Q.If $F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$, show that $F(x) F(y) = F(x+y)$.

Concept and Intuition

$G(x) = \begin{bmatrix}\cos x & -\sin x & 0\\ \sin x & \cos x & 0\\ 0&0&1\end{bmatrix}$ is exactly the matrix that rotates a vector by angle $x$ about the $z$-axis (the top-left $2\times2$ block is the familiar plane-rotation matrix; the third row/column just carries the $z$-coordinate through unchanged). Composing two rotations — rotate by $x$, then by $y$ — is the same as one rotation by $x+y$. This is a geometric fact, not something you need to multiply out by brute force.

Step-by-Step Solution

1. Recognise $G(x)$ as the rotation-by-angle-$x$ matrix in 3D (about the $z$-axis).
2. Composition of rotations adds angles: $G(x)G(y) = G(x+y)$. (You can verify this directly: multiplying the two matrices and using $\cos x\cos y - \sin x\sin y = \cos(x+y)$ and $\sin x\cos y+\cos x\sin y=\sin(x+y)$ reproduces $G(x+y)$.)
3. Given $x+y=0$, so $G(x)G(y) = G(0)$. …

We start with the expression:

The key insight is that each matrix resembles a composition of a rotation and a scaling, linked to the tangent half-angle substitution. In fact, the second matrix is the inverse of the first up to a sign change in the off-diagonal, but not exactly — we need to compute carefully.

1. Compute the inverse of the second matrix. Let

The determinant is $\det(B) = 1\cdot 1 - t(-t) = 1 + t^2$.

The inverse is

2. Multiply the first matrix by this inverse. The first matrix is

So

Notice: $A$ and the numerator of $B^{-1}$ are identical! So we are squaring that matrix.

3. Square the matrix.

Hence

4. Use the tangent half-angle identities. Recall: …

Concept and Intuition

Each named transformation has a standard coordinate-update rule: shifting the origin to $(h,k)$ subtracts $(h,k)$ from the coordinates; translating a point by a vector adds that vector to its coordinates; and rotating the reference axes by angle $\theta$ transforms coordinates via $X=x\cos\theta+y\sin\theta,\ Y=-x\sin\theta+y\cos\theta$. Applying these one after another (feeding each output into the next) tracks the point through the whole sequence.

Step-by-Step Solution

1. Start: $P=(4,1)$.
2. Shift origin to $(1,6)$: new coordinates $=(4-1,\,1-6)=(3,-5)$.
3. Translate 2 units along $+X$: the point itself is displaced by $(2,0)$: $(3+2,\,-5)=(5,-5)$. …

Concept and Intuition

This problem chains three standard coordinate transformations. Each has a clean formula: translation of origin to $(h,k)$ maps $(x,y)\to(x-h,y-k)$; rotation of axes by $\theta$ maps $(x,y)\to(x\cos\theta+y\sin\theta,\,-x\sin\theta+y\cos\theta)$; reflection through the line $y=x$ simply swaps the two coordinates.

Step-by-Step Solution

1. Translation: origin shifted to $(2,-1)$, so $(a,b)=(-1-2,\,2-(-1))=(-3,3)$.
2. Rotation by $45^\circ$: with $\cos45^\circ=\sin45^\circ=\tfrac{\sqrt2}{2}$, $c=a\cos45^\circ+b\sin45^\circ=(-3+3)\cdot\tfrac{\sqrt2}{2}=0$, $d=-a\sin45^\circ+b\cos45^\circ=(3+3)\cdot\tfrac{\sqrt2}{2}=3\sqrt2$. So $(c,d)=(0,3\sqrt2)$. …