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Row 2 is exactly twice row 1, so it contributes no new independent direction. The remaining rows (1,2,3) and (1,1,1) are not scalar multiples, so two independent rows remain and the rank is 2.
Eigenvalues λ satisfy Av = λv for a nonzero vector v, which is equivalent to (A - λI) being singular. Hence λ must make det(A - λI) vanish, and the roots of this characteristic polynomial are exactly the eigenvalues.
Setting f'(x) = 3x^2 - 3 = 0 gives critical points x = 1 and x = -1. The second derivative f''(x) = 6x is negative at x = -1, so that critical point is a local maximum, while x = 1 gives a local minimum.
Using the standard limit lim_{u→0} (sin u)/u = 1 with u = 3x, we get (sin 3x)/x = 3·(sin 3x)/(3x) → 3·1 = 3. The constant factor 3 passes through the limit.
Separating variables gives dy/y = 2x dx, and integrating both sides yields ln|y| = x^2 + C'. Exponentiating produces y = C e^{x^2}, where C absorbs the constant of integration.
By definition L{u(t)} = ∫_0^∞ e^{-st}·1 dt = [-e^{-st}/s]_0^∞ = 1/s for s > 0. The integral converges precisely in that half-plane.
Each face 1 through 6 has probability 1/6, so E[X] = (1+2+3+4+5+6)/6 = 21/6 = 3.5. The mean of a discrete uniform distribution on 1..n is (n+1)/2.
The determinant of an upper (or lower) triangular matrix equals the product of its diagonal entries. Here 2 × (-3) × 5 = -30, and the off-diagonal entries do not affect the result.
A p-series Σ 1/n^p converges exactly when p > 1. Since p = 2 > 1, convergence follows immediately from the p-series criterion, which is the standard and most direct justification.
The matrix is upper triangular, so both eigenvalues equal the repeated diagonal entry 2. Solving (A - 2I)v = 0 gives only the single eigenvector direction (1, 0), so the eigenspace is one-dimensional and the geometric multiplicity is 1 while the algebraic multiplicity is 2.
The absolute value function has limit 0 from both sides at x = 0, matching f(0) = 0, so it is continuous. However, the left derivative is -1 and the right derivative is +1, so the two-sided derivative does not exist and f is not differentiable at 0.
Newton-Raphson gives x1 = x0 - f(x0)/f'(x0). Here f(1) = -1 and f'(x) = 2x so f'(1) = 2, giving x1 = 1 - (-1)/2 = 1.5.
The single-panel trapezoidal rule gives (b - a)[f(a) + f(b)]/2 = (1)(0 + 1)/2 = 0.5. The exact integral is ∫_0^1 x^2 dx = 1/3 ≈ 0.333, so the rule overestimates because x^2 is convex.
Integrating the density gives P(X ≤ 0.5) = ∫_0^{0.5} 2x dx = [x^2]_0^{0.5} = 0.25. The cumulative distribution is F(x) = x^2 on [0,1].
For an odd function the cosine coefficients vanish because the product of an odd function with the even cosine integrates to zero over a symmetric interval. The sine coefficients generally survive, so the series reduces to sine terms only.
The standard pair L{sin ωt} = ω/(s^2 + ω^2) with ω = 2 gives L{sin 2t} = 2/(s^2 + 4). Dividing by 2 yields L^{-1}{1/(s^2+4)} = (1/2) sin 2t.
Adding the first two equations gives 2x + 3y + 4z = 9, which is exactly the third equation, so the third row is dependent and the coefficient matrix has rank 2 while the augmented rank is also 2. With three unknowns and rank 2, the system is consistent with one free parameter, giving infinitely many solutions.
The gradient is (2x, 2y), which at (1, 1) equals (2, 2). The directional derivative is the dot product of the gradient with the unit direction: (2, 2)·(1/√2, 1/√2) = (2 + 2)/√2 = 4/√2 = 2√2.
The characteristic equation is det(B - λI) = λ^2 + 1 = 0, whose roots are λ = ±i. This matrix represents a 90-degree rotation, which has no real eigenvalues but has the conjugate pair of imaginary eigenvalues.
The number of heads follows a binomial distribution with n = 4 and p = 1/2. The probability of exactly 2 heads is C(4,2)(1/2)^2(1/2)^2 = 6/16 = 3/8.
The gradient (2x, -2y) vanishes at the origin, so it is a critical point. The Hessian is diag(2, -2) with determinant -4 < 0, indicating a saddle: f increases along the x-axis and decreases along the y-axis.
Integrating in x first gives ∫_0^1 (x^2/2 + xy)|_{x=0}^{1} dy = ∫_0^1 (1/2 + y) dy = 1/2 + 1/2 = 1. By symmetry each of the two terms contributes 1/2.
The spectral theorem for real symmetric matrices guarantees real eigenvalues and an orthonormal basis of eigenvectors. In particular, eigenvectors corresponding to distinct eigenvalues are orthogonal, which follows from symmetry of the bilinear form.
Since x^3 is an odd function, its product with the even function cos(nx) is odd, and the integral of an odd function over the symmetric interval (-π, π) vanishes. Hence every cosine coefficient a_n equals zero, and the series contains only sine terms.
Separating variables gives dy/y^2 = dt, so -1/y = t + C, and y(0) = 1 yields C = -1, hence y = 1/(1 - t). The solution blows up as t approaches 1 from below, so the maximal interval of existence is t < 1.
The trace is the sum of the diagonal entries, 4 + 3 = 7. The determinant of a 2×2 matrix [[a, b], [c, d]] is ad − bc, giving 4·3 − 1·2 = 12 − 2 = 10. Hence trace 7 and determinant 10.
Differentiating with respect to y while holding x fixed gives ∂f/∂y = 3x^2 y^2 + x e^{xy}. At (1, 0) this is 3·1·0 + 1·e^0 = 0 + 1 = 1.
The probability the first ball is red is 3/8; given that, the probability the second is red is 2/7. Their product is 6/56 = 3/28. Equivalently, C(3,2)/C(8,2) = 3/28.
The gradient is (∂φ/∂x, ∂φ/∂y, ∂φ/∂z) = (2x, z, y). Evaluating at (1, 2, 3) gives (2, 3, 2). Note that the y-component is z = 3 and the z-component is y = 2.
Integrating by parts with u = x, dv = e^x dx gives ∫ x e^x dx = x e^x − e^x. Evaluating from 0 to 1 yields (e − e) − (0 − 1) = 1.
The second row is twice the first, so it adds no new independent information. The first row (1, 2, 3) and the third row (1, 1, 1) are not multiples of each other, so exactly two rows are linearly independent and the rank is 2.
The standard result is L{t^n} = n!/s^{n+1} for n a nonnegative integer. With n = 2 this gives 2!/s^3 = 2/s^3, valid for s > 0.
Setting f'(x) = 4x^3 − 8x = 4x(x^2 − 2) = 0 gives x = 0, ±√2. The second derivative f''(x) = 12x^2 − 8 is negative at x = 0 (local maximum) and positive at x = ±√2 (local minima).
Solving (A − 3I)v = 0 gives [[−2, 2], [2, −2]]v = 0, so v1 = v2. Thus any nonzero multiple of (1, 1) is an eigenvector for eigenvalue 3, and indeed A(1, 1) = (3, 3).
Newton-Raphson uses x_{n+1} = x_n − f(x_n)/f'(x_n). Here f'(x) = 3x^2, so x1 = 2 − (8 − 8)/(12) = 2, wait - with x0 = 2 the point is already the exact root. Recomputing with x0 = 3 as the intended guess is not the stated problem; for x0 = 2, f(2) = 0 and x1 = 2. Since the exact root is x = 2, the iteration stays at 2.
The function f(x) = x is odd on the symmetric interval (−π, π). For an odd function all cosine coefficients vanish and only sine terms (plus no constant term) remain in the Fourier series.
The integrand x dx + y dy is exact, equal to d((x^2 + y^2)/2). The integral therefore depends only on the endpoints: [(1^2 + 2^2) − (0 + 0)]/2 = 5/2.
For the exponential distribution with rate λ, the mean is E[X] = 1/λ and the variance is Var(X) = 1/λ^2. Both follow from integration by parts of the density, or from the moment generating function.
The second equation is exactly twice the first, so the two equations are dependent and represent the same line. There is one free variable, and the solution set is the infinite family x = 3 − 2y for any real y.
The inner integral over y gives xy + y^2/2 evaluated from 0 to x, which equals x^2 + x^2/2 = 3x^2/2. Integrating 3x^2/2 from 0 to 1 gives (3/2)(1/3) = 1/2. Rechecking: ∫_0^1 (3x^2/2) dx = [x^3/2]_0^1 = 1/2.
The alternating series with terms 1/n decreasing to zero converges by the alternating series test. However, the series of absolute values Σ 1/n is the harmonic series, which diverges, so the convergence is conditional.
The standard pair is L{cos ωt} = s/(s^2 + ω^2). Matching ω^2 = 9 gives ω = 3, so the inverse transform is cos 3t.
The gradient is (2x − 4, 2y), which vanishes at (2, 0). The Hessian is [[2, 0], [0, 2]], positive definite, so the critical point is a local minimum; completing the square gives f = (x − 2)^2 + y^2 − 4.
For the midpoint method, first take a half-step: y(0.25) ≈ 1 + 0.25(0 + 1) = 1.25. Then y(0.5) ≈ 1 + 0.5·f(0.25, 1.25) = 1 + 0.5(0.25 + 1.25) = 1 + 0.75 = 1.75. Rechecking with the standard predictor–corrector form gives 1.75, so the keyed 1.375 is not reproducible and the item is flagged.
The existence of n linearly independent eigenvectors is exactly the condition for A to be diagonalizable, since A = P D P^{−1} with P the eigenvector matrix. Symmetry is sufficient but not necessary for this property.
Near z = i, factor z^2 + 1 = (z − i)(z + i), so f(z) = 1/[(z − i)(z + i)]. The coefficient of 1/(z − i) is the value of 1/(z + i) at z = i, which is 1/(2i). This is the residue of f at the simple pole z = i.
Separating variables gives ∫ y^{−2} dy = ∫ dt, so −1/y = t + C. The condition y(0) = 1 gives C = −1, hence y = 1/(1 − t), which tends to infinity as t → 1^−. The maximal forward interval is therefore [0, 1).
The integrand has a pole of order 3 at z = 1 inside C. By the residue formula for a pole of order n, the residue is (1/2!)·d^2/dz^2[e^z] at z = 1, which equals e/2. Multiplying by 2πi gives π i e.
A repeated eigenvalue does not by itself decide diagonalizability; it depends on whether the eigenspace for eigenvalue 1 has dimension 2. If the geometric multiplicity is 2 the matrix is diagonalizable, and if it is 1 the matrix is defective.
The characteristic equation is (2−λ)^2 − 1 = 0, giving 2−λ = ±1, so λ = 1 or λ = 3. Equivalently, trace 4 equals the sum and determinant 3 equals the product of the eigenvalues.