Unit 4 · Topic 4.11 · about 30 minutes · 7 parts · 14 scored checks, 2 of them multiple choice

AP Precalculus 4.11: The Inverse and Determinant of a Matrix

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.

Two pictures of 2 by 2 matrices drawn as vectors from the origin. Left: the matrix with rows 4, 2 and 1, 3 has column vectors <4, 1> and <2, 3>. They are adjacent sides of a shaded parallelogram with corners (0, 0), (4, 1), (6, 4) and (2, 3), whose area is 10, the same as the determinant 4 times 3 minus 2 times 1. Right: the matrix with rows 2, 4 and 1, 2 has column vectors <2, 1> and <4, 2>. They lie along the same line, so the determinant is 0 and they span no parallelogram.

Part 1 Think first Can Leo read the message?

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

Predict first

Maya sends Leo pairs of secret numbers. She writes each pair as a column and multiplies it by the matrix $[\begin{matrix} 3 & 6 \\ 1 & 2 \end{matrix}]$ , then sends the result. Leo knows the matrix she used. Can he always work out the pair she started with?

In short. The inverse $A^{- 1}$ of a square matrix $A$ is the matrix whose product with $A$ is the identity matrix $I$ , and a $2 \times 2$ matrix has one exactly when its determinant, $\det (A) = a d - b c$ , is not 0.

The identity matrix $I$ has 1s on the diagonal from the top left to the bottom right and 0s everywhere else, and $A I = I A = A$ for a square matrix $A$ .
For $A = [\begin{matrix} a & b \\ c & d \end{matrix}]$ with $\det (A) \neq 0$ , $A^{- 1} = \frac{1}{a d - b c} [\begin{matrix} d & - b \\ - c & a \end{matrix}]$ , and $A A^{- 1} = A^{- 1} A = I$ .
With two vectors from $ℝ^{2}$ in its columns or rows, $| \det (A) |$ is the area of the parallelogram they span when it is not 0, and $\det (A) = 0$ means the vectors are parallel.

Part 2 Learn it Undoing a matrix

2a. The matrix that changes nothing

Multiplying a number by 1 leaves it alone. Matrices have their own version of 1. The identity matrix, $I$ , is a square matrix with 1s on the diagonal that runs from the top left to the bottom right and 0s everywhere else:

[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]

The first is the $2 \times 2$ identity and the second the $3 \times 3$ one. You use the one that matches the size of the square matrix it meets.

Multiply a square matrix by its identity, on either side, and you get the matrix you started with. Take $A = [\begin{matrix} 4 & - 1 \\ 3 & 2 \end{matrix}]$ . Row 1 of $A$ times column 1 of $I$ is $4 (1) + (- 1) (0) = 4$ , and row 1 times column 2 is $4 (0) + (- 1) (1) = - 1$ . Row 2 works the same way, so $A I = A$ . In the other order, $I A = A$ too.

The 1s have to sit on that diagonal. The matrix $[\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}]$ has them on the other one, and multiplying $A$ by it on the right swaps the two columns of $A$ instead of keeping them.

Quick check · Sort it

Let $A = [\begin{matrix} 3 & 1 \\ 2 & 5 \end{matrix}]$ . Each card is a product. Tap a card, then tap whether the product equals $A$ .

Equals $A$

Does not equal $A$

2b. One number from four components

Every $2 \times 2$ matrix has a number attached to it called its determinant. For $A = [\begin{matrix} a & b \\ c & d \end{matrix}]$ ,

\det (A) = a d - b c

Multiply along the diagonal from the top left, then subtract the product along the other diagonal. For $[\begin{matrix} 5 & 2 \\ 3 & 4 \end{matrix}]$ that is $5 \cdot 4 - 2 \cdot 3 = 20 - 6 = 14$ .

Negative components are where the slips happen. In $[\begin{matrix} 2 & - 3 \\ 4 & 1 \end{matrix}]$ the second product is $(- 3) (4) = - 12$ , and subtracting it adds 12: $2 \cdot 1 - (- 12) = 14$ . Write each product in parentheses before you subtract.

A graphing calculator can find a determinant too, which is a good way to check your arithmetic. Still, a $2 \times 2$ determinant by hand takes a few seconds, and you will want it that fast in the rest of this lesson.

Quick check · Multiple choice

What is $\det (A)$ for $A = [\begin{matrix} 3 & - 2 \\ 4 & 5 \end{matrix}]$ ?

2c. The inverse, and when there is one

The inverse of a square matrix $A$ , written $A^{- 1}$ , is the matrix whose product with $A$ is the identity:

A A^{- 1} = A^{- 1} A = I

For $A = [\begin{matrix} a & b \\ c & d \end{matrix}]$ , swap $a$ and $d$ , change the signs of $b$ and $c$ , and divide every component by the determinant:

A^{- 1} = \frac{1}{a d - b c} [\begin{matrix} d & - b \\ - c & a \end{matrix}]

Take $A = [\begin{matrix} 3 & 5 \\ 1 & 2 \end{matrix}]$ . Its determinant is $6 - 5 = 1$ , so $A^{- 1} = [\begin{matrix} 2 & - 5 \\ - 1 & 3 \end{matrix}]$ . Check one component: row 1 of $A$ times column 1 of $A^{- 1}$ is $3 (2) + 5 (- 1) = 1$ . The other three components come out 0, 0 and 1, so the product is $I$ .

The formula divides by $a d - b c$ , and nothing can be divided by 0. A square matrix has an inverse exactly when its determinant is not 0. Maya's matrix has determinant $3 (2) - 6 (1) = 0$ , which is why Leo could not undo it. A calculator's inverse command gives the same matrix as the formula, and gives an error for a determinant of 0.

Quick check · Spot the error

A student finds the inverse of $B = [\begin{matrix} 6 & 4 \\ 2 & 3 \end{matrix}]$ . Tap the first line that goes wrong.

2d. Area and parallel vectors

A $2 \times 2$ matrix can hold two vectors from $ℝ^{2}$ , one in each column. The matrix $[\begin{matrix} 4 & 2 \\ 1 & 3 \end{matrix}]$ holds $⟨ 4, 1 ⟩$ in its first column and $⟨ 2, 3 ⟩$ in its second. Draw both from the origin and complete the parallelogram, as in the left picture in the diagram.

When the determinant is not 0, its absolute value is the area of that parallelogram. Here $4 \cdot 3 - 2 \cdot 1 = 10$ , so the area is 10 square units. Put the same vectors in the rows, $[\begin{matrix} 4 & 1 \\ 2 & 3 \end{matrix}]$ , and the determinant is $12 - 2 = 10$ again.

The order of the vectors can flip the sign. With the columns swapped, $[\begin{matrix} 2 & 4 \\ 3 & 1 \end{matrix}]$ has determinant $2 - 12 = - 10$ . An area is never negative, so take the absolute value.

When the determinant is 0, the vectors are parallel. $⟨ 2, 1 ⟩$ and $⟨ 4, 2 ⟩$ give $[\begin{matrix} 2 & 4 \\ 1 & 2 \end{matrix}]$ , with determinant $4 - 4 = 0$ . The second vector is 2 times the first, so the parallelogram flattens onto a line, as in the right picture. Maya's columns, $⟨ 3, 1 ⟩$ and $⟨ 6, 2 ⟩$ , are parallel for the same reason.

Quick check · Calculate

What is the area, in square units, of the parallelogram spanned by the vectors $⟨ 1, 4 ⟩$ and $⟨ 3, 1 ⟩$ ?

Your answer

Part 3 Watch it done Find an inverse and check it

Watch each step, then do the last one yourself.

Find the inverse of $A = [\begin{matrix} 4 & 3 \\ 2 & 2 \end{matrix}]$ , then check that $A A^{- 1} = I$ .

Step 1 · Find the determinant
$\det (A) = 4 \cdot 2 - 3 \cdot 2 = 8 - 6 = 2$ . It is not 0, so $A$ has an inverse.
Why this step?
Why find the determinant before anything else?
The last step divides by it. If it were 0 there would be no inverse, and the swapping and sign changes would be wasted work. Here it is 2, so the inverse exists.
Step 2 · Swap and change signs
Swap $a = 4$ and $d = 2$ , and change the signs of $b = 3$ and $c = 2$ : $[\begin{matrix} 2 & - 3 \\ - 2 & 4 \end{matrix}]$ .
Step 3 · Divide by the determinant
Multiply that matrix by $\frac{1}{2}$ , which divides each component by 2. The top left component becomes $\frac{2}{2} = 1$ , and the other three are divided the same way.
Your step · Calculate

Finish it. For $A = [\begin{matrix} 4 & 3 \\ 2 & 2 \end{matrix}]$ , $A^{- 1} = \frac{1}{2} [\begin{matrix} 2 & - 3 \\ - 2 & 4 \end{matrix}]$ . What is the component in row 1, column 2 of $A^{- 1}$ ?

Your answer

Now check the product. $A A^{- 1} = [\begin{matrix} 4 & 3 \\ 2 & 2 \end{matrix}] [\begin{matrix} 1 & - 1.5 \\ - 1 & 2 \end{matrix}]$ . Row 1 times column 1 is $4 (1) + 3 (- 1) = 1$ , and row 1 times column 2 is $4 (- 1.5) + 3 (2) = 0$ . Row 2 gives $2 (1) + 2 (- 1) = 0$ and $2 (- 1.5) + 2 (2) = 1$ . The product is $I$ , so the inverse is right.

Part 4 Try it Choose a key the club can decode

A robotics club sends messages in a number code. Each letter becomes its place in the alphabet (A is 1, B is 2, and so on), the numbers are taken in pairs, and each pair is written as a column and multiplied by a key matrix. The receiver multiplies each coded column by the inverse of the key to get the letters back. Two keys were proposed. Work on paper.

Keys proposed for the club code
Key	Matrix
$K$	$[\begin{matrix} 2 & 1 \\ 5 & 3 \end{matrix}]$
$J$	$[\begin{matrix} 4 & 6 \\ 2 & 3 \end{matrix}]$

Try it · Decide and justify

A code multiplies each pair of numbers, written as a column, by a key matrix, and the receiver decodes with the inverse of the key. Can $J = [\begin{matrix} 4 & 6 \\ 2 & 3 \end{matrix}]$ be used as a key that can always be decoded?

Which reason supports your answer?

Try it · Calculate

A code multiplies each pair of numbers, written as a column, by the key $K = [\begin{matrix} 2 & 1 \\ 5 & 3 \end{matrix}]$ . A coded column arrives as $[\begin{matrix} 25 \\ 67 \end{matrix}]$ . Decode it by multiplying by $K^{- 1}$ . What is the top component of the decoded column, the first number of the original pair?

Your answer

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 $A = [\begin{matrix} 3 & - 1 \\ 4 & 2 \end{matrix}]$ and $B = [\begin{matrix} 2 & - 3 \\ - 4 & 6 \end{matrix}]$ .

(i) Determine $\det (A)$ and explain how it shows that $A$ has an inverse.

(ii) Determine $A^{- 1}$ , and justify your answer by showing that $A A^{- 1} = I$ .

(iii) Determine $\det (B)$ . Describe what it tells you about the column vectors of $B$ , and explain whether $B$ has an inverse.

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

Model answer

(i) $\det (A) = 3 \cdot 2 - (- 1) (4) = 6 + 4 = 10$ . A $2 \times 2$ matrix has an inverse exactly when its determinant is not 0, and $10 \neq 0$ . (ii) Swapping 3 and 2, changing the signs of $- 1$ and 4, and dividing by 10 gives $A^{- 1} = \frac{1}{10} [\begin{matrix} 2 & 1 \\ - 4 & 3 \end{matrix}] = [\begin{matrix} 0.2 & 0.1 \\ - 0.4 & 0.3 \end{matrix}]$ . In $A A^{- 1}$ , row 1 times column 1 is $3 (0.2) + (- 1) (- 0.4) = 1$ , row 1 times column 2 is $3 (0.1) + (- 1) (0.3) = 0$ , row 2 times column 1 is $4 (0.2) + 2 (- 0.4) = 0$ , and row 2 times column 2 is $4 (0.1) + 2 (0.3) = 1$ . So $A A^{- 1} = I$ . (iii) $\det (B) = 2 \cdot 6 - (- 3) (- 4) = 12 - 12 = 0$ . So the column vectors $⟨ 2, - 4 ⟩$ and $⟨ - 3, 6 ⟩$ are parallel; in fact $⟨ - 3, 6 ⟩ = - 1.5 ⟨ 2, - 4 ⟩$ , and the parallelogram they span has no area. $B$ has no inverse, because the formula would divide by $\det (B) = 0$ .

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

Finds $\det (A) = 10$ and explains that a determinant that is not 0 means $A$ has an inverse.
Writes $A^{- 1} = \frac{1}{10} [\begin{matrix} 2 & 1 \\ - 4 & 3 \end{matrix}]$ , with 3 and 2 swapped and the signs of $- 1$ and 4 changed.
Works out all four components of $A A^{- 1}$ and gets the identity matrix.
Finds $\det (B) = 0$ , says the column vectors are parallel, and explains that $B$ has no inverse.

Part 6 Check yourself Three matrices

The three case questions use the same table of three matrices. Answer them from its components.

Three $2 \times 2$ matrices, $P$ , $Q$ and $R$ , are given in the table.

The matrices P, Q and R
Name	Matrix
$P$	$[\begin{matrix} 2 & 5 \\ 1 & 3 \end{matrix}]$
$Q$	$[\begin{matrix} 4 & - 2 \\ - 6 & 3 \end{matrix}]$
$R$	$[\begin{matrix} 3 & - 2 \\ 1 & 4 \end{matrix}]$

Case question 1 · Multiple choice

Which matrix is $P^{- 1}$ , the inverse of $P = [\begin{matrix} 2 & 5 \\ 1 & 3 \end{matrix}]$ ?

Case question 2 · Calculate

The rows of $R = [\begin{matrix} 3 & - 2 \\ 1 & 4 \end{matrix}]$ are the vectors $⟨ 3, - 2 ⟩$ and $⟨ 1, 4 ⟩$ . What is the area, in square units, of the parallelogram they span?

Your answer

Case question 3 · Decide and justify

The columns of $Q = [\begin{matrix} 4 & - 2 \\ - 6 & 3 \end{matrix}]$ are the vectors $⟨ 4, - 6 ⟩$ and $⟨ - 2, 3 ⟩$ . Are the two vectors parallel?

Which reason supports your answer?

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

Check yourself · Select all

Which two of these matrices have an inverse? Pick two.

Select 2 answers, then check.

Check yourself · Fill the statement

Let $A = [\begin{matrix} 2 & 1 \\ 7 & 4 \end{matrix}]$ and $B = [\begin{matrix} 4 & - 1 \\ - 7 & 2 \end{matrix}]$ . Multiply $A$ by $B$ , then complete the statement. Tap a blank, then tap an answer.

$A B$ = , so $B$ = .

Part 7 Keep it Bring back earlier ideas

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

From 4.10 Matrices · Calculate

Let $A = [\begin{matrix} 3 & - 1 \\ 2 & 5 \end{matrix}]$ and $B = [\begin{matrix} 4 & 2 \\ 1 & - 3 \end{matrix}]$ . What is the component in row 2, column 2 of $A B$ ?

Your answer

From 2.8 Inverse Functions · Put it in order

Put the steps for finding the inverse of $g (x) = 2 (x - 1)^{3} + 5$ in order. Tap them in sequence.

In three lines

The identity matrix $I$ has 1s on the diagonal from the top left to the bottom right and 0s elsewhere, and $A I = I A = A$ .
$\det (A) = a d - b c$ , and $A^{- 1} = \frac{1}{a d - b c} [\begin{matrix} d & - b \\ - c & a \end{matrix}]$ exists exactly when $\det (A) \neq 0$ ; then $A A^{- 1} = A^{- 1} A = I$ .
With two vectors in its columns or rows, $| \det (A) |$ is the area of the parallelogram they span, and $\det (A) = 0$ means they are parallel.

Vocabulary

Square matrix: A matrix with as many rows as columns, such as a $2 \times 2$ matrix; identity matrices and inverses are square.
Identity matrix: The square matrix $I$ with 1s on the diagonal from the top left to the bottom right and 0s everywhere else, so that $A I = I A = A$ for a square matrix $A$ of the same size.
Inverse of a matrix: For a square matrix $A$ , the matrix $A^{- 1}$ with $A A^{- 1} = A^{- 1} A = I$ ; it exists exactly when $\det (A) \neq 0$ .
Determinant: For $A = [\begin{matrix} a & b \\ c & d \end{matrix}]$ , the number $\det (A) = a d - b c$ .
Parallelogram spanned by two vectors: The parallelogram that has the two vectors, drawn from one point, as adjacent sides; its area is $| \det (A) |$ for the matrix $A$ that holds the vectors in its columns or rows.

Course alignment, for teachers

AP Precalculus topic 4.11, 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 3.B: Apply numerical results in a given mathematical or applied context.

This lesson maps all 7 essential knowledge statements of the topic to the part that teaches each one: 7 are checked by a scored question. The map is in the lesson file.

Part 1 Think first Can Leo read the message?

Part 2 Learn it Undoing a matrix

2a. The matrix that changes nothing

2b. One number from four components

2c. The inverse, and when there is one

2d. Area and parallel vectors

Part 3 Watch it done Find an inverse and check it

Part 4 Try it Choose a key the club can decode

Part 5 Explain it Put it in writing

Part 6 Check yourself Three matrices

Part 7 Keep it Bring back earlier ideas

Vocabulary

Labs that use this topic