Given vectors 𝐀, 𝐁, and 𝐂, where
𝐀 has components one, five, negative five; 𝐁 has components two, four, three; and
𝐂 has components zero, five, negative four, find the scalar triple product of 𝐀,
𝐁, and 𝐂.
We’re asked to calculate the scalar
triple product of vectors 𝐀, 𝐁, and 𝐂. And we know that this is equivalent
to calculating the determinant of the matrix consisting of the components of the
three vectors as shown. So with our vectors 𝐀, 𝐁, and 𝐂,
we want to find the determinant of the matrix whose first line consists of the
components of vector 𝐀, whose second line has components of vector 𝐁, and whose
third line has the components of vector 𝐂.
And remember to calculate the
determinant of a three-by-three matrix using the first row as our pivot, we take the
first top-left element 𝐀 𝑥 and multiply this by the determinant of the two-by-two
matrix in the bottom-right corner. We then take the negative of the
second element in the top row and multiply this by the determinant shown. And finally, we add the third
element in the top row, multiply it by the determinant formed by the four elements
in the bottom-left-hand corner.
In our case, this translates into
one times the two-by-two matrix with elements four, three, five, negative four minus
five times the determinant of the matrix with elements two, three, zero, and
negative four plus a negative five times the determinant of the two-by-two matrix
with elements two, four, zero, five. And now remembering that the
determinant of a two-by-two matrix with elements 𝑎, 𝑏, 𝑐, 𝑑 is 𝑎𝑑 minus 𝑏𝑐,
we have one times four times negative four minus three times five minus five times
two times negative four minus three times zero minus five again times two times five
minus four times zero.
That is one times negative 16 minus
15 minus five times negative eight minus zero minus five times 10 minus zero. Evaluating this gives us negative
31 plus 40 minus 50, which is negative 41. The scalar triple product of
vectors 𝐀, 𝐁, and 𝐂 is therefore negative 41.