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The Cartesian product A × B contains every ordered pair (x, y) with x in A and y in B, so its size is |A| × |B|. Here 3 × 2 = 6. Order matters, so (1, a) and (a, 1) are not both elements unless a is in A and 1 is in B.
R is reflexive because each of 1, 2, 3 is related to itself, and symmetric because (1, 2) and (2, 1) both appear. However, 1 R 2 and 2 R 1 hold while 1 ≠ 2, which violates antisymmetry. Transitivity also holds for this relation.
To invert, write y = 3x − 5 and solve for x: x = (y + 5)/3. Swapping the variable name gives f⁻¹(x) = (x + 5)/3. The function is a bijection on ℝ, so this inverse is valid for all real x.
A relation is a function when every element of the domain A is paired with exactly one element of B. Only the set pairing 1→4, 2→5, 3→6 assigns a single image to each of 1, 2 and 3.
The composite (g ∘ f)(x) means g(f(x)), so substitute f(x) into g: g(x + 2) = (x + 2)². Expanding gives x² + 4x + 4. Note that composition is not commutative, so this differs from (f ∘ g)(x) = x² + 2.
Divisibility of a − b by 5 is reflexive (a − a = 0), symmetric (if 5 divides a − b then it divides b − a) and transitive (adding multiples of 5 gives a multiple of 5). The classes are the five residue classes modulo 5, so R is an equivalence relation.
Completing the square gives f(x) = (x − 2)² + 3, whose minimum value is 3 at x = 2, so the range is [3, ∞). Substituting x = 3 gives 9 − 12 + 7 = 4.
If f(a₁) = f(a₂), then applying g gives (g ∘ f)(a₁) = (g ∘ f)(a₂), and injectivity of g ∘ f forces a₁ = a₂, so f is injective. Nothing forces g to be injective, because g could identify elements of B that are not in the image of f.
The absolute value function is not injective because f(−2) = f(2) = 2, so distinct inputs give the same output. It is not surjective onto ℝ either, since |x| ≥ 0 never takes negative values.
Transitivity requires that whenever (a, b) and (b, c) are in R, the pair (a, c) is also present. From (1, 2) and (2, 3) we must add (1, 3); no other pair is forced, so the transitive closure is {(1, 2), (2, 3), (1, 3)}.
Set f(x) = 1 and solve: (2x − 3)/5 = 1 gives 2x − 3 = 5, so 2x = 8 and x = 4. Equivalently, f⁻¹(x) = (5x + 3)/2, which yields 4 when x = 1.
Here (f ∘ g)(x) = √((x + 3) − 2) = √(x + 1), which is defined only when x + 1 ≥ 0, that is x ≥ −1. The domain is therefore [−1, ∞).
Additivity plus continuity at one point forces f to be linear: f(x) = kx where k = f(1). Without a regularity condition such as continuity or boundedness, wild additive functions exist. Continuity at a single point is enough to rule them out.
Count pairs (a, b) with a | b: for a = 1 there are four (1,1),(1,2),(1,3),(1,4); for a = 2 there are two (2,2),(2,4); for a = 3 there is one (3,3); for a = 4 there is one (4,4). The total is 4 + 2 + 1 + 1 = 8.
The derivative 3x² + 1 is always positive, so f is strictly increasing and hence injective. As x → ±∞, f(x) → ±∞, so by continuity f takes every real value, making it surjective. Thus f is a bijection, and its inverse is increasing because f is.
Equivalence classes partition A, so two classes are either identical or disjoint. Since a and b are related, they lie in the same class, hence [a] = [b]. Sharing exactly one element would force the classes to be equal, not merely overlapping.
The function sin is one-to-one and onto [−1, 1] when restricted to [−π/2, π/2], so this closed interval is taken as the range of the principal branch of sin⁻¹. Endpoints are included because sin(±π/2) = ±1 are attained.
The principal range of cos⁻¹ is [0, π], and cos(2π/3) = −1/2, so the principal value is 2π/3. The value must lie in the second quadrant because the cosine is negative there and the range restricts to [0, π].
For positive x, y with xy > 1, tan⁻¹x + tan⁻¹y = π + tan⁻¹((x+y)/(1−xy)). Pairing tan⁻¹2 + tan⁻¹3 gives π + tan⁻¹(−1) = 3π/4, and adding tan⁻¹1 = π/4 yields π.
By definition of the inverse sine, y is the unique angle in the principal range [−π/2, π/2] whose sine equals x. Both conditions, the equation and the range restriction, are needed to make the inverse single-valued.
The principal range of sin⁻¹ is [−π/2, π/2], so 2π/3 is not admissible. Since sin(2π/3) = sin(π/3) = √3/2 and π/3 lies in the principal range, the value is π/3.
Apply the tangent sum formula: (2x+3x)/(1−6x²) = 1, giving 6x² + 5x − 1 = 0, i.e. x = 1/6 or x = −1. Both values keep 2x and 3x inside the domain and are rejected only if the product condition 6x² > 1 fails; for x = 1/6 and x = −1 it is satisfied, so both survive.
Cosine is even, so cos(−7π/6) = cos(7π/6) = −√3/2. The principal range of cos⁻¹ is [0, π], and the angle in that interval with cosine −√3/2 is 5π/6.
Let θ = tan⁻¹x, so tan θ = x and sin θ = x/√(1+x²), cos θ = 1/√(1+x²). Then sin 2θ = 2 sin θ cos θ = 2x/(1+x²), valid for every real x.
For x > 0, cot⁻¹x = tan⁻¹(1/x), and tan⁻¹x + tan⁻¹(1/x) = π/2 since x·(1/x) = 1 makes the tangent sum formula's denominator zero, forcing a right angle. Hence the two angles are complementary.
Testing x = 0 gives 0 + π/2 = π/2 on the left and sin⁻¹1 = π/2 on the right, so it works; x = 1 gives π/2 + 0 = π/2 and sin⁻¹0 = 0, which fails, so the valid solution set must be checked term by term. Careful substitution shows x = 0 satisfies the equation while x = 1 does not, leaving x = 0 as the sole solution.
A fundamental identity for square matrices is A·adj(A) = adj(A)·A = |A|·I. This holds for every square matrix, including singular ones; if A is singular then |A| = 0 and the product is the zero matrix. It is the basis for the formula A⁻¹ = adj(A)/|A| when |A| ≠ 0.
From A² = A we get A(A − I) = O. If A were invertible, multiplying by A⁻¹ would give A − I = O, i.e. A = I, contradicting the hypothesis. Hence A cannot be invertible and must be singular, so its determinant is zero.
For an n×n matrix, |adj(A)| = |A|^(n−1). Here n = 3, so |adj(A)| = 5^(3−1) = 5² = 25. This follows from taking determinants in A·adj(A) = |A|·I.
Scaling an n×n matrix by a scalar k multiplies its determinant by kⁿ. With n = 3 and k = 2, |2A| = 2³·|A| = 8·4 = 32. Each row is multiplied by 2, and each row scaling contributes a factor of 2.
The determinant of an inverse equals the reciprocal of the determinant: |A⁻¹| = 1/|A|. Since |A| = −2, we get |A⁻¹| = −1/2, and the sign is preserved through the reciprocal.
Taking determinants of Aᵀ = −A gives |A| = (−1)ⁿ|A|. For odd n, (−1)ⁿ = −1, so |A| = −|A|, which forces |A| = 0. A skew-symmetric matrix of odd order is therefore always singular.
The second equation is 2(x + 2y) = 7, which would require x + 2y = 3.5, contradicting x + 2y = 3. With |A| = 0 and inconsistent constants, the system has no solution; a linear system can never have exactly two solutions.
The coefficient determinant is (3)(−1) − (2)(1) = −5. Replacing the x-column with constants gives |Aₓ| = (7)(−1) − (2)(1) = −9, so x = |Aₓ|/|A| = −9/−5 = 9/5. Substituting back confirms y = 4/5.
For an n×n matrix, |adj(A)| = |A|^(n−1). With n = 2 the exponent is 1, so |adj(A)| = |A| = 6. For a 2×2 matrix the adjugate simply swaps diagonal entries and negates the others, which leaves the determinant unchanged.
Rewriting the equation as A² − 5A = −6I gives A(A − 5I) = −6I, hence A·[(5I − A)/6] = I. This exhibits an explicit inverse, so A is invertible with A⁻¹ = (5I − A)/6. The determinant of A is not fixed by the equation alone.
The determinant is multiplicative: |AB| = |A|·|B| for square matrices of the same order. The other expansions fail in general because matrix multiplication is not commutative, so cross terms such as AB and BA need not cancel or combine.
Interchanging any two rows of a matrix multiplies its determinant by −1. A single row swap therefore gives |B| = −|A| = −7. Two swaps would restore the original sign.
Applying |adj(M)| = |M|^(n−1) twice: first |adj(A)| = 2² = 4, then |adj(adj(A))| = 4² = 16. Equivalently, |adj(adj(A))| = |A|^((n−1)²) = 2⁴ = 16.
A square matrix has determinant zero exactly when it is singular, which occurs precisely when its rows (and columns) are linearly dependent. This is equivalent to saying the matrix has no inverse and its rank is less than its order.
First A·adj(A) = |A|·I = 3I. Then 2A·adj(A) = 2·(3I) = 6I, and |6I| for order 3 is 6³ = 216. Equivalently, |2A|·|adj(A)| = (8·3)·(3²) = 24·9 = 216.
For a 2×2 matrix, the adjugate is obtained by swapping the diagonal entries a and d while negating the off-diagonal entries b and c. This gives adj(A) = [[d, −b], [−c, a]], whose product with A equals (ad − bc)I.
Adding a multiple of one row to another is an elementary row operation that leaves the determinant unchanged. Hence |B| = |A| = 5. Only row swaps and row scalings alter the determinant value.
The identity A·adj(A) = |A|·I holds for every square matrix. Comparing with A·adj(A) = 4I gives |A| = 4 directly, with no further exponent needed.
The inverse of a product reverses the order of the factors: (AB)⁻¹ = B⁻¹A⁻¹. Checking, (AB)(B⁻¹A⁻¹) = A(BB⁻¹)A⁻¹ = AA⁻¹ = I. Order matters because matrix multiplication is not commutative.
The determinant is the product of eigenvalues, so |A| = 1·2·3 = 6. Then |adj(A)| = |A|^(n−1) = 6² = 36 for order 3. The eigenvalues of adj(A) are the products of pairs of eigenvalues of A, whose product is also 36.
The modulus function is continuous everywhere, so the left and right limits both equal f(3) = 0. However the left derivative is -1 and the right derivative is +1, so the two one-sided derivatives disagree and the derivative fails to exist at the corner point x = 3.
At every integer n the left limit of the greatest integer function is n - 1 while the right limit and value equal n, so the limit does not exist. Between consecutive integers the function is constant and therefore continuous.
By the chain rule, the derivative of sin(u) is cos(u) times u'. Here u = 3x so u' = 3, giving f'(x) = 3cos(3x).
Differentiating gives f'(x) = 2x - 4, which is positive exactly when x > 2. Hence the function is strictly increasing on (2, infinity) and decreasing on (-infinity, 2).