Determinant(not really but still) Problem

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

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

#MathProblem #Math

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

Hero Miles - 7 years, 9 months ago

Please see the edit now..And please could you give a solution other than hit and trial??Like using synthetic division??

Krishna Jha - 7 years, 9 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.

Hero Miles - 7 years, 9 months ago

@Hero Miles – 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.

Mark Hennings - 7 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$