Video Transcript
Consider the vector 𝐀 equals one,
two, three; 𝐁 equals negative four, negative five, three; and 𝐂 equals one, two,
negative two. Find the scalar triple product of
𝐀, 𝐁, and 𝐂. Find the scalar triple product of
𝐁, 𝐀, and 𝐂.
When we read these questions, we
defined this expression to be the scalar triple product of our three vectors. To find the scalar triple product
as represented, we find the dot product of the vector 𝐀 with the cross product of
𝐁 and 𝐂. So, of course, we could begin by
finding the cross product of 𝐁 and 𝐂 and then finding the dot product of that with
𝐀, or we can use a formula. The scalar triple product of 𝐀,
𝐁, and 𝐂, which doesn’t actually need these parentheses, where the vector 𝐀 has
elements 𝐴 sub 𝑥, 𝐴 sub 𝑦, and 𝐴 sub 𝑧 and the vector 𝐁 has elements 𝐵 sub
𝑥, 𝐵 sub 𝑦, 𝐵 sub 𝑧 and 𝐂 is 𝐶 sub 𝑥, 𝐶 sub 𝑦, 𝐶 sub 𝑧, is given by the
determinant of the three-by-three matrix containing these elements.
This means the scalar triple
product of 𝐀, 𝐁, and 𝐂 is going to be equal to the determinant of the
three-by-three matrix that contains the elements one, two, three; negative four,
negative five, three; and one, two, negative two. So, next, we’ll remind ourselves
how to find this determinant. We take the elements in the top row
and multiply them by the determinant of the two-by-two matrix that remains when we
eliminate the elements in that row and that column. So, in this case, we multiply one
by the determinant of the matrix negative five, three, two, negative two. And of course the determinant of
this two-by-two matrix is negative five times negative two minus three times
two. We then subtract two times the
determinant of the two-by-two matrix with elements negative four, three, one,
negative two.
Finally, we’re going to add three
times the determinant of the remaining two-by-two matrix. We’re multiplying that three by
negative four times two minus negative five times one. This all simplifies to four minus
10 plus negative nine. And that’s equal to negative
15. And so the scalar triple product of
𝐀, 𝐁, and 𝐂 is negative 15.
The second part of this question
wants us to find the scalar triple product of the matrices 𝐁, 𝐀, and 𝐂. And so we could repeat the process
from earlier, finding the determinant of negative four, negative five, three, one,
two, three, one, two, negative two. But actually, there are some
properties of the scalar triple product that can save us some time. For instance, let’s imagine what
happens if we switch two vectors in our scalar triple product. Swapping two horizontal rows in a
three-by-three determinant changes its sign. So, rather than working out this
full determinant, we can see that it contains the same values as the previous
determinant. It’s just that two rows are
swapped. This means we can change its
sign. And so the scalar triple product of
𝐁, 𝐀, and 𝐂 is positive 15.
