<s>[INST] 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?
 [/INST] 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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