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Easy: det(A)=0 <=> no iverse 
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#1
Nov1511, 12:29 PM

P: 18

My question is really simple, but I just have to get it confirmed:
for a matrix A, if det(A)=0 that means A has no inverse? (Edit: Just realised it should be in the homework section, my bad.) 


#2
Nov1511, 03:53 PM

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yes, for a square matrix. i.e. only a square matrix can hope to have an inverse. then an inverse exists if and only if the unit cube is mapped to a parallelepiped that ahs non zero n dimensional volume.
since the determinant measures the volume, this means the det is non zero. why don't you learn what this stuff means, and them you won't have to try to memorize these facts. 


#3
Nov1511, 04:24 PM

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P: 906

in the original poster's defense, it is not obvious that the determinant (especially defined abstractly as a certain arithmetic operation on numerical arrays) admits of a geometric interpretation.
volume is a topological (metric) concept, and matrices can be defined for commutative rings (like integers), and the determinant is still welldefined. someone studying linear algebra, may not realize there are connections with analysis (the fact that the determinant is an alternating tensor (which forms the basis of it being considered a volume element) is often delayed for some time, i was a junior in college before i knew anything about it, and so the jacobian in the changeofvariable theorem was mystifying to me, for a while). what the original poster should know (although perhaps not in such terms) is that the determinant is a semigroup homomorphism: End(V) → F. here, End(V) is more typically presented as F^{nxn}, the set of nxn matrices over F (often F is the real numbers). in practical terms: det(AB) = det(A)det(B). from this, it is clear that if A is invertible, det(A) ≠ 0, since 1 = det(I) = det(A)det(A^{1}), and 1 = 0x has no solution, in a field. it is not so obvious that if det(A) ≠ 0, A is invertible. however, it can be shown that if det(A) ≠ 0, then det(rref(A)) ≠ 0, so rref(A) has no 0 rows, so rank(A) = n, so nullity(A) = 0, and thus A is invertible (A represents a bijective linear map). 


#4
Nov1911, 02:49 PM

P: 18

Easy: det(A)=0 <=> no iverse
I'm in high school right now, so right now we only deal with 2 x 2 matrices. Of (2 x 1)matrices as well when talking about vectors and such, but we are mostly dealing with 2 x 2 for now. Larger matrices will come later on, when doing systems of equations etc.
I hate memorization, and am somewhat obsessed with insight when studying mathematics. But I didn't know that the area of TA(1,1) = det(A) is that was what you were saying. If so, why? 


#5
Nov1911, 04:38 PM

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Suppose
[tex]A= \begin{parray} a & b \\ c & d\end{parray}[/tex] Then [tex]Ae_1= \begin{parray} a & b \\ c & d\end{parray}\begin{parray}1 \\ 0\end{parray}= \begin{parray}a \\ c\end{\parray}[/tex] and [tex]Ae_1= \begin{parray} a & b \\ c & d\end{parray}\begin{parray}0 \\ 1\end{parray}= \begin{parray}b \\ d\end{\parray}[/tex] 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)". 


#6
Nov2011, 04:06 AM

P: 18

Have just read wiki on cross product now. Seems like just a definition: a x b = a*b*sin(α)
I can see, by sketching a parallelgram in R^2, that a*b*sin(α)=Area because the height=b*sin(α). But where is the connection between that and the cross product of <(0, 0), ac bd> ? 


#7
Nov2111, 12:17 PM

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P: 9,498

just write down the images of (1,0) and (0,1) under the matrix and compute the area by noticing it lies in a rectangle, and the complement is a union of triangles and rectangles. It's quite easy to do directly just from the area formulas for triangles and rectangles.



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