5.4 - Combining transformations with matrices

Replace successive transformations by one matrix product, keeping their order clear. You will compare reversed orders and use repeated multiplication to recognise cycles, so a long sequence of transformations can become a short calculation.

The first transformation goes on the right

How would you work out the effect of $2027$ anticlockwise quarter-turns without multiplying $2027$ matrices? We will reduce that sequence to one familiar matrix. First, get the order right for just two transformations.

You already know

A matrix acts on a point column from the left. The columns give the images of $A(1,0)$ and $C(0,1)$ ; $O(0,0)$ stays fixed. Multiply rows of the left matrix by columns of the right matrix.

Suppose $\mathbf{K}$ acts first on an input column $\mathbf{p}$ . Its image is $\mathbf{Kp}$ . Applying $\mathbf{L}$ next gives $\mathbf{L}(\mathbf{Kp})$ , so the combined matrix is $\mathbf{LK}$ .

Combining two transformations

For $2\times2$ matrices $\mathbf{K}$ , $\mathbf{L}$ , and a $2\times1$ input column $\mathbf{p}$ ,

\mathbf{L}(\mathbf{Kp})=(\mathbf{LK})\mathbf{p}

Thus “ $\mathbf{K}$ followed by $\mathbf{L}$ ” uses $\mathbf{LK}$ : the rightmost matrix acts first. Grouping a product does not change the result, but swapping its factors may.
Not on the formulae sheet: learn it.

Read the product from the input outwards. This is the same logic as applying one function and then another: the first operation sits closest to the input.

Worked example 1: reflection followed by a quarter-turn

Question. $\mathbf{F}=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ represents reflection in $y=0$ . $\mathbf{R}=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ represents rotation $90^\circ$ anticlockwise about $O$ . Work out the matrix for $\mathbf{F}$ followed by $\mathbf{R}$ , describe the single transformation and work out $A'$ , $B'$ , $C'$ for the unit square.

Plan. Calculate $\mathbf{RF}$ , recognise its image columns and multiply for $B(1,1)$ .

\begin{aligned} \mathbf{RF} &=\begin{pmatrix}0(1)+(-1)(0)&0(0)+(-1)(-1)\\1(1)+0(0)&1(0)+0(-1)\end{pmatrix} && \text{first matrix on the right}\\ &=\begin{pmatrix}0&1\\1&0\end{pmatrix} && \text{evaluate the product}\\ A'&=(0,1),\qquad C'=(1,0) && \text{read the columns}\\ \mathbf{RF}\begin{pmatrix}1\\1\end{pmatrix} &=\begin{pmatrix}1\\1\end{pmatrix} && \text{calculate }B' \end{aligned}

Why this step? $\mathbf{F}$ must act on the original point before $\mathbf{R}$ , so it must be closest to that point column.

Answer. The matrix is $\begin{pmatrix}0&1\\1&0\end{pmatrix}$ , a reflection in $y=x$ . The images are $A'(0,1)$ , $B'(1,1)$ , $C'(1,0)$ .

Check. Follow $A$ in two steps: reflection in $y=0$ keeps $(1,0)$ , then the quarter-turn sends it to $(0,1)$ , agreeing with the first product column.

Worked example 2 changes only the direction of the rotation. The order rule stays the same.

Worked example 2: reversing the turn's direction

Question. Keep $\mathbf{F}$ as above, but use $\mathbf{S}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ , rotation $90^\circ$ clockwise about $O$ . Work out the matrix for $\mathbf{F}$ followed by $\mathbf{S}$ , its single transformation and $A'$ , $B'$ , $C'$ .

Plan. Form $\mathbf{SF}$ , then interpret its columns and map $B$ .

\begin{aligned} \mathbf{SF} &=\begin{pmatrix}0(1)+1(0)&0(0)+1(-1)\\(-1)(1)+0(0)&(-1)(0)+0(-1)\end{pmatrix} && \text{rightmost acts first}\\ &=\begin{pmatrix}0&-1\\-1&0\end{pmatrix} && \text{evaluate}\\ A'&=(0,-1),\qquad C'=(-1,0) && \text{read the columns}\\ \mathbf{SF}\begin{pmatrix}1\\1\end{pmatrix} &=\begin{pmatrix}-1\\-1\end{pmatrix} && \text{calculate }B' \end{aligned}

Why this step? Changing the turn changes the combined matrix, but does not change which factor belongs on the right.

Answer. $\begin{pmatrix}0&-1\\-1&0\end{pmatrix}$ , a reflection in $y=-x$ , with $A'(0,-1)$ , $B'(-1,-1)$ , $C'(-1,0)$ .

Check. $C(0,1)$ first reflects to $(0,-1)$ , then turns clockwise to $(-1,0)$ , agreeing with the second column.

For three successive transformations $\mathbf{K}$ , then $\mathbf{L}$ , then $\mathbf{M}$ , the same reasoning gives $\mathbf{MLK}$ . All calculations here remain $2\times2$ products or $2\times1$ point columns.

Watch out

Wrong: “ $\mathbf{K}$ followed by $\mathbf{L}$ ” means $\mathbf{KL}$ . Right: write $\mathbf{L}(\mathbf{Kp})$ first, then remove the brackets to get $\mathbf{LKp}$ . The input shows the order.

Use a new point so that the two possible orders produce visibly different answers.

When changing the order changes the image

Two transformations generally cannot be swapped, because the second acts on a different intermediate image when the order changes.

Commuting transformations

Two matrix transformations commute when applying them in either order has exactly the same effect on every point; their matrices satisfy $\mathbf{MN}=\mathbf{NM}$ . Agreement at just one point does not prove this: every matrix here fixes $O$ , even when the transformations do not commute.

To compare orders, calculate both products and compare every entry. One point with different final images proves that two transformations are different; one shared image, especially the origin, is not enough to prove they are the same.

Worked example 3: a reflection and a rotation do not commute

Question. $\mathbf{F}=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ and $\mathbf{R}=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ . Compare $\mathbf{RF}$ with $\mathbf{FR}$ , and work out their effects on $P(2,1)$ .

Plan. Calculate the reversed product, compare it with the known product and map the same point under each.

\begin{aligned} \mathbf{RF}&=\begin{pmatrix}0&1\\1&0\end{pmatrix} && \text{reflection first}\\ \mathbf{FR} &=\begin{pmatrix}1(0)+0(1)&1(-1)+0(0)\\0(0)+(-1)(1)&0(-1)+(-1)(0)\end{pmatrix} && \text{rotation first}\\ &=\begin{pmatrix}0&-1\\-1&0\end{pmatrix} && \text{different entries}\\ \mathbf{RF}\begin{pmatrix}2\\1\end{pmatrix} &=\begin{pmatrix}1\\2\end{pmatrix} && \text{first order}\\ \mathbf{FR}\begin{pmatrix}2\\1\end{pmatrix} &=\begin{pmatrix}-1\\-2\end{pmatrix} && \text{reversed order} \end{aligned}

Why this step? Comparing the same input under both orders isolates the effect of reversing the transformations.

Answer. $\mathbf{RF}\ne\mathbf{FR}$ : the images are $(1,2)$ and $(-1,-2)$ , so these transformations do not commute.

Check. Two-step routes give $(2,1)\to(2,-1)\to(1,2)$ and $(2,1)\to(-1,2)\to(-1,-2)$ , respectively.

Worked example 4 replaces only the reflection with an enlargement of scale factor $2$ , centre $O$ . This time the two orders agree.

Worked example 4: a pair that does commute

Question. $\mathbf{E}=\begin{pmatrix}2&0\\0&2\end{pmatrix}$ , and $\mathbf{R}$ is unchanged. Compare $\mathbf{RE}$ with $\mathbf{ER}$ , and work out their effects on $P(2,1)$ .

Plan. Calculate both products rather than assuming that every pair behaves like the previous example.

\begin{aligned} \mathbf{RE} &=\begin{pmatrix}0(2)+(-1)(0)&0(0)+(-1)(2)\\1(2)+0(0)&1(0)+0(2)\end{pmatrix} && \text{enlargement first}\\ &=\begin{pmatrix}0&-2\\2&0\end{pmatrix} && \text{evaluate}\\ \mathbf{ER} &=\begin{pmatrix}2(0)+0(1)&2(-1)+0(0)\\0(0)+2(1)&0(-1)+2(0)\end{pmatrix} && \text{rotation first}\\ &=\begin{pmatrix}0&-2\\2&0\end{pmatrix} && \text{the same matrix}\\ \begin{pmatrix}0&-2\\2&0\end{pmatrix}\begin{pmatrix}2\\1\end{pmatrix} &=\begin{pmatrix}-2\\4\end{pmatrix} && \text{the common final image} \end{aligned}

Why this step? Doubling both coordinates scales both unit directions equally, so it does not matter whether that doubling happens before or after this rotation.

Answer. $\mathbf{RE}=\mathbf{ER}$ , so this pair commutes. Both orders send $P$ to $(-2,4)$ .

Check. Enlarge first: $(2,1)\to(4,2)\to(-2,4)$ . Rotate first: $(2,1)\to(-1,2)\to(-2,4)$ .

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Algebra

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Geometry

5.4 - Combining transformations with matrices

The first transformation goes on the right

When changing the order changes the image

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