Transcript — Introduction
It is not uncommon for students to think that the only point of the determinant is to determine if a matrix is invertible or not. Certainly, this is one of the most important uses of the determinant. However, we saw in the last lecture that the value of the determinant can mean more. In particular, we use determinants in Cramer’s Rule to solve systems of linear equations. In this lecture, we will see that the determinant also has a very nice geometric interpretation.
Area of a Parallelogram
Let u = [u1; u2] and v = [v1; v2] be two vectors in R2. We can then form a parallelogram in R2 with corner points (0, 0), (u1, u2), (v1, v2), and (u1 + v1, u2 + v2). This is called the parallelogram induced by u and v. We know the area A of a parallelogram is length times height. Observe from the diagram that the length of the parallelogram is the length of u, and the height is the length of the perpendicular of the projection of v onto u. So we get the area is the length of u times the length of the perpendicular of the projection of v onto u. Using trigonometry, we can see that the length of the perpendicular of v onto u equals the length of v times |sin theta|, where theta is the angle between u and v. Hence, A = the length of u times the length of v times |sin theta|.
Now recall from Module 1 that cos theta = (u dot v)/((length of u)(length of v)). Using this gives the Area-squared = ((length of u)-squared)((length of v)-squared)(sin-squared theta). Using a trig identity, this equals ((length of u)-squared)((length of v)-squared)(1 – cos-squared theta). Expanding with the distributive property, and applying the identity above, we get ((length of u)-squared)((length of v)-squared) – (u dot v)-squared. Simplifying this, we get the square of the determinant of the matrix whose columns are the vectors u and v. And hence, the area of the parallelogram is the absolute value of the determinant of the matrix whose columns are u and v.
Example: Determine the area of the parallelogram induced by the vectors u = [2; 5] and v = [3; 4]. This is really easy. We have the area A of the parallelogram is the absolute value of the determinant of the matrix whose first column is u and whose second column is v. That is, it is the absolute value of the determinant of |2, 3; 5, 4|, which is |-7|, and hence the area of the parallelogram is 7.
So we see that the geometric interpretation of the absolute value of a 2 by 2 determinant is the area of a parallelogram. But also notice that we do need to take the absolute value of the determinant to get the area, since area is positive, but, as in the case of the last example, a determinant might be less than 0. This makes us wonder what the geometric interpretation of a negative determinant is. Recall that swapping columns multiplies the determinant by -1. So that means the area of the parallelogram induced by v = [3; 4] and u = [2; 5] is, the Area = the absolute value of the determinant of |3, 2; 4, 5|, which is 7. Of course, the area is the same, since it is the same parallelogram, but in this case, we found that the determinant was positive. Notice that in the first diagram, the angle from u to v is negative, while the angle from v to u in the second diagram is positive. So the sign of the determinant is an indication of how the vectors are oriented compared to each other in the order that we pick them. So we really have a geometric view of why we multiply a determinant by -1 when swapping columns.
Of course, we can get geometric interpretations of the other column operations on determinants as well. If we multiply a column by a scalar, we multiply the determinant by a scalar. Geometrically, this is obvious. For example, if we multiply the first column of |2, 3; 5, 4| by 3, we get |6, 3; 15, 4|. This corresponds to multiplying u by the scalar 3. Hence, we are just stretching one side of the parallelogram by a factor of 3, and so we get 3 times the area. Indeed, the area of the parallelogram induced by 3u and v is the absolute value of the determinant of |6, 3; 15, 4|, which equals 21.
Adding a multiple of one column to another performs what is called a shear. It keeps a pair of parallel sides of the parallelogram the same, and rotates the other two sides by the same amount. Visually, we can see that this does not change the area. For example, if we use the column operations C2 – 2C1 on the matrix [2, 3; 5, 4], we get [2, -1; 5, -6]. This corresponds to finding the area of the parallelogram induced by [2; 5] and [-1; -6]. We find that the area of this parallelogram is also 7.
Volume of a Parallelepiped
We now look at this in 3 dimensions. Observe that the three vectors u, v, and w in R3 induce a parallelepiped. We can also use the determinant to find the volume of the parallelepiped. We know the volume of a parallelepiped is the area of base times height. Observe that the base of the parallelepiped is a parallelogram induced by the vectors u and v. Unfortunately, we can’t use the determinant to calculate the area of this parallelogram because we only have two vectors in R3, and a matrix has to be square to have a determinant. So instead, we use our knowledge of the cross product in R3.
As before, we get that the area of the parallelogram is (length of u)(length of v)|sin theta|. By Theorem 1.3.4, we get that, since u and v are vectors in R3, that this is equal to the length of (u cross v). So now we just need to calculate the height. We observe that the height is the projection of w onto the normal vector n of the plane induced by u and v. Thus, with some simplification, we find that the height is |w dot (u cross v)|/(length of (u cross v)). Hence, as expected, we find that the volume V of the parallelepiped is given by the absolute value of the determinant of the matrix whose columns are u, v, and w.
Example: Determine the volume of the parallelepiped induced by u = [2; -1; 1], v = [1; 1; 0], and w = [2; 3; 1]. We find the volume is the determinant of the matrix whose columns are u, v, and w, which we find equals 4.
n-Volume of a Parallelotope
Of course, this formula works in higher dimensions as well. In particular, if v1 to vn are vectors in Rn, then they induce an n-dimensional parallelotope—the n-dimensional version of a parallelogram or parallelepiped. We can calculate the n-volume of the parallelotope by taking the absolute value of the determinant of the matrix whose columns are v1 to vn.
This concludes this lecture and this module. In this module, we have looked at properties of invertible matrices and determinants. What we have done here will be extremely important and useful in the next module, titled “Diagonalization”.