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100+ Free Extremadura PAU Applied Mathematics for Social Sciences II Practice Questions

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2026 Statistics

Key Facts: Extremadura PAU Applied Mathematics for Social Sciences II Exam

EUR 78.26

Ordinary registration fee for PAU

Universidad de Extremadura (UEx)

90 Mins

Official examination duration

Comisión Organizadora de la PAU de Extremadura

Min 4.0

Minimum Access Phase score required to combine with Bachillerato GPA

UEx / Junta de Extremadura

3 Units

Main thematic curriculum blocks

2º Bachillerato LOMLOE Applied Mathematics Curriculum

100

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Sample Extremadura PAU Applied Mathematics for Social Sciences II Practice Questions

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1Given matrices A = [[2, -1], [3, 4]] and B = [[1, 5], [-2, 0]], calculate the matrix C = 2A - B.
A.[[3, -7], [8, 8]]
B.[[1, -7], [8, 8]]
C.[[3, 3], [4, 8]]
D.[[3, -7], [4, 4]]
Explanation: To compute C = 2A - B: 2A = 2 * [[2, -1], [3, 4]] = [[4, -2], [6, 8]]. Then C = [[4, -2], [6, 8]] - [[1, 5], [-2, 0]] = [[4-1, -2-5], [6-(-2), 8-0]] = [[3, -7], [8, 8]].
2Compute the determinant of the 2x2 matrix A = [[5, -2], [3, 4]].
A.14
B.26
C.-26
D.20
Explanation: For a 2x2 matrix [[a, b], [c, d]], det(A) = a*d - b*c. Here, det(A) = (5)(4) - (-2)(3) = 20 - (-6) = 20 + 6 = 26.
3Given matrices A = [[1, 2], [3, 0]] and B = [[4, 1], [-1, 2]], calculate the product matrix AB.
A.[[2, 3], [12, 5]]
B.[[6, 5], [12, 3]]
C.[[2, 5], [12, 3]]
D.[[4, 2], [-3, 0]]
Explanation: Matrix multiplication AB: Row 1 * Col 1: 1(4) + 2(-1) = 4 - 2 = 2. Row 1 * Col 2: 1(1) + 2(2) = 1 + 4 = 5. Row 2 * Col 1: 3(4) + 0(-1) = 12 + 0 = 12. Row 2 * Col 2: 3(1) + 0(2) = 3 + 0 = 3. Thus, AB = [[2, 5], [12, 3]].
4If matrix A has dimensions 3x2 and matrix B has dimensions 2x4, what are the dimensions of the transposed product matrix (AB)^T?
A.3x4
B.2x2
C.3x2
D.4x3
Explanation: The product AB has dimensions (3x2) * (2x4) = 3x4. Transposing a 3x4 matrix swaps its rows and columns, giving dimensions 4x3.
5Calculate the trace of the 3x3 matrix A = [[4, -1, 3], [2, 7, 0], [1, 5, -2]].
A.9
B.13
C.7
D.11
Explanation: The trace of a square matrix is the sum of its main diagonal elements: Trace(A) = 4 + 7 + (-2) = 9.
6If matrix A is a 3x3 matrix with determinant det(A) = 5, what is the value of det(2A)?
A.10
B.40
C.15
D.125
Explanation: For an n x n matrix A and scalar k, det(kA) = k^n * det(A). Here n = 3 and k = 2, so det(2A) = 2^3 * det(A) = 8 * 5 = 40.
7Find the inverse of the matrix A = [[3, 1], [5, 2]].
A.[[2, 1], [5, 3]]
B.[[-2, 1], [5, -3]]
C.[[2, -1], [-5, 3]]
D.[[3, -5], [-1, 2]]
Explanation: For A = [[a, b], [c, d]], det(A) = ad - bc = 3(2) - 1(5) = 6 - 5 = 1. The inverse is A^-1 = (1/det(A)) * [[d, -b], [-c, a]] = [[2, -1], [-5, 3]].
8Solve the system of linear equations: 2x + 3y = 12 and 4x - y = 10. What is the value of x?
A.1
B.2
C.4
D.3
Explanation: From the second equation, y = 4x - 10. Substitute into the first equation: 2x + 3(4x - 10) = 12 => 2x + 12x - 30 = 12 => 14x = 42 => x = 3. (Then y = 4(3) - 10 = 2).
9Determine whether the point (3, 2) belongs to the feasible region defined by constraints: x >= 0, y >= 0, x + y <= 6, and 2x + y <= 8.
A.Yes, (3, 2) satisfies all constraints and is a feasible point
B.No, it violates x + y <= 6
C.No, it violates 2x + y <= 8
D.No, it violates non-negativity constraints
Explanation: Check constraints for x = 3, y = 2: 1) x >= 0, y >= 0 -> 3 >= 0, 2 >= 0 (True) 2) x + y = 3 + 2 = 5 <= 6 (True) 3) 2x + y = 2(3) + 2 = 6 + 2 = 8 <= 8 (True). Since all inequalities hold, (3, 2) is a valid feasible point.
10Solve the matrix equation AX = B for X, where A = [[1, 2], [0, 1]] and B = [[5, 6], [3, 4]].
A.[[1, 2], [3, 4]]
B.[[-1, -2], [3, 4]]
C.[[5, 6], [1, 2]]
D.[[-1, 2], [-3, 4]]
Explanation: Since det(A) = 1, A^-1 = [[1, -2], [0, 1]]. Multiply on the left: X = A^-1 * B = [[1, -2], [0, 1]] * [[5, 6], [3, 4]]. Row 1: [1*5 - 2*3, 1*6 - 2*4] = [5 - 6, 6 - 8] = [-1, -2]. Row 2: [0*5 + 1*3, 0*6 + 1*4] = [3, 4]. Thus X = [[-1, -2], [3, 4]].

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