Matrix Function f(x) f(y) = f(x + y), Exercise 3.2

This lesson works the product question from Exercise 3.2. We are given a matrix-valued function $f(x)$ built from $\cos x$ and $\sin x$, and we show that $f(x)\,f(y) = f(x+y)$ by multiplying the matrices and applying the angle-addition formulas.

The given matrix

The function is $f(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$ Replacing $x$ by $y$ gives $f(y) = \begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix}$ We want to compute the product $f(x)\,f(y)$.

Multiplying f(x) by f(y)

Both matrices are $3 \times 3$, so the product is $3 \times 3$. Multiply row into column.

First row

Entry (1,1): first row of $f(x)$ into first column of $f(y)$: $\cos x\cos y + (-\sin x)(\sin y) + 0 = \cos x\cos y - \sin x\sin y$

Entry (1,2): first row into second column: $\cos x(-\sin y) + (-\sin x)\cos y + 0 = -\cos x\sin y - \sin x\cos y$

Entry (1,3): first row into third column: $0 + 0 + 0 = 0$.

Second row

Entry (2,1): second row into first column: $\sin x\cos y + \cos x\sin y + 0 = \sin x\cos y + \cos x\sin y$

Entry (2,2): second row into second column: $\sin x(-\sin y) + \cos x\cos y + 0 = \cos x\cos y - \sin x\sin y$

Entry (2,3): second row into third column: $0$.

Third row

The third row of $f(x)$ is $\begin{bmatrix} 0 & 0 & 1 \end{bmatrix}$, so it picks out the third row of $f(y)$: the entries are $0$, $0$, and $1$.

Simplifying with the addition formulas

Now apply the standard identities: $\cos x\cos y - \sin x\sin y = \cos(x+y)$ $\sin x\cos y + \cos x\sin y = \sin(x+y)$ So the off-diagonal entry $-\cos x\sin y - \sin x\cos y = -\sin(x+y)$. Substituting these into the product gives $f(x)\,f(y) = \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Conclusion

This is exactly the given matrix with $x$ replaced by $x+y$, that is $f(x)\,f(y) = f(x+y)$ which is what we set out to prove.

Key takeaways

• Write $f(y)$ by replacing the angle $x$ with $y$ in $f(x)$, then multiply the two matrices row into column.
• The diagonal entries give $\cos x\cos y - \sin x\sin y = \cos(x+y)$ and the lower entry gives $\sin x\cos y + \cos x\sin y = \sin(x+y)$.
• The product matches $f$ evaluated at $x+y$, so $f(x)\,f(y) = f(x+y)$.