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Here is the question: Find the set of all pairs $(a, b)$ of real numbers such that

$\det(A)=\left| \, \begin{matrix} a+1 & 3a & b+3a & b+1\\ 2b & b+1 & 2-b & 1\\ a +2 & 0 & 1 & a +3\\ b -1 & 1 & a + 2 & a +b\\ \end{matrix}\, \right|=0. $

I have reduced the matrix using column operations where the first column is $(a,0,0,0)^T$, but that still leaves me with a nasty polynomial $p(a,b)$ to solve. When I solve, I get something like this: $\{(0,b)\, |\, b\in \mathbb{R}\}$ or $\left\{(a,b)\,|\, b\in \mathbb{R}\setminus\{-1\},\, a=\frac{(5 b+2)\pm \sqrt{(-5 b-2)^2-4 (-b-1) (b^2-8 b-1)}}{-2(b+1)}\right\}$

I feel like there has to be a trick here that I'm missing.

By the way, this is for a graduate level linear algebra class.

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    $\begingroup$ This is a bad question to ask for a graduate level linear algebra class. $\endgroup$ – parsiad Sep 30 '16 at 19:07
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    $\begingroup$ ...or for a high school term paper (11th grade) in the poor East European country I grew up in. Graduate level?? $\endgroup$ – mathguy Sep 30 '16 at 19:13
  • $\begingroup$ @par , Are you insinuating that this is a bad question because there is no trick? :) $\endgroup$ – hungryformath Sep 30 '16 at 19:16
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    $\begingroup$ I mean, even if there was a "trick", what would be the point of this question? To test that someone can find "tricks"? ^_^ $\endgroup$ – parsiad Sep 30 '16 at 19:20
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    $\begingroup$ Note that you forgot a possible solution: $(a,b) = (-8/3,-1)$. $\endgroup$ – TastyRomeo Sep 30 '16 at 21:28

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