Section 4.1 Determinants and their Geometry
¶Subsection 4.1.1 Linear Distortions
Investigation 4.1.1. Distorted Area.
Suppose we form a parallelogram with the vectors \(\vec{v}=\left\lt3,1\right\gt\) and \(\vec{w}=\left\lt1,2\right\gt\text{.}\) To find the area we need to multiply the length of the base by the height, or (from Figure 4.1.1)
which carries the square of area 1 made by the standard basis to this parallelogram, then
This is because given a matrix
the value
called the determinant of \(A\) measures how much \(A\) distorts \(\mathbb{R}^2\) if we think of it as a transformation.
Investigation 4.1.2. Distorted Volume.
Suppose we have a parallelepiped as in the figure, for simplicity assume the base is a rectangle. The the volume will be area of the base times the height. The area of the base is easy, \(Area=||\vec{v}||\, ||\vec{w}||\text{.}\) To get the height we need to use projections, in particular:
So, that the magnitude will be
the expression inside the absolute value is the determinant of the matrix
which transforms the vectors of the standard basis into the vectors \(\vec{v}\text{,}\) \(\vec{w}\text{,}\) and \(\vec{u}\text{.}\)
Subsection 4.1.2 Calculating Determinants (Laplace's Formula)
Investigation 4.1.3. Two-By-Two.
Consider the matrix...
Investigation 4.1.4. Three-By-Three.
Consider the matrix...
Investigation 4.1.5. N-By-N.
Consider the matrix...
Properties of Determinants Summarized.
Theorem 4.1.3.
If \(A\) and \(B\) are both \(n\times n\) matrices and \(I\) is the identity matrix, then:
- \(det(I)=1\)
- \(det(AB)=det(A)det(B)\)
- \(det(A^{-1})=det(A)^{-1}\) provided \(A\) is invertible
- \(det(A^T)=det(A)\)
- if \(A\) is upper diagonal, lower diagonal, or diagonal, then \(det(A)\) is the product of the entries along the main diagonal.
- swapping two rows of \(A\) changes the sign of \(det(A)\)
- multiplying a row of \(A\) by \(k\) multiplies \(det(A)\) by \(k\)
- adding copies of one row of \(A\) to another does not change \(det(A)\)