Wikipedia:Reference desk/Archives/Mathematics/2009 February 18#Problem from Linear Algebra

If A= P(1+r)n^{Superscript text} Then make n the subject of formula —Preceding unsigned comment added by 66.36.215.172 (talk) 10:21, 18 February 2009 (UTC)

:Assuming you mean A = P(1+r)ⁿ look into power rules etc. 89.240.5.5 (talk) 10:38, 18 February 2009 (UTC)

:(added header) Consider taking logarithms of both sides. Confusing Manifestation(Say hi!) 22:34, 18 February 2009 (UTC)

Don't get them at all, no sir.

Find all x that satisfy:

Now I was told I have to consider two cases, x > 3 and x < 3. x = 3 is obviously undefined.

x > 3:

$x\; >\; \backslash frac\{2\}\{3\}$

So x > 3 and $x\; >\; \backslash frac\{2\}\{3\}$. I guess that means $x\; >\; \backslash frac\{2\}\{3\}$?

x < 3:

$-2x\; +\; 5\; >\; x\; +\; 3\; \backslash ,$

(Since x+3 must be negative)

$-3x\; >\; -2\; \backslash ,$

We get the same kind of logic as before. This is how it was explained to me. I won't even start on the fact that there's a numerator that has an x in it as well, not to mention that you can probably rearrange simple inequalities as rational ones and no doubt create a world of pain for yourself anyway. Someone is playing a joke on me.

Anyhow I was told that the method I'm using sucks and that I should solve the inequality by getting the right hand side to zero first.

Okay, so above is negative for x > 0, but also for x < -3. That's what I've ascertained from examining the equation with my head, which is not a nice algorithmic process for solving problems, and there's probably something I've overlooked. I'm skeptical now because all the other answers in this useless book are of the form a < x < b. But I wouldn't be surprised if it's part of the elaborate joke to make all the even-numbered questions of a completely different type to the odds. Ha. Ha. HA. Help me before I kill myself. 82.32.49.186 (talk) 10:56, 18 February 2009 (UTC)

:You are approaching the problem correctly. Both of the methods you described work and can be used to solve this problem, but unfortunately you made numerical mistakes in both approaches.

:In your first approach: under what conditions is x+3 positive? You use that x+3 is positive if and only if x > 3, but that is incorrect. Similarly for negative.

:Also in your first approach, you have inadvertantly switched "and" and "or". When is it true that both x > 3 and x > 2/3? The answer is x > 3, not x > 2/3. (However, it was supposed to be x > -3 anyhow.)

:Although your first approach does indeed work, I agree that the second approach is easier. You correctly reach the inequality , but then incorrectly state that that holds for x > 0 and x < -3. As you haven't explained your reasoning in detail, I can't help you find your mistake, but maybe you should try x = 1/2 as an example.

:Finally, the answer will not be of the form a < x < b. Please let us know if you would further help with this problem; it looks to me like you are understanding this material well, and just making easy-to-make numerical errors. Eric. 131.215.158.184 (talk) 11:41, 18 February 2009 (UTC)

::Through 5 or 6 mintues of tedious trial and error, I examined the points where the sign of the function (the rational expression, treat it as a function) switched. The function, according to these calculations is negative for all x < -3 and all x > (2/3), if I got it right. I suspect that the trick is to examine the cases less than or equal to where the denominator is undefined, and where the numerator is = 0. Does this trick hold for every rational, polynomial expression? Brute force is not my idea of a good time 82.32.49.186 (talk) 12:10, 18 February 2009 (UTC)

:::You are (roughly) correct: for any rational expression a / b (whether polynomial or otherwise), the sign of a / b can be determined by knowing the signs of a and b. Thus the "critical points" to examine are when the numerator a or denominator b are zero, for those are the only points where the expression a / b can change sign. I am sure others can pick it up from here if you have further questions, I am going to bed. (And by the way, your answer is correct.) Eric. 131.215.158.184 (talk) 12:34, 18 February 2009 (UTC)

::::My book gives -3 < x < 2/3 as the answer. What the hell? 82.32.49.186 (talk) 12:48, 18 February 2009 (UTC)

:::::Did you give us the problem correctly? Then the book is wrong. Try x = 0 if you have doubts. Eric. 131.215.158.184 (talk) 18:54, 18 February 2009 (UTC)

:An obvious thing to do to get a quick feel whether your results are the right ones is to use a spreadsheet to calculate and graph f(x) against x. In your case maybe plot values from x = -5 to +5 in steps of 0.1? and just see which values of x have f(x) < 1. That will help you spot gross blunders (eg +3 instead of -3 and so on easily. -- SGBailey (talk) 12:46, 18 February 2009 (UTC)

If you say x > 2/3 AND x > 3, then that means x > 3. You cannot ignore the fact that the word AND means AND, nor the fact that 3 > 2/3.

:

Next, where you say x < 2/3 AND x < 3, then that means x < 2/3.

:

So the solution is x > 3 OR x < 2/3. Any number greater than 3 is a solution; any number less than 2/3 is a solution; no other numbers are solutions.

:

Your other method also works. You have

:

and then you can divide both sides by −3, which requires changing "<" to ">" since what you're dividing by is negative:

: $\backslash frac\{x\; -\; (2/3)\}\{x\; +\; 3\}\; >\; 0.$

Now either x > 3 or x < 2/3 or x is either between those two or equal to one of them. If x > 3 then both the numerator and denominator are positive, so you've got a solution. If x < 2/3 then both are negative, so you've got a solution. If x is between the two, then the denominator is negative and the numerator is positive, so you haven't got a solution. Michael Hardy (talk) 21:38, 18 February 2009 (UTC)

An alternative approach involves the rule that "multiplying both sides of an inequality by a strictly positive value does not affect the inequality". Instead of testing individual cases, multiply everything by $(x+3)^2$. For $x\; \backslash neq\; 3$, this is guaranteed to be positive, and you can then rearrange the inequality to give you (quadratic) > 0. Then a factorisation of the quadratic and a quick check of where it's positive should spit out the answer (just remember to check for any funny business around the x=3 point as well). Confusing Manifestation(Say hi!) 22:32, 18 February 2009 (UTC)

The following question is from Kluwer's Foundations of Linear Algebra and has been driving me crazy for the last day. I imagine the solution is simple and that I am just missing it, whatever the case, any help would be deeply appreciated:

Let A be a nonsingular 4 x 4 matrix so and B, C, D, and E be 2 x 2 matrices with A = (B C)

(D E)

can BE - DC, BE - CD, EB - DC, and EB - CD all be singular?

The best result I can get is that if B and D commute or C and E, then no; and if the commutator of B and E or

of D and C is singular, then no (not 100% sure that I didn't make a mistake on the last one though.) Again, thanks in advance for any help:) Phoenix1177 (talk) 11:18, 18 February 2009 (UTC)

:With the help of the partial results you provided, I found a counterexample with a little trial and error (i.e., a nonsingular four by four matrix A such that BE - DC etc. are all singular). However, in my example the commutator of B and E and of D and C are singular, so I'd suggest checking your proof of that fact. I'll leave the joy of hunting for counterexamples to you. Eric. 131.215.158.184 (talk) 11:51, 18 February 2009 (UTC)

::I assumed that I made a mistake there, it's late here. At any rate, I was hoping that it wouldn't come down to hunting out a counterexample...66.202.66.78 (talk) 12:16, 18 February 2009 (UTC)

:::I (or someone else) can give you a hint in the morning if you want one. As usual, when there exist counterexamples, there is often a very simple counterexample. Eric. 131.215.158.184 (talk) 12:38, 18 February 2009 (UTC)

::::Thank you, I actually found a counterexample as soon as you mentioned there was one. This is more a case of bad textbook logic, on my part, than anything else; since this was in a chapter on determinants, and finding a counterexample didn't seem to have much to do with determinants, I assumed that a proof was being looked for...obviously, the answer is in the negative, so no proof. Thank you :) P.S. The stuff about the commutators was assuming one of them was invertible. Phoenix1177 (talk) 13:15, 18 February 2009 (UTC)