09-240/Classnotes for Tuesday December 1: Difference between revisions

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MAT240 – December 1st

Basic Properties of det: M_nxn→F: 0 det(I) = 1

1. $det(E'_{i,j\,\!}A)=-det(A);|E'_{i,j\,\!}|=-1.[Note:det(EA)=|E||A|]$

2. $det(E_{i,c\,\!}^{2}A)=det(A);|E_{i,j,c\,\!}^{2}|=1$

3. $det((E_{i,j,c\,\!}A)=det(A);|E_{i,j,c\,\!}^{3}|=1$

The determinant of any matrix can be calculated using the properties above.

Theorem:

If $det':M_{nxn\,\!}$ →F satisfies properties 0-3 above, then $det'=det$

$det(A)=det'(A)$

Philosophical remark: Why not begin our inquiry with the properties above?

We must find an implied need for their use; thus, we must know whether a function $det$ exists first.

@@ Line 23: / Line 23: @@
 * Also, note that exchanging two rows flips the sign.
-. <math>det(E^2_{i,c\,\!}A) = det(A) ; |(E^2_{i,j,c\,\!}| = 1</math>
+. <math>det(E^2_{i,c\,\!}A) = det(A) ; |E^2_{i,j,c\,\!}| = 1</math>
 * These are "enough"!
-. <math>det((E_{i,j,c\,\!}A) = det(A) ; |(E^3_{i,j,c\,\!}| = 1</math>
+. <math>det((E_{i,j,c\,\!}A) = det(A) ; |E^3_{i,j,c\,\!}| = 1</math>
 * Adding a multiple of one row to another does not change the determinant.
@@ Line 34: / Line 34: @@
 <b>Theorem:</b>
+If <math> det' : M_{nxn\,\!}</math>→F  satisfies properties 0-3 above, then <math>det' = det</math>
+<math>det(A) = det'(A)</math>
+Philosophical remark: Why not begin our inquiry with the properties above?
+We must find an implied need for their use; thus, we must know whether a function <math>det</math> exists first.