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100+ Free PAU Mathematics II (Cantabria) Practice Questions

Cantabria PAU Mathematics II 2026 (Pruebas de Acceso a la Universidad, Universidad de Cantabria) practice questions are available now; exam metadata is being verified.

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Key Facts: PAU Mathematics II (Cantabria) Exam

90 min

Exam time duration for the PAU specific paper

University of Cantabria (UNICAN) PAU Organising Commission

4 x 2.5 pts

Four compulsory exercises (Algebra, Analysis, Geometry, Probability), each worth 2.5 of 10 points

UNICAN Matemáticas II Criterios de Corrección, June 2026

0–10 scale

Scoring system (minimum 4.0 in Access Phase to combine with Bachillerato)

Cantabria PAU Regulations

4 Blocks (25% each)

Official content blocks weighted equally: Algebra, Mathematical Analysis, Geometry in 3D, Probability & Statistics

UNICAN Matemáticas II Criterios de Corrección, June 2026

100

High-quality practice questions in this OpenExamPrep subject bank

OpenExamPrep

Master the 2026 Cantabria PAU Mathematics II exam with 100 practice questions covering matrix algebra and linear systems, calculus and optimization, 3D vector geometry, and probability and statistics, each weighted equally at 25% based on the official exam structure. This English-language MCQ bank is a study adaptation of the knowledge behind the official written exam, not a format simulation.

Sample PAU Mathematics II (Cantabria) Practice Questions

Try these sample questions to test your PAU Mathematics II (Cantabria) exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1Given A = [[4, -1], [2, 3]] and B = [[1, 2], [-3, 0]], compute A + 2B.
A.[[6, 3], [-4, 3]]
B.[[5, 1], [-1, 3]]
C.[[9, 0], [1, 6]]
D.[[6, 3], [8, 3]]
Explanation: First scale B by 2: 2B = [[2, 4], [-6, 0]]. Then add A entrywise: A + 2B = [[4+2, -1+4], [2-6, 3+0]] = [[6, 3], [-4, 3]].
2Let A = [[2, 0], [1, -1]] and B = [[3, 1], [2, 4]]. Compute the matrix product AB.
A.[[7, -1], [8, -4]]
B.[[6, 2], [1, -3]]
C.[[6, 0], [2, -4]]
D.[[6, 2], [1, 5]]
Explanation: Multiply row by column: row 1 gives [2*3+0*2, 2*1+0*4] = [6, 2]; row 2 gives [1*3+(-1)*2, 1*1+(-1)*4] = [1, -3]. So AB = [[6, 2], [1, -3]].
3Find the determinant of the matrix A = [[5, 3], [2, 4]].
A.26
B.-14
C.14
D.-2
Explanation: For a 2x2 matrix [[a,b],[c,d]], det(A) = ad - bc. Here det(A) = 5*4 - 3*2 = 20 - 6 = 14.
4Calculate the determinant of the matrix A = [[1, 2, 3], [0, 1, 4], [5, 6, 0]] using Sarrus's rule.
A.16
B.25
C.40
D.1
Explanation: Sarrus's rule: positive diagonals sum to (1*1*0)+(2*4*5)+(3*0*6) = 0+40+0 = 40, and negative diagonals sum to (3*1*5)+(1*4*6)+(2*0*0) = 15+24+0 = 39. det(A) = 40 - 39 = 1.
5If A is a 3x3 matrix with det(A) = 5, what is det(2A)?
A.40
B.10
C.20
D.13
Explanation: For an n x n matrix and scalar k, det(kA) = k^n * det(A). Here n=3 and k=2, so det(2A) = 2^3 * 5 = 8 * 5 = 40.
6If det(A) = -3 for a square matrix A, what is det(A^T), the determinant of its transpose?
A.3
B.-3
C.-1/3
D.0
Explanation: A fundamental determinant property states det(A^T) = det(A) for any square matrix, since transposing does not change the value of the determinant. So det(A^T) = -3.
7If det(A) = 4 and det(B) = -2 for two 3x3 matrices, what is det(AB)?
A.2
B.8
C.-8
D.6
Explanation: A key property of determinants is det(AB) = det(A) * det(B). Here det(AB) = 4 * (-2) = -8.
8Find the inverse of the matrix A = [[3, 1], [5, 2]].
A.[[3, -1], [-5, 2]]
B.[[2, 1], [5, 3]]
C.[[3, 5], [1, 2]]
D.[[2, -1], [-5, 3]]
Explanation: det(A) = 3*2 - 1*5 = 1. For a 2x2 matrix [[a,b],[c,d]], A^-1 = (1/det(A)) * [[d,-b],[-c,a]] = [[2,-1],[-5,3]].
9Use Cramer's rule to solve the system x + 2y = 5, 3x + 5y = 11 for (x, y).
A.x = -3, y = 4
B.x = 3, y = -4
C.x = 4, y = -3
D.x = 3, y = 4
Explanation: The determinant D = |1,2;3,5| = 1*5-2*3 = -1. Dx = |5,2;11,5| = 25-22 = 3, so x = Dx/D = 3/(-1) = -3. Dy = |1,5;3,11| = 11-15 = -4, so y = Dy/D = -4/(-1) = 4.
10For which value of k does the system x + y + z = 1, 2x + y - z = 0, x - y + kz = 2 fail to have a unique solution?
A.5
B.-5
C.-3
D.3
Explanation: The coefficient determinant is |1,1,1;2,1,-1;1,-1,k| = (k-1) - (2k+1) + (-3) = -k-5. Setting -k-5 = 0 gives k = -5, the value at which the coefficient matrix becomes singular.

About the PAU Mathematics II (Cantabria) Practice Questions

Verified exam format metadata for Cantabria PAU Mathematics II 2026 (Pruebas de Acceso a la Universidad, Universidad de Cantabria) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.