AP PreCalculus Practice Quiz: The Inverse and Determinant of a Matrix
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 15 questions to check your progress.
Question 1 of 15
All Questions (15)
A) 14
B) 26
C) -14
D) 20
Correct Answer: A
The determinant of a $2 \\times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is calculated as $ad - bc$. For matrix A, this is $(5)(4) - (2)(3) = 20 - 6 = 14$.
A) $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$
B) $\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$
C) $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
D) $\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$
Correct Answer: C
The identity matrix, $I$, is a square matrix with 1s on the main diagonal (from the top left to bottom right) and 0s everywhere else.
A) The zero matrix $\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$
B) The matrix $A$ itself
C) The identity matrix $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
D) A matrix with all entries equal to 1
Correct Answer: C
By definition, the product of a square matrix and its inverse, when it exists, is the identity matrix of the same size. So, $A \\cdot A^{-1} = I$.
A) $\det(A) = 0$
B) $\det(A) = 1$
C) $\det(A) \ne 0$
D) The matrix contains only positive numbers.
Correct Answer: C
A square matrix $A$ is invertible, meaning it has an inverse, if and only if its determinant is non-zero.
A) 0
B) 1
C) 2
D) 4
Correct Answer: B
A matrix is not invertible if its determinant is 0. The determinant of M is $\det(M) = (k)(8) - (4)(2) = 8k - 8$. Setting the determinant to zero gives $8k - 8 = 0$, which solves to $k=1$.
A) $\begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}$
B) $\begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}$
C) $\begin{pmatrix} -2 & 1 \\ 5 & -3 \end{pmatrix}$
D) $\begin{pmatrix} 1 & -1 \\ -5 & 3 \end{pmatrix}$
Correct Answer: A
First, find the determinant: $\det(A) = (3)(2) - (1)(5) = 6 - 5 = 1$. The inverse of $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is $\\frac{1}{\det(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$. So, $A^{-1} = \\frac{1}{1} \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}$.
A) $A$ is invertible because the vectors are non-zero.
B) $A$ is not invertible because its determinant is zero.
C) $A$ is invertible because its determinant is non-zero.
D) Invertibility cannot be determined from the given vectors.
Correct Answer: B
Let $A = \begin{pmatrix} -6 & 4 \\ 3 & -2 \end{pmatrix}$. The determinant is $\det(A) = (-6)(-2) - (4)(3) = 12 - 12 = 0$. Since the determinant is 0, the matrix is not invertible. This also implies the column vectors are linearly dependent (collinear).
A) 19
B) -23
C) 23
D) -19
Correct Answer: A
The determinant is $ad-bc = (-1)(2) - (7)(-3) = -2 - (-21) = -2 + 21 = 19$.
A) $x = 5$
B) $x = -1$
C) $x = 0$
D) $x = 5$ and $x = -1$
Correct Answer: C
The matrix is invertible if its determinant is not zero. $\det(C) = x(x-4) - (5)(1) = x^2 - 4x - 5$. We find when the determinant is zero: $x^2 - 4x - 5 = 0 \implies (x-5)(x+1) = 0$. The matrix is not invertible when $x=5$ or $x=-1$. Any other value, such as $x=0$, will make it invertible.
A) If $\det(A) > 0$, the matrix has an inverse.
B) If $\det(A) < 0$, the matrix does not have an inverse.
C) A matrix has an inverse only if $\det(A) = 1$.
D) A matrix has an inverse if and only if its determinant is any non-zero number.
Correct Answer: D
The only condition for a square matrix to be invertible is that its determinant is non-zero. The sign of the determinant (positive or negative) does not affect the existence of the inverse.
A) $\begin{pmatrix} 3 & -4 \\ -1 & 2 \end{pmatrix}$
B) $\\frac{1}{2} \begin{pmatrix} 3 & -4 \\ -1 & 2 \end{pmatrix}$
C) $\\frac{1}{10} \begin{pmatrix} 3 & -4 \\ -1 & 2 \end{pmatrix}$
D) $\begin{pmatrix} 2 & -4 \\ -1 & 3 \end{pmatrix}$
Correct Answer: B
First, calculate the determinant: $\det(P) = (2)(3) - (4)(1) = 6 - 4 = 2$. The inverse is $\\frac{1}{\det(P)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \\frac{1}{2} \begin{pmatrix} 3 & -4 \\ -1 & 2 \end{pmatrix}$.
A) $a / \det(A)$
B) $d / \det(A)$
C) $-c / \det(A)$
D) $1 / \det(A)$
Correct Answer: B
The formula for the inverse is $A^{-1} = \\frac{1}{\det(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$. The entry in the first row and first column is the top-left element of the adjusted matrix, which is $d$, multiplied by the scalar $\\frac{1}{\det(A)}$. This gives $d / \det(A)$.
A) 3
B) 4
C) 5
D) 6
Correct Answer: C
The determinant is $\det(M) = (x)(x) - (2)(8) = x^2 - 16$. We are given that $\det(M) = 9$. So, $x^2 - 16 = 9 \implies x^2 = 25$. The possible values for $x$ are 5 and -5. Of the options provided, 5 is a possible value.
A) The vectors are perpendicular.
B) The vectors are parallel (collinear).
C) The vectors have the same magnitude.
D) The vectors are unit vectors.
Correct Answer: B
A determinant of zero for a matrix formed by two vectors as its columns means the vectors are linearly dependent. In a 2D space, this geometric interpretation is that the vectors are collinear; they lie on the same line or are parallel to each other.
A) $B$ is the inverse of $A$.
B) $A$ is not invertible.
C) The product $AB$ is the zero matrix.
D) The determinant of $A$ is equal to the determinant of $B$.
Correct Answer: A
To check if $B$ is the inverse of $A$, we can multiply them and see if the result is the identity matrix. $AB = \begin{pmatrix} 4 & 7 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & -7 \\ -1 & 4 \end{pmatrix} = \begin{pmatrix} (4)(2)+(7)(-1) & (4)(-7)+(7)(4) \\ (1)(2)+(2)(-1) & (1)(-7)+(2)(4) \end{pmatrix} = \begin{pmatrix} 8-7 & -28+28 \\ 2-2 & -7+8 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. Since $AB=I$, $B$ is the inverse of $A$.