Is bmatrix 2 & 4 \\ 1 & 2 bmatrix invertible? [1 point]

Find the determinant of bmatrix 5 & 0 \\ 0 & 3 bmatrix. [1 point]

United StatesPrecalculusSyllabus dot point

What are the determinant and inverse of a matrix, and what do they tell you?

Topic 4.11 The Inverse and Determinant of a Matrix: compute the determinant and inverse of a 2x2 matrix, and use them to determine invertibility and solve matrix equations.

A focused answer to AP Precalculus Topic 4.11, covering the determinant of a 2x2 matrix, what it measures, the inverse formula, when a matrix is invertible, and using the inverse to solve a matrix equation.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

The determinant of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is $ad - bc$ , a single number. It measures the factor by which the matrix scales area, and its sign records whether orientation is preserved or flipped. A matrix is invertible exactly when its determinant is non-zero; a zero determinant means the matrix collapses the plane and has no inverse. When the determinant is non-zero, the inverse is $\frac{1}{ad - bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ : swap the main diagonal, negate the off-diagonal, and divide by the determinant. The inverse undoes the matrix, so $A^{-1}A = I$ , the identity. This lets you solve a matrix equation $A\mathbf{x} = \mathbf{b}$ by multiplying both sides by $A^{-1}$ , giving $\mathbf{x} = A^{-1}\mathbf{b}$ , the matrix analogue of dividing to solve a linear equation.

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What this topic is asking
The determinant
Invertibility
The inverse formula
The inverse as an undoing transformation
Try this

What this topic is asking

The College Board (Topic 4.11) wants you to compute the determinant and inverse of a $2 \times 2$ matrix. The determinant is a single number that measures how the matrix scales area and tells you whether the matrix is invertible; the inverse, when it exists, undoes the matrix and lets you solve matrix equations.

The determinant

A determinant of $1$ preserves area; a determinant of $0$ flattens the plane onto a line (or a point), which is why such a matrix cannot be undone.

Invertibility

So the determinant is a quick invertibility test: compute $ad - bc$ and check whether it is zero.

The inverse formula

Determinant, inverse and solving an equation

Let $A = \begin{bmatrix} 4 & 3 \\ 2 & 2 \end{bmatrix}$ . Find $\det A$ , then $A^{-1}$ , then solve $A\mathbf{x} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ .

step 1 Compute the determinant

$\det A = 4 \cdot 2 - 3 \cdot 2 = 8 - 6 = 2$ . Non-zero, so $A$ is invertible.

step 2 Apply the inverse formula

$A^{-1} = \frac{1}{2}\begin{bmatrix} 2 & -3 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} 1 & -\frac{3}{2} \\ -1 & 2 \end{bmatrix}$ .

step 3 Solve by multiplying by the inverse

$\mathbf{x} = A^{-1}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \cdot 1 + \left(-\frac{3}{2}\right)\cdot 0 \\ -1 \cdot 1 + 2 \cdot 0 \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$ .

step 4 Check

$A\mathbf{x} = \begin{bmatrix} 4 & 3 \\ 2 & 2 \end{bmatrix}\begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 4 - 3 \\ 2 - 2 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ , matching the right-hand side. The solution is correct.

The inverse as an undoing transformation

A point worth stating once is that the inverse matrix is the transformation that reverses what the original matrix does, exactly as an inverse function reverses a function (Topic 2.8). If $A$ rotates and stretches the plane, $A^{-1}$ rotates and stretches it back, and applying both in succession returns every vector to where it started ( $A^{-1}A = I$ ). This is why a matrix with determinant zero has no inverse: it collapses the plane, destroying the information needed to undo it, just as a non-one-to-one function cannot be inverted. Seeing the determinant as "does this transformation lose area, and therefore information?" explains both the invertibility test and the geometry of the matrices-as-functions topic that follows.

Try this

Q1. Is $\begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix}$ invertible? [1 point]

Cue. $\det = 2\cdot2 - 4\cdot1 = 4 - 4 = 0$ , so it is not invertible.

Q2. Find the determinant of $\begin{bmatrix} 5 & 0 \\ 0 & 3 \end{bmatrix}$ . [1 point]

Cue. $5 \cdot 3 - 0 \cdot 0 = 15$ .

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2025 (style)1 marksSection I, Part A (multiple choice, no calculator). What is the determinant of

\begin{bmatrix} 3 & 5 \\ 2 & 4 \end{bmatrix}

? (A)

2

(B)

12

(C)

22

(D)

-2

Show worked answer →

The correct answer is (A), $2$ .

For $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , the determinant is $ad - bc$ . Here $3 \cdot 4 - 5 \cdot 2 = 12 - 10 = 2$ . A non-zero determinant means the matrix is invertible.

AP 2025 (style)4 marksSection II (free response, no calculator). Let

A = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}

. (a) Find the determinant and explain why

A

is invertible. (b) Find

A^{-1}

Show worked answer →

A 4-point question on the determinant and inverse.

(a) Determinant (2 points): $\det A = 2 \cdot 1 - 1 \cdot 1 = 2 - 1 = 1$ . Since the determinant is non-zero, $A$ is invertible.
(b) Inverse (2 points): $A^{-1} = \frac{1}{\det A}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} = \frac{1}{1}\begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}$ .

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Sources & how we know this

AP Precalculus Course and Exam Description — College Board (2023)