Linear Algebra: solving for x I was wondering why in this exercise there isn't a minus $(x^2-2x-3)$. I thought that for this type of exercise you must use $(ad-bc)$.


Problem 50 (p 131) Solve for $x$.
  $\begin{vmatrix} x-2& -1\\ -3 & x \end{vmatrix}=0$
  $(x-2)\cdot x + (-3)\cdot(-1)=0$
  $x^2-2x+3=0$
  $x-3=0$ or $x+1=0$
  $x=3$ or $x=-1$


This is a mistake. To calculate the determinant of a $2\times 2$ matrix you use $ad-bc$ which gives you $(x-2)x-(-1)(-3)=x^2-2x-3=(x+1)(x-3)=0$. Hence $x=3$ or $x=-1$.
