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Mathematical Expression Editor
Determinants, Areas, and Volumes
Determinant and the Area of a Parallelogram
Consider the parallelogram determined by vectors and in .
Recall that the area of a parallelogram is given by the product of the length of the
base and the height. As shown in the diagram below, the length of the base is the
magnitude of . The height, , can be found using trigonometry
Using the area of a parallelogram formula together with Theorem th:crossproductsin we get
We have established the following formula.
The area of a parallelogram determined by vectors and in is given by
We can start by visualizing the parallelogram in GeoGebra. RIGHT-CLICK and
DRAG to rotate the image below. The area of the parallelogram, rounded to two
decimal places, is displayed inside the parallelogram.
To find the exact area we compute
Formula form:areaofparallelogram can be easily adapted to parallelograms determined by vectors in , as
illustrated by the following example.
Find the area of the parallelogram in the diagram.
The vectors that determine the parallelogram are
The problem we run into is that these vectors are in , whereas the cross product is
defined only for vectors in . We will get around this difficulty by “padding” our
vectors with zeros on the bottom. In other words, we will consider them as vectors
sitting in the -coordinate plane in . This allows us to compute the cross
product
The area of the parallelogram is then given by
Example ex:areaofparallelogram illustrates an important phenomenon. Observe that the zeros in the last
column of the determinant ensure that the and components of the cross product are
zero, while the last component is the determinant of the matrix whose rows (or
columns) are the two vectors that determine the parallelogram in . In general, if the
parallelogram is determined by vectors
then the area of the parallelogram can be computed as follows:
So the area of the parallelogram turns out to be the absolute value of the
determinant of the matrix whose rows (or columns) are the two vectors
that determine the parallelogram. The following formula summarizes our
discussion.
Let and be vectors of . The area of the parallelogram determined by and is given
by
We will start by splitting this region into triangles.
We can find the total area of the polygon by finding the area of each triangle. The
area of each triangle is half of the area of the corresponding parallelogram. For
instance, is half of the area of the parallelogram depicted below.
We compute
The total area of the polygon is .
Determinant and the Volume of a Parallelepiped
Our next goal is to find the volume of a three-dimensional figure called a
parallelepiped. A parallelepiped is a six-faced figure whose opposite faces are
congruent parallelograms located in parallel planes. A parallelepiped is a
three-dimensional counterpart of a parallelogram, and is determined by three
non-coplanar vectors in . The figure below shows a parallelepiped determined by
three vectors.
Consider a parallelepiped determined by vectors , and , as shown below.
The volume of a parallelepiped is given by
We will consider the parallelogram determined by and to be the base of the
parallelepiped. Thus, the area of the base is given by
The height of the parallelepiped is measured along a line perpendicular to the base.
By Theorem th:crossproductorthtouandv, lies on such a line. Let be the angle between and , . Then the height,
, of the parallelepiped is given by
It may be difficult to visualize this in two dimensions. Below is a replica of of
the above diagram in GeoGebra. RIGHT-CLICK and DRAG to rotate the
image.
This gives us the following formula for the volume of the parallelepiped
We have established the following formula.
The volume of a parallelepiped determined by vectors , and in is given
by
Our next goal is to show that this expression for the volume is equal to the
determinant of a matrix whose rows are the vectors that determine the
parallelepiped.
Let
then \begin{align} \label{eq:boxproduct}(\vec{u}\times \vec{v})\dotp \vec{w}=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\u_1&u_2&u_3\\v_1&v_2&v_3\end{vmatrix}\dotp \begin{bmatrix}w_1\\w_2\\w_3\end{bmatrix}=\begin{vmatrix}w_1&w_2&w_3\\u_1&u_2&u_3\\v_1&v_2&v_3\end{vmatrix} \end{align}
The expression in (eq:boxproduct) is sometimes referred to as the box product or the scalar triple
product.
Recall that (Theorem th:detoftrans). Therefore, the three vectors that determine the
parallelogram can be used to form rows or columns of the determinant on the right
side of (eq:boxproduct). This gives us the following formula.
Let be vectors in . Then the volume of the parallelepiped determined by , and is
given by
Determinants and Linear Transformations
We will now turn our attention to the determinant of a matrix of a linear
transformation.
The following GeoGebra interactive shows a polygon located in the domain
of a linear transformation induced by the matrix . The right-hand side
shows the image of under . The number inside each polygon indicates its
area.
Let . Find the determinant of .
Drag the vertices of to change the polygon. Make a note of how the area
of and the area of the image change. How are the areas related to each
other?
Change the matrix to a matrix whose determinant is 1. Compare the areas of and .
Try matrices whose determinant is 0 or negative. What do you observe about the
areas?
Formulate a conjecture about the relationship between the area of the polygon and
the area of its image under a linear transformation.
We will not prove your conjecture in Exploration exp:LinTransAreaDet for arbitrary figures as it is
beyond the scope of this text. However, we can tackle the problem of how linear
transformations affect areas of parallelograms. This is the topic of our next
example.
Let be a linear transformation induced by matrix . Suppose and are vectors in .
Let be a parallelogram determined by and . Show that
Let and , and let . By Formula form:areaofparallelogramdeterminant, . Applying to and we get
Using Formula form:areaofparallelogramdeterminant, we compute \begin{align*} \mbox{Area}(T(P))=&|(am+bn)(cp+dq)-(ap+bq)(cm+dn)|\\ =&|(mq-np)(ad-bc)|\\ =&|\det{M}|\mbox{Area}(P) \end{align*}
Practice Problems
Let be a square determined by and . Let be a parallelogram determined by vectors
and .
(a)
Sketch both figures in the same coordinate plane, and use geometry to
explain why and have the same area. Compute the area of using Formula
form:areaofparallelogramdeterminant.
(b)
Suppose is the standard matrix of a linear transformation such that .
Find .
