Given that matrix 𝐴 is equal to cos 𝜃, sin 𝜃, sin 𝜃, negative cos 𝜃 and matrix 𝐵 is equal to sin 𝜃, sin 𝜃, negative cos 𝜃, negative cos 𝜃, find 𝐴𝐵 if possible.
We recall that matrix multiplication is defined if the number of columns in the first matrix is equal to the number of rows in the second matrix. In this question, the matrices 𝐴 and 𝐵 are both two-by-two square matrices. This means that the resulting matrix 𝐴𝐵 will also be a two-by-two square matrix. When performing matrix multiplication, we multiply the elements in each row of the first matrix by each column of the second matrix. Multiplying the elements of the first row of matrix 𝐴 by the first column of matrix 𝐵 gives us cos 𝜃 sin 𝜃 minus sin 𝜃 cos 𝜃.
Multiplying the first row of matrix 𝐴 by the second column of matrix 𝐵 gives us the same expression. Repeating this process for the second row of matrix 𝐴 gives us the elements sin squared 𝜃 plus cos squared 𝜃. The elements in the top row of matrix 𝐴𝐵 are both equal to zero. We know from the Pythagorean identity that sin squared 𝜃 plus cos squared 𝜃 is equal to one. This means that the matrix 𝐴𝐵 is equal to zero, zero, one, one.