# Easy: det(A)=0 <=> no iverse

by vilhelm
Tags: <>, deta0, iverse
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,338 Suppose $$A= \begin{parray} a & b \\ c & d\end{parray}$$ Then $$Ae_1= \begin{parray} a & b \\ c & d\end{parray}\begin{parray}1 \\ 0\end{parray}= \begin{parray}a \\ c\end{\parray}$$ and $$Ae_1= \begin{parray} a & b \\ c & d\end{parray}\begin{parray}0 \\ 1\end{parray}= \begin{parray}b \\ d\end{\parray}$$ So that the rectangle with adjacent sides from (0,0) to (1,0) and from (0,0) to (1, 0) is mapped into the parallelgram with adjacent sides from (0,0) to (a, c) and from (0,0) to (b,d). But the area of such a parallelogram is just the length of the cross product of those two sides: here that cross product is <0, 0, ac- bd> so its length is ac- bd= det(A). That is, T maps all of the points in a rectangle of area A to a parallelogram of area det(A). That is what is meant by "area of TA(1,1)= det(A)".