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100+ Free Valencian Community PAU Applied Mathematics for Social Sciences II Practice Questions
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Sample Valencian Community PAU Applied Mathematics for Social Sciences II Practice Questions
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1Matrix A has dimensions 2 x 3 and Matrix B has dimensions 3 x 4. What are the dimensions of the product matrix C = A · B?
A.2 x 4
B.3 x 3
C.4 x 2
D.The matrix product A · B is undefined
Explanation: Matrix multiplication A · B is defined when the number of columns in A (3) equals the number of rows in B (3). The resulting product matrix C has dimensions equal to the number of rows of A and the number of columns of B, which is 2 x 4.
2Given the matrix A = [[3, -2], [4, 5]], calculate the determinant det(A).
A.7
B.15
C.23
D.-7
Explanation: For a 2 x 2 matrix [[a, b], [c, d]], the determinant is det(A) = ad - bc. Here, det(A) = (3)(5) - (-2)(4) = 15 - (-8) = 15 + 8 = 23.
3Compute the inverse of matrix A = [[2, 1], [4, 3]].
A.[[3, -1], [-4, 2]]
B.[[1.5, -0.5], [-2, 1]]
C.[[-1.5, 0.5], [2, -1]]
D.[[0.5, 1], [0.25, 0.33]]
Explanation: The determinant det(A) = (2)(3) - (1)(4) = 6 - 4 = 2. The formula for a 2 x 2 inverse is A^(-1) = (1/det(A)) * [[d, -b], [-c, a]] = (1/2) * [[3, -1], [-4, 2]] = [[1.5, -0.5], [-2, 1]].
4Solve the matrix equation A · X = B for X = [[x], [y]], where A = [[1, 2], [3, 4]] and B = [[5], [11]].
A.x = 1, y = 2
B.x = 2, y = 1
C.x = 3, y = 1
D.x = 5, y = 0
Explanation: det(A) = (1)(4) - (2)(3) = -2. A^(-1) = (-1/2) * [[4, -2], [-3, 1]] = [[-2, 1], [1.5, -0.5]]. Multiplying X = A^(-1) · B gives [[-2(5) + 1(11)], [1.5(5) - 0.5(11)]] = [[-10 + 11], [7.5 - 5.5]] = [[1], [2]]. Thus x = 1, y = 2.
5Which of the following matrix identity properties is correct for conformable matrices A and B?
A.(A · B)^T = A^T · B^T
B.A · B = B · A for all square matrices
C.det(A + B) = det(A) + det(B)
D.(A · B)^T = B^T · A^T
Explanation: The transpose of a matrix product reverses the order of multiplication: (A · B)^T = B^T · A^T. Matrix multiplication is generally non-commutative (A · B != B · A), and determinants are non-linear under addition (det(A + B) != det(A) + det(B)).
6Calculate the determinant of the 3 x 3 matrix A = [[1, 0, 2], [2, -1, 3], [4, 1, 8]].
A.1
B.0
C.-5
D.12
Explanation: Expanding along the first row: det(A) = 1*((-1)(8) - (3)(1)) - 0 + 2*((2)(1) - (-1)(4)) = 1*(-8 - 3) + 2*(2 + 4) = 1*(-11) + 2*(6) = -11 + 12 = 1.
7For what values of parameter k is the matrix A = [[k, 4], [2, k]] singular (non-invertible)?
A.k = 2 only
B.k = +/- 2*sqrt(2)
C.k = +/- 8
D.k = 0
Explanation: A matrix is singular when its determinant is zero: det(A) = k^2 - (4)(2) = k^2 - 8 = 0. Solving k^2 = 8 yields k = +/- sqrt(8) = +/- 2*sqrt(2).
8If A is a 3 x 3 square matrix with det(A) = 5, what is the determinant of 2A?
A.10
B.15
C.40
D.125
Explanation: For an n x n matrix and scalar c, det(c A) = c^n * det(A). For n = 3 and c = 2: det(2A) = 2^3 * det(A) = 8 * 5 = 40.
9A system of 3 linear equations in 3 unknowns has coefficient matrix A with rank(A) = 2 and augmented matrix (A|B) with rank(A|B) = 3. According to the Rouché-Capelli Theorem, how many solutions does the system have?
A.Zero solutions (Inconsistent system / Sistema incompatible)
B.Exactly one unique solution (Consistent determined / Sistema compatible determinado)
C.Infinitely many solutions depending on 1 parameter (Consistent indeterminate)
D.Infinitely many solutions depending on 2 parameters
Explanation: According to the Rouché-Capelli Theorem, a system is consistent if and only if rank(A) = rank(A|B). When rank(A) = 2 != rank(A|B) = 3, the system has no solution and is classified as an inconsistent system (sistema incompatible).
10Consider the system of linear equations: x + y + z = 1, x + 2y + 3z = 2, 2x + 3y + k z = 3. For what value of parameter k does the system have infinitely many solutions?
A.k = 0
B.k = 2
C.k = 4
D.k = 6
Explanation: The determinant of A is det(A) = 1(2k - 9) - 1(k - 6) + 1(3 - 4) = k - 4. Setting det(A) = 0 yields k = 4. At k = 4, rank(A) = 2. Notice that Equation 3 = Equation 1 + Equation 2 (2x + 3y + 4z = 3), so rank(A|B) is also 2. Since rank(A) = rank(A|B) = 2 < 3 variables, k = 4 gives infinitely many solutions.
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