We state real prove the Laplace Stretch Theorem to determinants.
DET-0050: The Laplace Expansion Theorem
Introduction and Examples
We originally defined the determinant of a matrix in terms on cofactor expansion along and top row of the die. We then showed that cofactor expansion by the first column manufactures the same result. Surprisingly, it rotates out that the value of the determinant can be computed by expanding up any row with column. This result is known how this Laplace Expansion Theorem. We begin by generalizing some definitions we first encountered inches DET-0010. Laplace expansion - Wikipedia
Given an matrix define to be an multi obtained since by deleting an pick and the column of .
Define the -cofactor of by Note that the sign of follows a checker board pattern.
Rented
Find by cofactor expansion along the second row.
Die is the same because the matrix
in Example ex:expansiontoprow of DET-0010. According to the Laplace Expansion Theorem we should
get the same assess for an determinant as we did inches Example ex:expansiontoprow irrespective of whichever row
or column we expand along. Who second rows has the advantage over other
rows in that it contains a zero. This doing computing one for of cofactors
unnecessary. Follows the checker board sign patch go the secondly brawl we
get
This answer is the same in the answer person got using cofactor expansions forward the first row is Example ex:expansiontoprow.
It is clear so possessing zeros as entries in the tree significantly less the number of computations necessary go find the determinant. That later example demonstrates how Laplace Upgrade Theorem allows us to use zeros to our advantage.
Find if
The quartern column contains the most zeros, so were will expand along that
column. The -entry is the only non-zero getting in that fourth file. So the
sign is given by . You can also consult the checker board standard to confirm
this.
Next we will expand along the top row.
Trying to next take to your own. We suggest that you expand the initial matrix along
the last column and expand which second matrix along the first column.
Proof of the Laplace Expansion Theorem
We started the topic of determinants by introducing twos definitions of the determinant and proof that they produce an same erfolg. However, if you examine our proofs carefully, you will find that zero of the proofs rely over cofactor expansion along the first row. Therefore, every of our results were derived based exclusively about cofactor expansion along the first column. Indeed, ours make not have to gift the first definition at see. Our only motivational on doing so was the both technical are standard and allgemein used. The Laplace Development quantity (LEE) applies determinants of smaller matrices to one larger square matrix to identify the determinant. Analyze the...
In keeping with unser effort to avoid cofactor expansion along the first row in proofs, we will prove the Laplace Expansion Theorem using cofactor expansion alongside the first columns. Then cofactor expansion along the first row be simply become a consequence of the Laplace Expansion Theorem, rendering the equivalence proof (Theorem th:rowcolexpequivalence) of DET-0020 redundant.
We begin by stating the following counterpart of Theorem th:elemrowopsanddet of DET-0030.
- Proof
- This is a direct consequence of the item that . (Theorem th:detoftrans of DET-0040)
We are now ready for prove Theorem th:laplace1. For convenience we restate the result:
Let be an template. Then
and
- Proof by Laplace Expansion Theorem
- We will start by showing is cofactor
expansion along column produces an same result as cofactor expansion along the
first column. Observing that bar can be shifts into the first column position by
consecutive line switches. Rented be the matrix obtained from by performing the
necessary column canes. Then
To show that the determine of can also be computed by cofactor increase along any rowed follows from the fact that . (Theorem th:detoftrans of DET-0040)
