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# Determinant(not really but still) Problem

What is the sum of the non-integral solutions of the equation

$$\left| \begin{array}{ccc} x & 3 & 4 \\ 5 & x & 5 \\ 4 & 2 & x \end{array} \right|=0\hspace{2mm}?$$

Note by Krishna Jha
3 years, 11 months ago

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Reduce to a $$2 \times 2$$ matrix: $$\left[ \begin{array}{cc} 10-4x & x^2-10 \\ 12-2x & 8-3x \end{array} \right]= 0$$

Then to a $$1 \times 1$$ matrix: $$(8-3x)(10-4x)-(12-2x)(x^2-10) = 0$$

Which then becomes $$x^3 - 41x + 100 = 0$$

Then after trial and error, x = 4 is one of the non-zero solutions · 3 years, 11 months ago

Please see the edit now..And please could you give a solution other than hit and trial??Like using synthetic division?? · 3 years, 11 months ago

This solution was created using the Carroll-Dodson Condensation Method. Synthetic division in this case has no meaningful purpose. However, I will see what I can do as far as posting a better solution. · 3 years, 11 months ago

Why not just calculate the $$3\times3$$ determinant directly? The C-D method requires the calculation of five $$2\times2$$ determinants, when only three are necessary: $0 \; = \; x(x^2-10) - 3(5x-20) + 4(10-4x) \; = \; x^3 - 41x + 100$ About the only short-cut to looking for the integer root of this cubic is to remember that any integer root $$n$$ has to be a factor of $$100$$. Testing $$\pm1,\pm2,\pm4,\pm5,\ldots$$ we find $$n=4$$ quite quickly. We could trim the testing a little by observing that $$|x^3-41x| \le 8 + 82 = 90 < 100$$ for $$|x| \le 2$$, and hence $$|n| \ge 4$$. Thus we can start testing with $$n=\pm4$$, and hit paydirt first time. We might not have been so lucky, and it isn't really necessary. Testing on a predetermined finite set (the factors of $$100$$) is not really trial and error. · 3 years, 11 months ago