Find the multiplicative inverse of
sin 𝜃, negative cos 𝜃, cos 𝜃, sin 𝜃.
Remember, for a two-by-two matrix
𝐴 with elements 𝑎, 𝑏, 𝑐, 𝑑, its inverse is given by one over the determinant of
𝐴 multiplied by 𝑑, negative 𝑏, negative 𝑐, 𝑎, where its determinant can be
found by multiplying the elements in the top left and the bottom right and then
subtracting the product of the elements on the top right and the bottom left.
Notice this means if the
determinant is zero, the inverse cannot exist since one over the determinant of 𝐴
will be one over zero. And we know that to be
Let’s begin by finding the
determinant of the given matrix. We begin by finding the product of
the top left and bottom right elements. That’s sin 𝜃 multiplied by sin
𝜃. We then subtract the product of the
top right and bottom left elements. That’s negative cos 𝜃 multiplied
by cos 𝜃. Sin 𝜃 multiplied by sin 𝜃 can be
simply written as sin squared 𝜃. And then, we subtract negative cos
squared 𝜃, which becomes plus cos squared 𝜃.
Recall the trigonometric
identity. Sin squared 𝜃 plus cos squared 𝜃
is equal to one. And we can replace sin squared 𝜃
plus cos squared 𝜃 with one. And the determinant of this matrix
is simply one.
So let’s substitute what we know
into the formula for the inverse of the matrix. One over the determinant is one
over one. We then swap the elements in the
top left and the bottom right. Sin 𝜃 swaps with sin 𝜃. So these actually remain
unchanged. We multiply the elements on the top
right and bottom left by negative one, essentially changing their sin. And we get cos 𝜃 on the top right
and negative cos 𝜃 on the bottom left. One over one is simply one.
So the multiplicative inverse of
this matrix is sin 𝜃, cos 𝜃, negative cos 𝜃, sin 𝜃.