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100+ Free Canary Islands PAU Mathematics II / Matemáticas II COPAU ULPGC ULL Practice Questions
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Sample Canary Islands PAU Mathematics II / Matemáticas II COPAU ULPGC ULL Practice Questions
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1Given the matrices A = [[2, 1], [-1, 3]] and B = [[1, -2], [0, 4]], what is the result of the matrix operation 2A - 3B?
A.[[1, 8], [-2, -6]]
B.[[1, 8], [-2, 18]]
C.[[3, -4], [-2, -6]]
D.[[1, 4], [-2, -6]]
Explanation: To calculate 2A - 3B, multiply matrix A by scalar 2: 2A = [[4, 2], [-2, 6]], and multiply matrix B by scalar 3: 3B = [[3, -6], [0, 12]]. Subtracting elementwise gives: 2A - 3B = [[4-3, 2-(-6)], [-2-0, 6-12]] = [[1, 8], [-2, -6]].
2Let A = [[1, 2], [3, 4]] and B = [[2, 0], [1, 5]]. Compute the matrix product A * B.
A.[[4, 10], [10, 20]]
B.[[2, 0], [3, 20]]
C.[[4, 10], [6, 20]]
D.[[3, 10], [10, 20]]
Explanation: Matrix multiplication is computed by taking row-column dot products. Row 1: [1*2 + 2*1, 1*0 + 2*5] = [4, 10]. Row 2: [3*2 + 4*1, 3*0 + 4*5] = [10, 20]. Thus, A * B = [[4, 10], [10, 20]].
3If A is a 3x3 matrix with determinant |A| = -4, what is the determinant of the matrix 3A?
A.-108
B.-36
C.-12
D.108
Explanation: For an n x n matrix A and scalar k, |k A| = k^n * |A|. For a 3x3 matrix (n = 3) and scalar k = 3: |3A| = 3^3 * |A| = 27 * (-4) = -108.
4Calculate the determinant of the matrix A = [[1, 0, 2], [3, -1, 1], [2, 0, 4]].
A.-4
B.4
C.0
D.2
Explanation: Expand along the second column, whose only non-zero entry is a₂₂ = -1 with cofactor sign (+1)^(2+2) = +1: |A| = (-1) · |[1, 2], [2, 4]| = (-1) · (1·4 - 2·2) = (-1) · 0 = 0. The quickest route is to notice that Row 3 = 2 · Row 1, so the rows are linearly dependent and the determinant is automatically 0.
5A sample of n = 64 light bulbs is drawn from a normal population with known standard deviation σ = 40 hours, and the sample mean lifetime is x̄ = 1200 hours. Construct the 95% confidence interval for the population mean μ. (Use z_(α/2) = 1.96.)
A.(1160.0, 1240.0)
B.(1190.2, 1209.8)
C.(1198.8, 1201.2)
D.(1191.8, 1208.2)
Explanation: For a normal population with known σ, the confidence interval for μ is x̄ ± z_(α/2) · σ/√n. The standard error is σ/√n = 40/√64 = 40/8 = 5, so the margin of error is 1.96 · 5 = 9.8. The interval is 1200 ± 9.8 = (1190.2, 1209.8).
6Given A = [[2, 1], [5, 3]], calculate its inverse matrix A^(-1).
A.[[2, -1], [-5, 3]]
B.[[-3, 1], [5, -2]]
C.[[3, -1], [-5, 2]]
D.[[3, 5], [1, 2]]
Explanation: For a 2x2 matrix A = [[a, b], [c, d]], |A| = ad - bc = 2*3 - 1*5 = 1. The inverse is A^(-1) = (1/|A|) * [[d, -b], [-c, a]] = [[3, -1], [-5, 2]].
7Solve for the matrix X in the equation A * X + B = C, assuming matrix A is invertible.
A.X = (B - C) * A^(-1)
B.X = A^(-1) * (C - B)
C.X = A^(-1) * C - B
D.X = (C - B) * A^(-1)
Explanation: Subtract B from both sides: A * X = C - B. Left-multiply both sides by A^(-1): A^(-1) * (A * X) = A^(-1) * (C - B) ⇒ X = A^(-1) * (C - B). Left-multiplication is required because matrix multiplication is non-commutative.
8What is the rank of the matrix M = [[1, 2, 3], [2, 4, 6], [3, 6, 9]]?
A.0
B.2
C.3
D.1
Explanation: Row 2 is 2 * Row 1 and Row 3 is 3 * Row 1. All 2x2 and 3x3 minors equal 0. Since M contains non-zero entries, rank(M) = 1.
9According to the Rouché-Capelli Theorem, a system of 3 linear equations with 3 unknowns is consistent and dependent (compatible indeterminado) if and only if:
A.rank(A) = rank(A|B) = 3
B.rank(A) = 3 and rank(A|B) = 2
C.rank(A) = rank(A|B) < 3
D.rank(A) < rank(A|B)
Explanation: By Rouché-Capelli: a system is consistent if rank(A) = rank(A|B). If this common rank r is strictly less than the number of unknowns n (n = 3), the system is compatible indeterminado (infinitely many solutions).
10Consider the system: x + y + z = 6, 2x - y + z = 3, x + 2y - z = 2. Find the value of x using Cramer's rule.
A.1
B.2
C.3
D.4
Explanation: Coefficient matrix A = [[1, 1, 1], [2, -1, 1], [1, 2, -1]]. |A| = 1(1 - 2) - 1(-2 - 1) + 1(4 - (-1)) = -1 + 3 + 5 = 7. Replacing column 1 with B = [6, 3, 2]^T: |A_x| = |[6, 1, 1], [3, -1, 1], [2, 2, -1]| = 6(1 - 2) - 1(-3 - 2) + 1(6 - (-2)) = -6 + 5 + 8 = 7. Thus x = |A_x| / |A| = 7 / 7 = 1.
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