Linear transformations as matricesEdexcel International A Level Further Maths: Revision notes

A point $(x,y)$ is written as the column vector $\begin{pmatrix}x\\ y\end{pmatrix}$. A linear transformation sends it to $\begin{pmatrix}ax+by\\ cx+dy\end{pmatrix}$ for constants $a,b,c,d$. Two properties follow: $T(\mathbf p+\mathbf q)=T(\mathbf p)+T(\mathbf q)$ and $T(k\mathbf p)=kT(\mathbf p)$. In particular the origin always maps to itself. The image of a point is where the transformation sends it.

The transformation $(x,y)\to(ax+by,\,cx+dy)$ is represented by the matrix $\begin{pmatrix}a&b\\ c&d\end{pmatrix},\qquad\begin{pmatrix}a&b\\ c&d\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}ax+by\\ cx+dy\end{pmatrix}.$ The first column is the image of $(1,0)$ and the second column is the image of $(0,1)$. Example: if $T$ maps $(1,0)\to(2,3)$ and $(0,1)\to(-1,4)$, its matrix is $\begin{pmatrix}2&-1\\ 3&4\end{pmatrix}$.

To find the image of a point, multiply its column vector by the matrix, with the matrix on the left: $\begin{pmatrix}2&-1\\ 3&4\end{pmatrix}\begin{pmatrix}-2\\ 5\end{pmatrix}=\begin{pmatrix}2(-2)+(-1)(5)\\ 3(-2)+4(5)\end{pmatrix}=\begin{pmatrix}-9\\ 14\end{pmatrix}.$ You can also use linearity: $(-2,5)=-2(1,0)+5(0,1)$ maps to $-2(2,3)+5(-1,4)=(-9,14)$.

If the images of $(1,0)$ and $(0,1)$ are not given, use any two points. Suppose $\mathbf M=\begin{pmatrix}a&b\\ c&d\end{pmatrix}$ maps $(1,2)\to(5,3)$ and $(2,-1)\to(5,-4)$. Then $a+2b=5$, $c+2d=3$, $2a-b=5$ and $2c-d=-4$. These split into two pairs of simultaneous equations: $a=3$, $b=1$ and $c=-1$, $d=2$. Keep the unknowns for each row together: the first row $(a,b)$ uses the first coordinates of the images and the second row $(c,d)$ uses the second coordinates.

If $A$ is the transformation with matrix $\mathbf A$ and $B$ has matrix $\mathbf B$, then $B$ followed by $A$ has matrix $\mathbf{AB}$. The first transformation applied is the matrix nearest the vector, so it is written on the right: $\mathbf{AB}\mathbf p=\mathbf A(\mathbf B\mathbf p)$. Example: $\mathbf T=\begin{pmatrix}2&1\\ -1&3\end{pmatrix}$ and $\mathbf U=\begin{pmatrix}1&2\\ 0&-1\end{pmatrix}$. $T$ followed by $U$ is $\mathbf{UT}=\begin{pmatrix}0&7\\ 1&-3\end{pmatrix}$. $U$ followed by $T$ is $\mathbf{TU}=\begin{pmatrix}2&3\\ -1&-5\end{pmatrix}$. These differ, because matrix multiplication is not commutative.