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2026 Statistics
Key Facts: Valencian PAU Mathematics II Exam
90 Minutes
Exam duration
Universitats GVA / UPV / UV
EUR 78.37
Base registration fee for Access Phase
Generalitat Valenciana
0–10 Scale
Grading scale for Bachillerato and PAU exams
Spanish Ministry of Education
4 Core Blocks
Linear Systems, Calculus, 3D Geometry, Probability
LOMLOE Mathematics II Syllabus
100 Questions
Practice bank size in OpenExamPrep
OpenExamPrep
Valencian Community PAU Mathematics II (Universitats GVA 2026) is a 90-minute university admission exam covering 2º Bachillerato Science & Technology Mathematics. This bank provides an English-language MCQ study adaptation.
Sample Valencian PAU Mathematics II Practice Questions
Try these sample questions to test your Valencian PAU Mathematics II exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1Given the matrices A = [[1, 2], [0, 1]] and B = [[2, 0], [1, 3]], what is the result of the matrix square (A + B)^2?
A.[[11, 14], [7, 18]]
B.[[9, 12], [5, 16]]
C.[[11, 12], [6, 18]]
D.[[10, 14], [7, 16]]
Explanation: First calculate A + B = [[1+2, 2+0], [0+1, 1+3]] = [[3, 2], [1, 4]]. Next compute (A + B)^2 = [[3, 2], [1, 4]] * [[3, 2], [1, 4]] = [[3*3 + 2*1, 3*2 + 2*4], [1*3 + 4*1, 1*2 + 4*4]] = [[11, 14], [7, 18]]. Note that in general matrix algebra, (A+B)^2 = A^2 + AB + BA + B^2, which equals [[11, 14], [7, 18]] because matrix multiplication is non-commutative.
2For which values of the real parameter 'a' does the matrix A = [[a, 1, 0], [1, a, 1], [0, 1, a]] fail to be invertible?
A.a = 0, a = sqrt(2), a = -sqrt(2)
B.a = 0, a = 1, a = -1
C.a = 1, a = 2, a = -2
D.a = 0, a = 2, a = -2
Explanation: A square matrix fails to be invertible if and only if its determinant is zero. Computing det(A) by expanding along the first row yields det(A) = a*(a^2 - 1) - 1*(a - 0) + 0 = a^3 - a - a = a^3 - 2a = a*(a^2 - 2). Setting det(A) = 0 gives a*(a^2 - 2) = 0, which has solutions a = 0, a = sqrt(2), and a = -sqrt(2).
3Compute the inverse matrix A^(-1) for A = [[1, 0, 2], [2, -1, 3], [4, 1, 8]].
A.[[-11, 2, 2], [-4, 0, 1], [6, -1, -1]]
B.[[-11, -4, 6], [2, 0, -1], [2, 1, -1]]
C.[[11, -2, -2], [4, 0, -1], [-6, 1, 1]]
D.[[-8, 2, 2], [-4, 1, 1], [6, -1, 0]]
Explanation: First compute det(A) expanding by row 1: det(A) = 1*((-1)*8 - 3*1) + 2*(2*1 - (-1)*4) = 1*(-11) + 2*(6) = 1. Since det(A) = 1, A^(-1) equals the transpose of the cofactor matrix adj(A). Cofactors: C11 = -11, C12 = -4, C13 = 6, C21 = 2, C22 = 0, C23 = -1, C31 = 2, C32 = 1, C33 = -1. Transposing the cofactor matrix yields A^(-1) = [[-11, 2, 2], [-4, 0, 1], [6, -1, -1]].
4What is the rank of the matrix A = [[1, 2, 3], [2, 4, k], [3, 6, 9]] as a function of the parameter 'k'?
A.Rank is 1 if k = 6, and rank is 2 if k != 6.
B.Rank is 1 for all real values of k.
C.Rank is 2 if k = 6, and rank is 3 if k != 6.
D.Rank is 3 for all k != 6, and rank is 0 if k = 6.
Explanation: Observe that Row 3 = 3 * Row 1, so Row 3 is always linearly dependent on Row 1. Performing row operations R2 -> R2 - 2*R1 yields the second row [0, 0, k - 6]. If k = 6, the second row becomes [0, 0, 0], leaving only 1 non-zero row, so rank(A) = 1. If k != 6, the second row has a non-zero third entry, giving 2 linearly independent rows, so rank(A) = 2. Rank can never be 3 because all 3x3 minors have determinant zero (since col 2 = 2 * col 1).
5Find the matrix X that satisfies the matrix equation A * X = B, where A = [[2, 1], [5, 3]] and B = [[4, 1], [1, 2]].
A.[[11, 1], [-18, -1]]
B.[[7, 5], [-19, -3]]
C.[[-11, -1], [18, 1]]
D.[[11, -18], [1, -1]]
Explanation: Since det(A) = 2*3 - 1*5 = 1 != 0, A is invertible and X = A^(-1) * B. The inverse A^(-1) = (1/det(A)) * [[3, -1], [-5, 2]] = [[3, -1], [-5, 2]]. Multiplying A^(-1) * B = [[3, -1], [-5, 2]] * [[4, 1], [1, 2]] = [[3*4 + (-1)*1, 3*1 + (-1)*2], [(-5)*4 + 2*1, (-5)*1 + 2*2]] = [[11, 1], [-18, -1]].
6If A is a 3x3 square matrix with determinant det(A) = 4, what is the determinant of the matrix 2 * A^(-1)?
A.2
B.1/2
C.8
D.1
Explanation: Using determinant properties for an n x n matrix, det(k * M) = k^n * det(M). For a 3x3 matrix (n=3), det(2 * A^(-1)) = 2^3 * det(A^(-1)) = 8 * (1 / det(A)). Substituting det(A) = 4 yields det(2 * A^(-1)) = 8 * (1/4) = 2.
7Solve for matrix X in X * A + B = C, where A = [[1, 2], [0, 1]], B = [[1, 0], [2, 3]], and C = [[3, 4], [2, 5]].
A.[[2, 0], [0, 2]]
B.[[2, 4], [0, 2]]
C.[[2, -4], [0, 2]]
D.[[1, 0], [0, 1]]
Explanation: Isolate X * A: X * A = C - B = [[3-1, 4-0], [2-2, 5-3]] = [[2, 4], [0, 2]]. Multiply on the right by A^(-1): X = (C - B) * A^(-1). The inverse of A is A^(-1) = [[1, -2], [0, 1]]. Thus X = [[2, 4], [0, 2]] * [[1, -2], [0, 1]] = [[2*1 + 4*0, 2*(-2) + 4*1], [0*1 + 2*0, 0*(-2) + 2*1]] = [[2, 0], [0, 2]] = 2 * I_2.
8If A is an idempotent square matrix (A^2 = A) and A is neither the zero matrix nor the identity matrix, what are all possible eigenvalues of A?
A.0 and 1
B.1 and -1
C.0 and -1
D.Only 1
Explanation: Let lambda be an eigenvalue of A with non-zero eigenvector v, so A * v = lambda * v. Multiplying by A gives A^2 * v = A * (lambda * v) = lambda * (A * v) = lambda^2 * v. Since A^2 = A, we have A^2 * v = A * v = lambda * v. Therefore lambda^2 * v = lambda * v, which implies (lambda^2 - lambda) * v = 0. Since v != 0, lambda^2 - lambda = 0, giving lambda * (lambda - 1) = 0. The only possible eigenvalues are lambda = 0 and lambda = 1.
9If A and B are 3x3 matrices with det(A) = -3 and det(B) = 5, what is the value of det(A^T * B)?
A.-15
B.15
C.-45
D.-2
Explanation: By properties of determinants, det(M * N) = det(M) * det(N) and det(A^T) = det(A). Therefore, det(A^T * B) = det(A^T) * det(B) = det(A) * det(B) = (-3) * 5 = -15.
10Any square matrix A can be uniquely written as A = S + K, where S is symmetric and K is skew-symmetric. If A = [[2, 5], [1, 4]], find the skew-symmetric component K.
A.[[0, 2], [-2, 0]]
B.[[2, 3], [3, 4]]
C.[[0, 3], [-3, 0]]
D.[[0, 1], [-1, 0]]
Explanation: The unique decomposition of a square matrix A into symmetric part S and skew-symmetric part K is given by S = (1/2)*(A + A^T) and K = (1/2)*(A - A^T). Here A^T = [[2, 1], [5, 4]]. Thus A - A^T = [[2-2, 5-1], [1-5, 4-4]] = [[0, 4], [-4, 0]]. Multiplying by 1/2 yields K = [[0, 2], [-2, 0]].
About the Valencian PAU Mathematics II Practice Questions
Verified exam format metadata for Valencian Community PAU Mathematics II Examination — Universitats GVA (Matemáticas II 2026) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.