(New page: =Properties of the Determinant= The determinant is a fundamental property of any square matrix. It is therefore important to know how the determinant is affected by various ope...) |
|||
Line 50: | Line 50: | ||
This is an incredible thing. | This is an incredible thing. | ||
+ | ----- | ||
+ | [[MA351|Back to MA351:"Elementary Linear Algebra"]] | ||
+ | |||
+ | [[Category:MA351]] |
Revision as of 06:12, 14 April 2010
Contents
Properties of the Determinant
The determinant is a fundamental property of any square matrix. It is therefore important to know how the determinant is affected by various operations
Row Operations
This section outlines the effect that elementary row operations on a matrix have on the determinant
Row Switching
If $ B $ is a square matrix formed from swapping one row of $ A $ with another, then
$ \text{det}(B)=-\text{det}(A) $
If n number of row swaps have been made, then
$ \text{det}(B)=(-1)^n\text{det}(A) $
Row Scaling
If $ B $ is a square matrix formed from dividing one row of $ A $ by a number, k, then
$ \text{det}(B)=\frac{1}{k}\text{det}(A) $
Row Addition
If $ B $ is a square matrix formed from adding one row of $ A $ to another, then
$ \text{det}(B)=\text{det}(A) $
Calculating Determinant from RREF
By combining the previous three properties and tracing the math you use to get to the reduced row echelon form you can easily calculate the determinant. Just keep track of how many row swaps and scalings you make. If you use n row swaps and p divisions then
$ \text{det}(A)=(-1)^n*\text{det}(\text{rref}(A))*\prod_{i=1}^{p}\frac{1}{k_i} $
where $ \frac{1}{k_i} $ is the ith scaling factor
One thing to note is that if you go all the way to RREF, then the determinant of the reduced row echelon form must either be one or zero because the reduced row echelon form is either the identity or has a row of zeros. From this we get our next property
Invertibility
A is invertible if and only if
$ \text{det}(A)\ne 0 $
Matrix Multiplication
This is best explained simply with an equation.
$ \text{det}(AB)=\text{det}(A)\text{det}(B) $
And since the right hand side is simply multiplication of scalars, it is commutative, and therefore we can show that
$ \begin{align} \text{det}(AB)&=&\text{det}(A)\text{det}(B)\\ &=&\text{det}(A)\text{det}(B)\\ &=&=\text{det}(BA)\\\end{align} $
This is an incredible thing.