# Linear Algebra: different determinant answers

I'm having a problem verifying my answer to this question: Solve for x:

$$\left| \begin{array}{cc} x+3 & 2 \\ 1 & x+2 \end{array} \right| = 0$$

I get:

$(x+3)(x+2)-2=0$
$(x+3)(x+2)=2$ thus:
$x+3=2$ and $x+2=2$
$x=-1$ and $x=0$

The book says that $x=-1$ and $x=-4$ is the correct answer.

I tried doing it a different way by expanding and got totally different answers:

$x^2+5x=-4$
$x(x+5)=-4$
$x=-4$ and $x=-9$

What is going on here?

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## 4 Answers

No. What you have done is $$(x+3) \cdot (x+2)=2$$ does not imply $x+3=2$ or $x+2=2$. For example, you can have $x+3 = \frac{1}{2}$ and $x+2 = 4$ which on multiplying gives $2$. Or you can have $x+3 = \frac{1}{8}$ and $x+2 =16$ which on multiplying gives $2$. So there are infinitely many possibilities which you can choose. So this is not the correct way to do.

You have the determinant as $(x+3) \cdot (x+2) -2 = x^{2} +5x+4$. This can be written as $(x+1)(x+4)=0$ which says that $x=-1,-4$.

If you have an equation of the form $f(x) \cdot g(x) =0$, then only either $f(x)=0$ or $g(x)=0$. I think this where you have got confused. But if $f(x) \cdot g(x)= K$, for $K \in \mathbb{R}$, then you cant have $f(x)=K$, or $g(x)=K$, because their product when multiplied gives $K^{2}$. You can have $f(x)= \frac{1}{K}$ and $g(x)=K^{2}$, but then this would keep on continuing. And you can't get all values by doing this.

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The step in your reasoning $$(x+3)(x+2)=2$$ $$\Longrightarrow$$ $$x+3=2\text{ and }x+2=2$$ is incorrect - if $x+2=2$, then $x=0$, hence $x+3=0+3=3$, hence $(x+3)(x+2)=3\cdot 2=6$, not $2$. Also the step in your second approach $$x^2+5x=-4$$ $$\Longrightarrow$$ $$x(x+5)=-2$$ is incorrect. $x^2+5x$ and $x(x+5)$ are the same thing, so it can't equal both $-4$ and $-2$.

Your second approach is on the right track though; you are correct in multiplying $(x+3)$ and $(x+2)$ together, but now bring everything over to one side, leaving $0$ on the other side: $$x^2+5x+4=0$$ Now use the quadratic formula to get that the solutions are $x=-1$ and $x=-4$.

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@ Zev: I believe there is no reason to go through using the quadratic formula when this problem has perfect factors. Being, $(\rm x+1)(\rm x+4)=0$. –  night owl Jul 21 '11 at 8:33

The problem here is that if both $(x+3)$ and $(x+2$ are equal to $2$, then $(x+3)(x+2) = (2)(2) = 4 \not = 2$. So you must first distribute, recollect the terms, and factor it again.

So $(x+2)(x+3) - 2= x^2 + 5x + 4 = x^2 + 4x + x + 4 = (x+1)(x+4) = 0$.

I don't know how you did the second way, so I can't comment on that. But I can say that I appreciate you showing your work. Were only everyone to do so...

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Let me point out where I think your error of reasoning arises: it comes from drawing an invalid analogy to the case where we have an equation of the form $$(x-a)(x-b) = 0.$$ In that case, the standard next step is to say "therefore $x-a=0$ or $x-b=0$"; you are trying to do the same thing with $0$ replaced by $2$.

The reason this works for $0$ is that in the real numbers we have:

If $r$ and $s$ are real numbers, and $rs=0$, then either $r=0$ or $s=0$.

That is: the only way for a product to be equal to $0$ is if (at least) one of the factors is already equal to $0$.

This does not hold for any number other than $0$. There are many ways in which a product can equal $2$; in fact, infinitely many different ways. So you cannot simply break up the product by cases in this situation (or in any situation in which you have a product of real numbers equal to something other than $0$).

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