View Full Version : Determinant=0 and invertibility
Jin314159
Mar23-04, 06:14 AM
Can someone provide an intuitive understanding of why a matrix is not invertible when it's determinant is zero?
matt grime
Mar23-04, 06:23 AM
The determinant measures how the volume of the unit box changes. Unit box here means all the points
{(a,b,c...,d) | 0<= a,b, ..d <=1
Determinant zero means that it gets squished into smaller dimenisions:
eg, for 2x2, the unit square gets sent to a line segment, in 3x3 the unit cube gets sent to either a 2-d or 1-d figure
you can't undo these operations, because infinitely many points get sent to the same place.
eg
|1 0|
|0 0|
sends all the points with the same y coordinate to the same place, and it squashes the unit square to the unit interval.
Is that ok? That's the geometry, we can talk algebraic reasons too.
HallsofIvy
Mar23-04, 10:21 AM
A very good "intuitive reason" is that det(AB)= det(A)det(B).
If AB= I then det(A)det(B)= 1 not 0 so neither det(A) nor det(B) can be 0.
Jin314159
Mar23-04, 08:27 PM
Thanks guys for both the geometric and algebraic intuition.
PRodQuanta
Apr3-04, 12:34 AM
To find a inverse matrix, you must take 1/det. If your det is equal to zero, it is undifined.
Paden Roder
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