we will get the linear equation:

h = 18(4-a)/(17-4a).

When substituting 1 to 8 integer for the h-value, only h = 4, a = 2 and h = 6, a = 5 work. However, if a=5, then b=4 and d=4, which doesn't correlate with the problem (must be distinct digits).

Thus, a=2, then b=1, c=6, d=7, e=8, f=3, g=5, h=4, and i=9. There shouldn't be other answer.

Can 325674189 be a solution?

Jasraj Singh - 5 years, 7 months ago

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Determinant of your matrix equals 198, so it's invertible and so not the solution. The singular matrix in the question has det S=0. :)

Worranat Pakornrat - 5 years, 7 months ago

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