How do I find a missing matrix when M equals B A?

Undo the known matrix with its inverse on the correct side: B equals M times A inverse if A is known, or A equals B inverse times M if B is known.

How do I find a missing matrix when M equals B A?

Undo the known matrix with its inverse on the correct side: B equals M times A inverse if A is known, or A equals B inverse times M if B is known.

Year 11 Specialist (Unit 1 & 2) Transformations of the plane

Composition of transformations

Q: If I apply A then B, is the single matrix AB or BA?

It is B A. The transformation applied second is written on the left, because the image is B of A v, which equals B A times v.

Q: Why does the order of transformations matter?

Matrix multiplication is not commutative, so in general B A is not equal to A B. Applying the transformations in a different order usually maps a point to a different place.

Q: How do I compose three transformations?

For A then B then C the single matrix is C B A: each new transformation multiplies on the left. Evaluate the product from the right.

Q: How does a composition change area?

Area is multiplied by the absolute value of the determinant of B A. Because determinants multiply, this equals the absolute value of det B times the absolute value of det A, the product of the individual area factors.

20 practice questions 0 video lessons Theory + worked examples

Learn composition of transformations for Year 11 Specialist Mathematics in Queensland (QCAA). When you apply one linear transformation and then another, a single matrix does the whole job — and it is the matrix product of the two, taken in the right order.

You will learn why applying a transformation and then a second one has single matrix the second times the first, how to find the image of a point or polygon, how to compose three transformations, and how determinants give the area factor — with a clear focus on why order matters. This links transformations to the matrix algebra used later for inverses and determinants.

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Practice questions

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Theory

Composition of transformations combines two or more linear transformations into one. In Year 11 Specialist Mathematics (QCAA, Queensland), if you apply the transformation with matrix \(A\) and then the one with matrix \(B\), the single matrix for the whole sequence is the product \(BA\) — the second matrix on the left. Order matters: in general \(BA\ne AB\).

A linear transformation of the plane (a dilation, a rotation about the origin, or a reflection in a line through the origin) is represented by a \(2\times2\) matrix. It sends a point \(\mathbf{v}\) to the point \(A\mathbf{v}\).

A composition applies one transformation and then another. If you first apply \(A\) and then apply \(B\), the image is \(B(A\mathbf{v})\). Because matrix multiplication is associative, \(B(A\mathbf{v})=(BA)\mathbf{v}\), so the single matrix for “\(A\) then \(B\)” is the product \(BA\).

The order rule is the key idea: the second transformation is written on the left. Reading “\(A\) then \(B\)” left to right and writing \(AB\) is the classic mistake — it usually gives a different result.

Composition is generally not commutative: \(BA\ne AB\). A useful special case is that rotations about the same centre commute, because their angles simply add.

Apply \(A\) then \(B\): the image is \(B(A\mathbf{v})=(BA)\mathbf{v}\), so the single matrix is \(BA\).

Reflect triangle \(T\) in the \(x\)-axis, then rotate \(90^\circ\) anticlockwise: the image \(T'\) is a reflection in \(y=x\).

Applying \(A\) first and then \(B\) has the single matrix

\[ M = BA \]

M = B A

The image of a point \(\mathbf{v}=\begin{pmatrix}x\\y\end{pmatrix}\) under the composition is

\[ \mathbf{v}\;\longrightarrow\; B(A\mathbf{v}) = (BA)\mathbf{v} \]

B (A v) = (B A) v

A composition of three steps “\(A\) then \(B\) then \(C\)” has single matrix \(CBA\) (each new matrix multiplies on the left):

\[ M = C\,B\,A \]

M = C B A

Area scales by the absolute value of the determinant, and determinants multiply, so a composite scales area by

\[ |\det(BA)| = |\det B|\,|\det A| \]

| \det (B A) | = | \det B | | \det A |

Order matters. “\(A\) then \(B\)” is \(BA\), not \(AB\). In general \(BA\ne AB\); the exception you can rely on is that rotations about the same centre commute (their angles add).

How to compose transformations

Write each matrix: read off the \(2\times2\) matrix for every transformation in the sequence (rotation, reflection or dilation).
Order right to left: for “\(A\) then \(B\)” form \(M=BA\) — put the last transformation on the left. For three steps, \(M=CBA\).
Multiply: evaluate the product with the row-by-column rule to get the single matrix \(M\).
Apply or interpret: use \(M\mathbf{v}\) to find an image point, \(|\det M|\) for the area factor, or recognise \(M\) as a single standard transformation.

Example 1 — Single matrix for “A then B”

Transformation \(A\) is a reflection in the \(y\)-axis and transformation \(B\) is a rotation of \(90^\circ\) anticlockwise about \(O\). Find the single matrix for \(A\) then \(B\).

Solution

Write each matrix, then form \(M=BA\) (the second transformation, \(B\), on the left):

\(A\)	\(=\)	\(\begin{pmatrix}-1&0\\0&1\end{pmatrix}, \quad B=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\)
\(M\)	\(=\)	\(BA\)
	\(=\)	\(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}-1&0\\0&1\end{pmatrix}\)

Multiply row-by-column:

\(M\)	\(=\)	\(\begin{pmatrix}0\times(-1)+(-1)\times0 & 0\times0+(-1)\times1\\ 1\times(-1)+0\times0 & 1\times0+0\times1\end{pmatrix}\)
	\(=\)	\(\begin{pmatrix}0&-1\\-1&0\end{pmatrix}\)

\(M=\begin{pmatrix}0&-1\\-1&0\end{pmatrix}\), a reflection in the line \(y=-x\).

Example 2 — Image of a point, and why order matters

The point \(P(3,2)\) is rotated \(90^\circ\) anticlockwise about \(O\) (transformation \(A\)) and then reflected in the \(x\)-axis (transformation \(B\)). Find the image of \(P\), and check the reverse order gives a different result.

Solution

Form \(M=BA\) with \(B=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\) and \(A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\):

\(M\)	\(=\)	\(BA=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}0&-1\\1&0\end{pmatrix}\)
	\(=\)	\(\begin{pmatrix}0&-1\\-1&0\end{pmatrix}\)

Apply \(M\) to the column vector of \(P\):

\(M\begin{pmatrix}3\\2\end{pmatrix}\)	\(=\)	\(\begin{pmatrix}0&-1\\-1&0\end{pmatrix}\begin{pmatrix}3\\2\end{pmatrix}\)
	\(=\)	\(\begin{pmatrix}0\times3+(-1)\times2\\ -1\times3+0\times2\end{pmatrix}\)
	\(=\)	\(\begin{pmatrix}-2\\-3\end{pmatrix}\)

Reverse the order (\(B\) then \(A\)) to see it differs:

\(AB\)	\(=\)	\(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\)
	\(=\)	\(\begin{pmatrix}0&1\\1&0\end{pmatrix}\ne BA\)

The image is \((-2,-3)\); since \(BA\ne AB\), the order of the two transformations matters.

Example 3 — Three transformations in sequence

A shape is transformed in order: first a reflection in \(y=x\) (\(A\)), then an enlargement of factor \(2\) about \(O\) (\(B\)), then a rotation of \(90^\circ\) anticlockwise about \(O\) (\(C\)). Find the single matrix for the whole sequence.

Solution

Three steps “\(A\) then \(B\) then \(C\)” give \(M=CBA\). Multiply the right-hand pair \(BA\) first:

\(BA\)	\(=\)	\(\begin{pmatrix}2&0\\0&2\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}\)
	\(=\)	\(\begin{pmatrix}0&2\\2&0\end{pmatrix}\)

Now multiply on the left by \(C=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\):

\(M\)	\(=\)	\(C(BA)=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}0&2\\2&0\end{pmatrix}\)
	\(=\)	\(\begin{pmatrix}0\times0+(-1)\times2 & 0\times2+(-1)\times0\\ 1\times0+0\times2 & 1\times2+0\times0\end{pmatrix}\)
	\(=\)	\(\begin{pmatrix}-2&0\\0&2\end{pmatrix}\)

\(M=\begin{pmatrix}-2&0\\0&2\end{pmatrix}\).

Example 4 — Area under a composite (determinants multiply)

A logo of area \(5\) square cm is transformed by a dilation \((x,y)\to(3x,2y)\) (transformation \(A\)) and then a rotation of \(90^\circ\) anticlockwise about \(O\) (transformation \(B\)). Find the area of the image.

Solution

Area scales by \(|\det(BA)|\), and determinants multiply, so find each determinant:

\(\det A\)	\(=\)	\(\det\begin{pmatrix}3&0\\0&2\end{pmatrix}=3\times2-0\times0=6\)
\(\det B\)	\(=\)	\(\det\begin{pmatrix}0&-1\\1&0\end{pmatrix}=0\times0-(-1)\times1=1\)

Combine, then scale the original area:

\(\|\det(BA)\|\)	\(=\)	\(\|\det B\|\,\|\det A\|\)
	\(=\)	\(1\times6=6\)
\(\text{image area}\)	\(=\)	\(6\times5\)
	\(=\)	\(30\)

The image has area \(30\) square cm.

Common pitfalls

Writing \(AB\) for “\(A\) then \(B\)”. Watch the order: the transformation applied second goes on the left, so “\(A\) then \(B\)” is \(BA\). Multiplying in reading order usually gives the wrong matrix.

Assuming the order does not matter. In general \(BA\ne AB\). Reversing the two transformations normally changes the image — do not swap them to make the arithmetic easier.

Multiplying determinants but forgetting the modulus. The area factor is \(|\det(BA)|\). A determinant can be negative (an orientation flip), but area is always scaled by its absolute value.

Solving \(M=BA\) for a missing matrix incorrectly. To find \(B\) from \(M=BA\), right-multiply by \(A^{-1}\): \(B=MA^{-1}\). To find \(A\), left-multiply: \(A=B^{-1}M\). Keep the inverse on the correct side.

Frequently asked questions

If I apply A then B, is the single matrix AB or BA?

It is \(BA\). The transformation applied second is written on the left, because the image is \(B(A\mathbf{v})=(BA)\mathbf{v}\).

Why does the order of transformations matter?

Matrix multiplication is not commutative, so in general \(BA\ne AB\). Applying the transformations in a different order usually maps a point to a different place.

Do any transformations commute?

Yes. Rotations about the same centre commute, because composing them just adds the angles. Most other pairs, such as a reflection and a rotation, do not commute.

How do I compose three transformations?

For “\(A\) then \(B\) then \(C\)” the single matrix is \(CBA\): each new transformation multiplies on the left. Evaluate the product from the right.

How does a composition change area?

How do I find a missing matrix in \(M=BA\)?

Undo the known matrix with its inverse on the correct side: \(B=MA^{-1}\) if \(A\) is known, or \(A=B^{-1}M\) if \(B\) is known.

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Inverse transformations