Unit 4 · Topic 4.13 · about 30 minutes · 7 parts · 14 scored checks, 2 of them multiple choice

AP Precalculus 4.13: Matrices as Functions

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Not assessed on the AP Exam. Unit 4 is part of the course, but nothing in it is on the exam.

The unit square, with sides <1, 0> and <0, 1>, drawn dashed at the origin. The matrix A with rows 2, -1 and 1, 1 sends <1, 0> to <2, 1>, its first column, and <0, 1> to <-1, 1>, its second column. Those two vectors are the sides of a shaded parallelogram with corners (0, 0), (2, 1), (1, 2) and (-1, 1). The square has area 1 and the parallelogram has area 3, which is det(A) = 2 times 1 minus (-1) times 1.

Part 1 Think first Which button first?

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

A photo app has two buttons. Rotate turns the picture a quarter turn counterclockwise. Stretch doubles every width and leaves heights alone. Mia and Jon each start with the same photo, 4 inches wide and 3 inches tall. Mia presses Rotate, then Stretch. Jon presses Stretch, then Rotate. How do their pictures compare?

In short. A linear transformation $L (\vec{v}) = A \vec{v}$ is a function on vectors in $ℝ^{2}$ , and its $2 \times 2$ matrix $A$ records where it sends $⟨ 1, 0 ⟩$ and $⟨ 0, 1 ⟩$ .

The images of $⟨ 1, 0 ⟩$ and $⟨ 0, 1 ⟩$ are the first and second columns of $A$ , and rotating by $θ$ counterclockwise about the origin has the matrix $[\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}]$ .
The transformation multiplies the area of every region by $| \det (A) |$ .
Applying $L (\vec{v}) = A \vec{v}$ and then $M (\vec{v}) = B \vec{v}$ is the linear transformation with matrix $B A$ , and when $A^{- 1}$ exists, $L^{- 1} (\vec{v}) = A^{- 1} \vec{v}$ undoes $L$ .

Part 2 Learn it Matrices that move the plane

2a. Read the matrix off the unit vectors

You have used a matrix $A$ to define a transformation $L (\vec{v}) = A \vec{v}$ . Now go the other way. The linear transformation that maps $⟨ x, y ⟩$ to $⟨ a_{11} x + a_{12} y, a_{21} x + a_{22} y ⟩$ is associated with the matrix

[\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}]

The coefficients of the first output component make row 1, and those of the second make row 2. For $L (⟨ x, y ⟩) = ⟨ 4 x - 3 y, x + 2 y ⟩$ the matrix is $[\begin{matrix} 4 & - 3 \\ 1 & 2 \end{matrix}]$ .

Now put in the unit vectors. $L (⟨ 1, 0 ⟩) = ⟨ 4, 1 ⟩$ , which is the first column, and $L (⟨ 0, 1 ⟩) = ⟨ - 3, 2 ⟩$ , the second column. That always happens, so knowing where $⟨ 1, 0 ⟩$ and $⟨ 0, 1 ⟩$ land tells you the whole matrix. It tells you every output too: $⟨ 5, 2 ⟩ = 5 ⟨ 1, 0 ⟩ + 2 ⟨ 0, 1 ⟩$ , so $L (⟨ 5, 2 ⟩) = 5 ⟨ 4, 1 ⟩ + 2 ⟨ - 3, 2 ⟩ = ⟨ 14, 9 ⟩$ . The formula agrees: $4 (5) - 3 (2) = 14$ and $5 + 2 (2) = 9$ .

Written as rows, the images give a different matrix and a different transformation.

Quick check · Fill the statement

A linear transformation is given by $L (⟨ x, y ⟩) = ⟨ 6 x - y, 2 x + 5 y ⟩$ . Tap a blank, then tap an answer.

$L (⟨ 1, 0 ⟩)$ = and $L (⟨ 0, 1 ⟩)$ = . In the matrix associated with $L$ , these two vectors are the .

2b. Rotations, and what happens to area

Rotate every vector by an angle $θ$ counterclockwise about the origin. The unit vector $⟨ 1, 0 ⟩$ lands on the unit circle at $⟨ \cos θ, \sin θ ⟩$ . The vector $⟨ 0, 1 ⟩$ starts a quarter turn further round, so it lands at $⟨ \cos (θ + \frac{π}{2}), \sin (θ + \frac{π}{2}) ⟩ = ⟨ - \sin θ, \cos θ ⟩$ . Those images are the columns:

[\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}]

For $θ = \frac{π}{2}$ that is $[\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}]$ , and it sends $⟨ 3, 1 ⟩$ to $⟨ - 1, 3 ⟩$ , a quarter turn.

Now area. The unit square has sides $⟨ 1, 0 ⟩$ and $⟨ 0, 1 ⟩$ and area 1. Under $A$ it becomes the parallelogram spanned by the columns of $A$ , whose area is $| \det (A) |$ . Every region is dilated by that same factor: under a matrix with determinant $- 3$ , a shape of area 5 becomes a shape of area 15. The sign does not shrink anything; the factor is the absolute value.

A rotation has determinant $\cos^{2} θ + \sin^{2} θ = 1$ , so it moves a region without changing its area. The matrix $[\begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix}]$ triples every length, and its determinant is 9, so areas grow by a factor of 9, not 3.

Quick check · Match

Each matrix is associated with a rotation about the origin, with angles in radians. Tap a matrix, then the counterclockwise angle it rotates every vector through.

2c. One transformation after another

Apply $L (\vec{v}) = A \vec{v}$ first, then $M (\vec{v}) = B \vec{v}$ . A vector $\vec{v}$ goes to $A \vec{v}$ and then to $B (A \vec{v})$ , and that is the same as $(B A) \vec{v}$ . So the composition is itself a linear transformation, and its matrix is the product $B A$ . The transformation applied first sits on the right, next to $\vec{v}$ .

The photo app shows why the order matters. Rotate is $A = [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}]$ and Stretch is $B = [\begin{matrix} 2 & 0 \\ 0 & 1 \end{matrix}]$ . Mia's Stretch after Rotate has matrix

B A = [\begin{matrix} 0 & - 2 \\ 1 & 0 \end{matrix}]

Jon's Rotate after Stretch has matrix $A B = [\begin{matrix} 0 & - 1 \\ 2 & 0 \end{matrix}]$ . Test $⟨ 1, 0 ⟩$ : Mia's app turns it to $⟨ 0, 1 ⟩$ , and stretching leaves that vertical vector alone, so it ends at $⟨ 0, 1 ⟩$ , the first column of $B A$ . Jon's stretches it to $⟨ 2, 0 ⟩$ first and then turns it to $⟨ 0, 2 ⟩$ , the first column of $A B$ . Different products, different pictures.

Quick check · Calculate

Let $L (\vec{v}) = A \vec{v}$ with $A = [\begin{matrix} 2 & - 1 \\ 1 & 3 \end{matrix}]$ and $M (\vec{v}) = B \vec{v}$ with $B = [\begin{matrix} 1 & 3 \\ 0 & 2 \end{matrix}]$ . The transformation that applies $L$ first and then $M$ is linear. What is the component in row 1, column 2 of its matrix?

Your answer

2d. Undoing a transformation

Two linear transformations are inverses if their composition maps any vector to itself. If $L (\vec{v}) = A \vec{v}$ and $A$ has an inverse, then

L^{- 1} (\vec{v}) = A^{- 1} \vec{v}

because $A^{- 1} (A \vec{v}) = I \vec{v} = \vec{v}$ for every $\vec{v}$ .

Take $L (⟨ x, y ⟩) = ⟨ x + 2 y, y ⟩$ , with $A = [\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}]$ . Its determinant is 1, so $A^{- 1} = [\begin{matrix} 1 & - 2 \\ 0 & 1 \end{matrix}]$ and $L^{- 1} (⟨ x, y ⟩) = ⟨ x - 2 y, y ⟩$ . Check it on a general vector: $L^{- 1} (L (⟨ x, y ⟩)) = ⟨ (x + 2 y) - 2 y, y ⟩ = ⟨ x, y ⟩$ . Checking one vector is not enough; the composition has to return every vector.

A rotation by $θ$ is undone by a rotation by $- θ$ , the same angle clockwise. And a matrix with determinant 0 flattens the plane onto a line or a point, so different vectors land in the same place and nothing can send them back. That is the transformation version of having no inverse matrix.

Quick check · Spot the error

A student finds the inverse of the transformation $L$ that rotates every vector $\frac{π}{6}$ radians counterclockwise about the origin. Tap the first line that goes wrong.

Part 3 Watch it done Find an inverse transformation and use it

Watch each step, then do the last one yourself.

A linear transformation is $L (⟨ x, y ⟩) = ⟨ 4 x + 3 y, 3 x + 2 y ⟩$ . Find a formula for $L^{- 1}$ , then find the vector that $L$ sends to $⟨ 23, 16 ⟩$ .

Step 1 · Write the matrix
The coefficients of each output component make a row: $A = [\begin{matrix} 4 & 3 \\ 3 & 2 \end{matrix}]$ .
Step 2 · Find the determinant and the inverse
$\det (A) = 4 \cdot 2 - 3 \cdot 3 = - 1$ , which is not 0. Swap 4 and 2, change the signs of 3 and 3, and divide by $- 1$ : $A^{- 1} = \frac{1}{- 1} [\begin{matrix} 2 & - 3 \\ - 3 & 4 \end{matrix}] = [\begin{matrix} - 2 & 3 \\ 3 & - 4 \end{matrix}]$ .
Step 3 · Write the inverse transformation
$L^{- 1} (\vec{v}) = A^{- 1} \vec{v}$ , so $L^{- 1} (⟨ x, y ⟩) = ⟨ - 2 x + 3 y, 3 x - 4 y ⟩$ .
Why this step?
Why does multiplying by $A^{- 1}$ undo $L$ ?
For any vector, $A^{- 1} (A \vec{v}) = I \vec{v} = \vec{v}$ . Applying $L$ and then $L^{- 1}$ maps every vector to itself, which is what makes two transformations inverses. So whatever $L$ produced, $L^{- 1}$ returns the vector it came from.
Step 4 · Apply it to the output
The input is $L^{- 1} (⟨ 23, 16 ⟩)$ . Its first component is $- 2 (23) + 3 (16) = 2$ .
Your step · Calculate

Finish it. For $L (⟨ x, y ⟩) = ⟨ 4 x + 3 y, 3 x + 2 y ⟩$ , the inverse is $L^{- 1} (⟨ x, y ⟩) = ⟨ - 2 x + 3 y, 3 x - 4 y ⟩$ . What is the second component of the vector that $L$ sends to $⟨ 23, 16 ⟩$ ?

Your answer

So $L$ sends $⟨ 2, 5 ⟩$ to $⟨ 23, 16 ⟩$ . Check it forward: $4 (2) + 3 (5) = 23$ and $3 (2) + 2 (5) = 16$ .

Part 4 Try it Two moves in a drawing app

Warm up in the lab. It is set to the composition $B$ after $A$ with $A = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}]$ and $B = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}]$ , and it draws only the unit square, $\vec{i}$ , $\vec{j}$ and $\vec{v}$ . The questions after it use different matrices.

Lab

Vectors and Matrices

On paper, find where $\vec{i}$ and $\vec{j}$ land under $A$ , then where those images land under $B$ . Those two final vectors should be the columns of $B A$ , so multiply $B A$ out and compare. Then drag the head of $\vec{v}$ to a new point and find $B (A \vec{v})$ both ways.

Open the full Vectors and Matrices lab

Now the app. A designer applies $S (⟨ x, y ⟩) = ⟨ x + 2 y, y ⟩$ , which slides each point sideways by twice its height, and then $R$ , a rotation of $\frac{3 π}{2}$ radians counterclockwise about the origin, which is a quarter turn clockwise. Work on paper.

Try it · Calculate

A drawing app applies $S (⟨ x, y ⟩) = ⟨ x + 2 y, y ⟩$ first and then $R$ , the rotation of $\frac{3 π}{2}$ radians counterclockwise about the origin. What is the second component of the image of $⟨ 2, 3 ⟩$ ?

Your answer

Try it · Multiple choice

A drawing app applies $S (⟨ x, y ⟩) = ⟨ x + 2 y, y ⟩$ and then $R$ , the rotation of $\frac{3 π}{2}$ radians counterclockwise, so the combined move has matrix $[\begin{matrix} 0 & 1 \\ - 1 & - 2 \end{matrix}]$ . Which matrix is associated with the transformation that undoes the combined move?

Part 5 Explain it Put it in writing

Nothing you type leaves this page. Check your answer against the model and each point on the list.

Unit 4 is not on the AP Exam, so this is written practice for the course.

Written practice

Let $T$ be the linear transformation that rotates every vector $\frac{3 π}{4}$ radians counterclockwise about the origin, with $T (\vec{v}) = A \vec{v}$ .

(i) Determine $T (⟨ 1, 0 ⟩)$ and $T (⟨ 0, 1 ⟩)$ , and use them to determine $A$ with exact values.

(ii) Determine $\det (A)$ and explain what it tells you about the area of a region after $T$ is applied.

(iii) Determine $A^{- 1}$ . Describe the transformation $T^{- 1} (\vec{v}) = A^{- 1} \vec{v}$ as a rotation, and justify that $T^{- 1} (T (\vec{v})) = \vec{v}$ for every vector $\vec{v}$ .

Your answer stays on this page. It is never sent or saved.

Model answer

(i) $⟨ 1, 0 ⟩$ lands on the unit circle at angle $\frac{3 π}{4}$ , so $T (⟨ 1, 0 ⟩) = ⟨ - \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} ⟩$ . $⟨ 0, 1 ⟩$ starts a quarter turn further on and lands at angle $\frac{5 π}{4}$ , so $T (⟨ 0, 1 ⟩) = ⟨ - \frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2} ⟩$ . These are the columns: $A = [\begin{matrix} - \frac{\sqrt{2}}{2} & - \frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & - \frac{\sqrt{2}}{2} \end{matrix}]$ . (ii) $\det (A) = (- \frac{\sqrt{2}}{2}) (- \frac{\sqrt{2}}{2}) - (- \frac{\sqrt{2}}{2}) (\frac{\sqrt{2}}{2}) = \frac{1}{2} + \frac{1}{2} = 1$ . The area of every region is multiplied by $| \det (A) | = 1$ , so $T$ moves regions without changing their area. (iii) Since $\det (A) = 1$ , swapping the diagonal components and changing the signs of the other two gives $A^{- 1} = [\begin{matrix} - \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ - \frac{\sqrt{2}}{2} & - \frac{\sqrt{2}}{2} \end{matrix}]$ . Its first column is $⟨ \cos (- \frac{3 π}{4}), \sin (- \frac{3 π}{4}) ⟩$ , so $T^{- 1}$ rotates every vector $\frac{3 π}{4}$ radians clockwise. In $A^{- 1} A$ , row 1 times column 1 is $\frac{1}{2} + \frac{1}{2} = 1$ , row 1 times column 2 is $\frac{1}{2} - \frac{1}{2} = 0$ , row 2 times column 1 is $\frac{1}{2} - \frac{1}{2} = 0$ , and row 2 times column 2 is $\frac{1}{2} + \frac{1}{2} = 1$ . So $A^{- 1} A = I$ , and $T^{- 1} (T (\vec{v})) = A^{- 1} A \vec{v} = I \vec{v} = \vec{v}$ for every $\vec{v}$ .

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Finds the images $⟨ - \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} ⟩$ and $⟨ - \frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2} ⟩$ and places them as the columns of $A$ .
Finds $\det (A) = 1$ and explains that areas are multiplied by $| \det (A) | = 1$ , so they do not change.
Gives $A^{- 1} = [\begin{matrix} - \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ - \frac{\sqrt{2}}{2} & - \frac{\sqrt{2}}{2} \end{matrix}]$ and describes it as a rotation of $\frac{3 π}{4}$ clockwise, an angle of $- \frac{3 π}{4}$ .
Shows $A^{- 1} A = I$ and concludes that the composition maps every vector to itself.

Part 6 Check yourself Three moves in a sign shop

The three case questions use the same table of transformations. Answer them from its formulas.

A sign shop designs logos on a coordinate grid measured in inches. Its software has three linear transformations, given in the table.

Transformations F, G and H in the sign software
Name	Formula
$F$	$F (⟨ x, y ⟩) = ⟨ 3 x + 2 y, x + 4 y ⟩$
$G$	$G (⟨ x, y ⟩) = ⟨ - y, x ⟩$
$H$	$H (⟨ x, y ⟩) = ⟨ 0.4 x - 0.2 y, - 0.1 x + 0.3 y ⟩$

Case question 1 · Calculate

A logo with an area of 6 square inches is transformed by $F (⟨ x, y ⟩) = ⟨ 3 x + 2 y, x + 4 y ⟩$ . What is the area of the new logo, in square inches?

Your answer

Case question 2 · Multiple choice

In sign software, $F (⟨ x, y ⟩) = ⟨ 3 x + 2 y, x + 4 y ⟩$ and $G (⟨ x, y ⟩) = ⟨ - y, x ⟩$ . A designer applies $F$ to a logo and then applies $G$ . Which matrix is associated with the combined transformation?

Case question 3 · Decide and justify

In sign software, $F (⟨ x, y ⟩) = ⟨ 3 x + 2 y, x + 4 y ⟩$ and $H (⟨ x, y ⟩) = ⟨ 0.4 x - 0.2 y, - 0.1 x + 0.3 y ⟩$ . Is $H$ the inverse transformation of $F$ ?

Which reason supports your answer?

The rest stand on their own and do not use the case.

Check yourself · Put it in order

Each matrix is associated with a linear transformation. Tap them in order of the factor by which they multiply the area of a region, from smallest to largest.

Check yourself · Spot the error

$L$ sends $⟨ 1, 0 ⟩$ to $⟨ 2, 1 ⟩$ and $⟨ 0, 1 ⟩$ to $⟨ 0, 3 ⟩$ , and $M (⟨ x, y ⟩) = ⟨ x - y, 2 y ⟩$ . A student finds the image of $⟨ 1, 2 ⟩$ when $L$ is applied first and then $M$ . Tap the first line that goes wrong.

Part 7 Keep it Bring back earlier ideas

Pulling an older idea back up is what makes it last.

From 4.12 Linear Transformations and Matrices · Calculate

Let $A = [\begin{matrix} 2 & - 3 \\ 1 & 4 \end{matrix}]$ , and let $V$ be the $2 \times 3$ matrix whose columns are the vectors $⟨ 3, 1 ⟩$ , $⟨ - 2, 5 ⟩$ and $⟨ 1, 2 ⟩$ , in that order. What is the component in row 2, column 3 of $A V$ ?

Your answer

From 3.3 Sine and Cosine Function Values · Sort it

Each card is an angle, in radians, in standard position. Tap a card, then the signs of its cosine and sine.

$\cos θ > 0$ and $\sin θ > 0$

$\cos θ < 0$ and $\sin θ > 0$

$\cos θ < 0$ and $\sin θ < 0$

$\cos θ > 0$ and $\sin θ < 0$

In three lines

The matrix of a linear transformation has the images of $⟨ 1, 0 ⟩$ and $⟨ 0, 1 ⟩$ as its columns, and a rotation by $θ$ counterclockwise has matrix $[\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}]$ .
A transformation with matrix $A$ multiplies the area of every region by $| \det (A) |$ .
Applying $A$ and then $B$ has matrix $B A$ , and $L (\vec{v}) = A \vec{v}$ is undone by $L^{- 1} (\vec{v}) = A^{- 1} \vec{v}$ , whose composition with $L$ maps every vector to itself.

Vocabulary

Linear transformation: A function that maps an input vector to an output vector whose components are each a sum of constant multiples of the input components, such as $L (⟨ x, y ⟩) = ⟨ a_{11} x + a_{12} y, a_{21} x + a_{22} y ⟩$ .
Matrix associated with a linear transformation: The matrix $A = [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}]$ with $L (\vec{v}) = A \vec{v}$ ; its columns are the images of $⟨ 1, 0 ⟩$ and $⟨ 0, 1 ⟩$ .
Rotation: The linear transformation that turns every vector through the same angle $θ$ counterclockwise about the origin, with matrix $[\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}]$ .
Dilation of regions: How a transformation with $2 \times 2$ matrix $A$ stretches or shrinks regions of the plane; its magnitude, the factor every area is multiplied by, is $| \det (A) |$ .
Composition of linear transformations: Applying one linear transformation to the output of another; it is itself a linear transformation, and applying $A$ and then $B$ has the matrix $B A$ .
Inverse transformation: The transformation $L^{- 1}$ whose composition with $L$ maps every vector to itself; for $L (\vec{v}) = A \vec{v}$ it is $L^{- 1} (\vec{v}) = A^{- 1} \vec{v}$ .

Course alignment, for teachers

AP Precalculus topic 4.13, Unit 4: Functions Involving Parameters, Vectors, and Matrices.

Skill 1.B: Express functions, equations, or expressions in analytically equivalent forms that are useful in a given mathematical or applied context.
Skill 2.A: Identify information from graphical, numerical, analytical, and verbal representations to answer a question or construct a model, with and without technology.
Skill 3.A: Describe the characteristics of a function with varying levels of precision, depending on the function representation and available mathematical tools.

This lesson maps all 8 essential knowledge statements of the topic to the part that teaches each one: 8 are checked by a scored question. The map is in the lesson file.

Part 1 Think first Which button first?

Part 2 Learn it Matrices that move the plane

2a. Read the matrix off the unit vectors

2b. Rotations, and what happens to area

2c. One transformation after another

2d. Undoing a transformation

Part 3 Watch it done Find an inverse transformation and use it

Part 4 Try it Two moves in a drawing app

Vectors and Matrices

Part 5 Explain it Put it in writing

Part 6 Check yourself Three moves in a sign shop

Part 7 Keep it Bring back earlier ideas

Vocabulary

Labs that use this topic