Linear transformations and matrices — AP Precalculus Study Guide

For: AP Precalculus candidates sitting AP Precalculus.

Covers: Definition of linear transformations, 2x2 matrix representation of plane transformations, matrix multiplication for transformation compositions, common geometric transformations, and inverse 2x2 matrices for inverse linear transformations.

You should already know: Basic matrix addition and multiplication, 2D vector arithmetic, function composition and inverse properties.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Precalculus style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Linear transformations and matrices?

A linear transformation is a function that maps input vectors in 2D space to output vectors also in 2D space, that preserves vector addition and scalar multiplication. Every 2D linear transformation can be represented as a 2x2 matrix, where multiplying the matrix by an input vector gives the corresponding output vector. Linear transformations are also called linear maps, and they are the core link between algebraic matrix operations and geometric changes to shapes in the plane.

This subtopic falls within Unit 4: Functions Involving Parameters, Vectors, and Matrices, which makes up 25-30% of the total AP Precalculus exam score. This specific topic accounts for roughly 8-10% of total exam points, and it appears regularly in both multiple-choice (MCQ) and free-response (FRQ) sections. Exam questions often test identification of linear transformations, matrix representation, composition of geometric transformations, and inverse transformations.

2. Definition and Matrix Representation of Linear Transformations

A transformation $T$ is linear if and only if it satisfies two core properties for all vectors $\vec{u},\vec{v}$ and all scalars $c$:

1. Additivity: $T(\vec{u}+\vec{v})=T(\vec{u})+T(\vec{v})$
2. Homogeneity: $T(c\vec{v})=cT(\vec{v})$

Any transformation that fails either property is not linear. For example, translation $T(x,y)=(x+h,y+k)$ is non-linear because $T(\vec{0})=(h,k)\ne \vec{0}$, which violates homogeneity (since $T(0\cdot \vec{v})=0\cdot T(\vec{v})=\vec{0}$ must hold for linear transformations).

To find the matrix representation of a linear transformation $T$, we use the fact that any input vector $\left[\begin{array}{c}x\\ y\end{array}\right]=x\left[\begin{array}{c}1\\ 0\end{array}\right]+y\left[\begin{array}{c}0\\ 1\end{array}\right]$. By linearity: $$T\left(\left[\begin{array}{c}x\\ y\end{array}\right]\right)=xT\left(\left[\begin{array}{c}1\\ 0\end{array}\right]\right)+yT\left(\left[\begin{array}{c}0\\ 1\end{array}\right]\right)$$ This means the output of $T$ on the standard basis vectors becomes the columns of the transformation matrix. For transformation $T(x,y)=(T_x(x,y),T_y(x,y))$, the matrix is: $$A=\left[\begin{array}{cc}{T}_x(1,0)& T_x(0,1)\\ T_y(1,0)& T_y(0,1)\end{array}\right]$$

Worked Example

Find the matrix representation of the linear transformation $T(x,y)=(4x-3y,2x+y)$.

1. First, evaluate $T$ at the first standard basis vector $\left[\begin{array}{c}1\\ 0\end{array}\right]$ by substituting $x=1,y=0$: $T(1,0)=(4(1)-3(0),2(1)+0)=(4,2)$. This is the first column of the matrix.
2. Next, evaluate $T$ at the second standard basis vector $\left[\begin{array}{c}0\\ 1\end{array}\right]$ by substituting $x=0,y=1$: $T(0,1)=(4(0)-3(1),2(0)+1)=(-3,1)$. This is the second column of the matrix.
3. Assemble the matrix from the columns: $A=\left[\begin{array}{cc}4& -3\\ 2& 1\end{array}\right]$.
4. Verify by multiplying $A$ by a general input vector: $\left[\begin{array}{cc}4& -3\\ 2& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}4x-3y\\ 2x+y\end{array}\right]$, which matches the given transformation, so the matrix is correct.

Exam tip: If you need to confirm if a transformation is linear, test it on the zero vector first. If $T(\vec{0})\ne \vec{0}$, it is automatically non-linear, no further testing needed.

3. Composition of Linear Transformations and Matrix Multiplication

If you apply two linear transformations in sequence — first $T_1$, then $T_2$ — the combined transformation $T_2\circ T_1=T_2(T_1(\vec{v}))$ is also linear. If $T_1$ has matrix $A_1$ and $T_2$ has matrix $A_2$, the matrix for the combined transformation is the product $A_2{A}_1$. This comes from the associative property of matrix multiplication: $T_2(T_1(\vec{v}))=A_2(A_1\vec{v})=(A_2{A}_1)\vec{v}$.

Order is critical here: matrix multiplication is not commutative, so $A_2{A}_1\ne A_1{A}_2$ in most cases. The first transformation applied always goes on the right side of the product, and the second transformation on the left. For two 2x2 matrices $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ and $B=\left[\begin{array}{cc}e& f\\ g& h\end{array}\right]$, the product $BA$ is calculated by taking the dot product of each row of $B$ with each column of $A$: $$BA=\left[\begin{array}{cc}ea+fc& eb+fd\\ ga+hc& gb+hd\end{array}\right]$$

Worked Example

Let $T_1$ be reflection over the y-axis (matrix $A=\left[\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right]$) and $T_2$ be dilation by a factor of 3 (matrix $B=\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right]$). Find the matrix for the transformation that first reflects over the y-axis, then dilates by 3.

1. Confirm the order of operations: we apply $T_1$ first, then $T_2$, so the composition is $T_2\circ T_1$, which means the matrix product is $BA$, not $AB$.
2. Calculate the product step-by-step: first entry (row 1 of B, column 1 of A) = $(3)(-1)+(0)(0)=-3$. Second entry (row 1 of B, column 2 of A) = $(3)(0)+(0)(1)=0$.
3. Third entry (row 2 of B, column 1 of A) = $(0)(-1)+(3)(0)=0$. Fourth entry (row 2 of B, column 2 of A) = $(0)(0)+(3)(1)=3$.
4. The product matrix is $BA=\left[\begin{array}{cc}-3& 0\\ 0& 3\end{array}\right]$. Verify with input $\vec{v}=\left[\begin{array}{c}2\\ 4\end{array}\right]$: reflect first to get $\left[\begin{array}{c}-2\\ 4\end{array}\right]$, dilate to get $\left[\begin{array}{c}-6\\ 12\end{array}\right]$, which matches $BA\vec{v}=\left[\begin{array}{cc}-3& 0\\ 0& 3\end{array}\right]\left[\begin{array}{c}2\\ 4\end{array}\right]=\left[\begin{array}{c}-6\\ 12\end{array}\right]$.

Exam tip: Always write your order explicitly before multiplying: "First transformation = [name], matrix = [A], second = [name], matrix = [B] = product $BA$". This eliminates almost all order errors on the exam.

4. Inverse Linear Transformations and Determinants

A linear transformation is invertible if it is one-to-one and onto, meaning every output vector comes from exactly one input vector. For a 2x2 transformation matrix $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$, invertibility is determined by the determinant $\det (A)=ad-bc$. If $\det (A)\ne 0$, $A$ is invertible; if $\det (A)=0$, $A$ is not invertible (it collapses the plane to a line or point, so the transformation cannot be reversed).

The inverse matrix $A^{-1}$ represents the inverse transformation $T^{-1}$ that undoes $T$, satisfying $A{A}^{-1}=A^{-1}A=I$, where $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ is the identity matrix (the transformation that leaves all vectors unchanged). The formula for the inverse of a 2x2 matrix is: $$A^{-1}=\frac{1}{\det (A)}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$$

Common geometric transformations have easily interpreted inverses: the inverse of rotation by $\theta$ counterclockwise is rotation by $\theta$ clockwise, the inverse of dilation by factor $k$ is dilation by factor $1/k$, and the inverse of any reflection is the reflection itself (reflecting twice returns you to the original position).

Worked Example

Find the inverse of the transformation matrix $A=\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]$, then verify that $A{A}^{-1}=I$.

1. First calculate the determinant: $\det (A)=(2)(4)-(1)(3)=8-3=5$, which is non-zero, so the inverse exists.
2. Apply the inverse formula: swap the diagonal entries ($a=2$ and $d=4$), flip the sign of the off-diagonal entries ($b=1\to -1$, $c=3\to -3$), then divide by the determinant: $A^{-1}=\frac{1}{5}\left[\begin{array}{cc}4& -1\\ -3& 2\end{array}\right]=\left[\begin{array}{cc}\frac{4}{5}& -\frac{1}{5}\\ -\frac{3}{5}& \frac{2}{5}\end{array}\right]$.
3. Verify the product $A{A}^{-1}$: $$\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]\left[\begin{array}{cc}\frac{4}{5}& -\frac{1}{5}\\ -\frac{3}{5}& \frac{2}{5}\end{array}\right]=\left[\begin{array}{cc}\frac{8-3}{5}& \frac{-2+2}{5}\\ \frac{12-12}{5}& \frac{-3+8}{5}\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I$$ This confirms the inverse is correct.

Exam tip: Memorize the standard geometric transformation matrices (rotation, reflection, dilation) to save time on the exam. They are tested frequently, and memorization avoids re-deriving them under time pressure.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Swapping the order of matrix multiplication for composition, writing $A_1{A}_2$ for a transformation that applies $T_1$ first then $T_2$. Why: Students confuse function notation: $f\circ g=f(g(x))$, so the first function applied is the inner function, which ends up on the right in matrix form. Correct move: Always label the order explicitly: "First $T_1$, then $T_2$ = $T_2(T_1(\vec{v}))=A_2{A}_1\vec{v}$, so matrix product is $A_2{A}_1$".
• Wrong move: Forgetting to divide the inverse matrix by the determinant when calculating $A^{-1}$. Why: Students remember to swap $a$ and $d$ and flip signs of $b$ and $c$, but omit the scaling factor of $1/\det (A)$. Correct move: After swapping and changing signs, always write the $1/(ad-bc)$ factor outside the matrix before simplifying entries.
• Wrong move: Calling translation a linear transformation. Why: Translations are affine transformations, which look linear, but they do not satisfy the linearity properties. Correct move: Always check $T(\vec{0})$: if $T(\vec{0})\ne \vec{0}$, mark the transformation as non-linear immediately.
• Wrong move: Placing the outputs of the standard basis vectors as rows instead of columns in the transformation matrix. Why: Students mix up the standard convention for matrix representation. Correct move: Always remember: output of $(1,0)$ is the first column, output of $(0,1)$ is the second column. Test with a general vector to confirm if unsure.
• Wrong move: Using the wrong sign for $\sin \theta$ in the rotation matrix. Why: Students mix up the direction of rotation. Correct move: The standard matrix is defined for counterclockwise $\theta$; for clockwise rotation, substitute $-\theta$, which flips the sign of the two $\sin$ entries.

6. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

Which of the following transformations is a linear transformation? A) $T(x,y)=(3x,y^2)$ B) $T(x,y)=(2x+5y,x-4y)$ C) $T(x,y)=(x+2,y-3)$ D) $T(x,y)=(xy,3x)$

Worked Solution: First apply the zero vector test to eliminate obviously non-linear transformations: for option C, $T(0,0)=(2,-3)\ne (0,0)$, so C is eliminated immediately. Next, test homogeneity for the remaining options: for A, $T(2(1,1))=T(2,2)=(6,4)$, while $2T(1,1)=2(3,1)=(6,2)\ne (6,4)$, so A is wrong. For D, test additivity: $T((1,0)+(0,1))=T(1,1)=(1,3)$, while $T(1,0)+T(0,1)=(0,3)+(0,0)=(0,3)\ne (1,3)$, so D is wrong. Option B satisfies both additivity and homogeneity, so it is linear. Correct answer: B

Question 2 (Free Response)

Let $A=\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]$ be the matrix for linear transformation $T_A$, and $B=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]$ be the matrix for $T_B$. (a) Find the matrix for the composition $S=T_B\circ T_A$, which applies $T_A$ first, then $T_B$. (b) Calculate the determinant of $S$, and state whether $S$ is invertible. (c) Find the matrix for $S^{-1}$, the inverse of $S$.

Worked Solution: (a) The composition $T_B\circ T_A$ has matrix $BA$. Calculate the product: $$BA=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}(1)(2)+(0)(1)& (1)(1)+(0)(1)\\ (0)(2)+(-1)(1)& (0)(1)+(-1)(1)\end{array}\right]=\left[\begin{array}{cc}2& 1\\ -1& -1\end{array}\right]$$

(b) Calculate the determinant of $S=\left[\begin{array}{cc}2& 1\\ -1& -1\end{array}\right]$: $$\det (S)=(2)(-1)-(1)(-1)=-2+1=-1$$ Since $\det (S)=-1\ne 0$, $S$ is invertible.

(c) Apply the 2x2 inverse formula: swap diagonal entries, flip off-diagonal signs, divide by determinant: $$S^{-1}=\frac{1}{-1}\left[\begin{array}{cc}-1& -1\\ 1& 2\end{array}\right]=\left[\begin{array}{cc}1& 1\\ -1& -2\end{array}\right]$$ Verification confirms $S{S}^{-1}=I$, so this is correct.

Question 3 (Application / Real-World Style)

A graphic designer is editing a logo centered at the origin of a 2D coordinate plane. The designer first rotates the logo 90 degrees counterclockwise, then dilates it by a factor of 2 to make it larger. A key corner of the original logo is at coordinates $(3,1)$. What are the coordinates of this corner after both transformations? Interpret your result in context.

Worked Solution: First, write the matrices for each transformation: rotation 90° counterclockwise is $A=\left[\begin{array}{cc}\cos 90^{\circ }& -\sin 90^{\circ }\\ \sin 90^{\circ }& \cos 90^{\circ }\end{array}\right]=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$, dilation by factor 2 is $B=\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]$. The order of transformations is rotation first, then dilation, so the combined matrix is $BA$. The input vector is $\vec{v}=\left[\begin{array}{c}3\\ 1\end{array}\right]$. Calculate the output: $$BA\vec{v}=\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]\left[\begin{array}{c}3\\ 1\end{array}\right]=\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]\left[\begin{array}{c}-1\\ 3\end{array}\right]=\left[\begin{array}{c}-2\\ 6\end{array}\right]$$ Interpretation: After the two transformations, the corner of the logo originally at (3, 1) is located at (-2, 6) on the graphic designer's coordinate plane.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundation for the remaining content of Unit 4, including using matrices to solve systems of linear equations and model multi-variable contextual relationships. Next, you will apply linear transformation and matrix conventions to solve linear systems and model real-world processes with multiple input variables. Without mastering the order rules, matrix construction, and inverse formulas covered here, you will constantly make avoidable sign and order errors when working with matrix systems. This topic also provides the foundational knowledge for college-level linear algebra, computer graphics, and data science, where linear transformations are core analytical tools.

• Matrix methods for linear systems
• Vectors in two-dimensional space
• Parametric equations and planar motion