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For a square matrix of order n, the determinant of the adjoint equals |A| raised to the power n minus 1. Here n = 3 and |A| = 5, so |adj A| = 5^(3-1) = 5^2 = 25.
The fundamental adjoint identity states A(adj A) = (adj A)A = |A|·I for any square matrix A. Comparing with kI gives k = |A|, independent of the order of the matrix.
A square matrix has an inverse precisely when it is non-singular, that is, when its determinant is nonzero. The formula A⁻¹ = (1/|A|)(adj A) fails if |A| = 0 because division by zero is undefined.
Taking determinants in AA⁻¹ = I gives |A|·|A⁻¹| = |I| = 1, so |A⁻¹| = 1/|A|. This requires |A| ≠ 0, which holds because A is non-singular.
The coefficient determinant is (1)(4) - (2)(2) = 0, so the system is not uniquely solvable. Checking consistency, the second equation is 2 times the first left side but not 2 times the right side (6 ≠ 7), so the lines are parallel and the system is inconsistent.
For a square matrix of order n, multiplying every entry by a scalar k multiplies the determinant by k^n. Here k = 2 and n = 3, so |2A| = 2^3 · |A| = 8 · 4 = 32.
For order n, |adj A| = |A|^(n-1), so |adj A| = 2^2 = 4. Applying the rule again to the order-3 matrix adj A gives |adj(adj A)| = 4^2 = 16, equivalently |A|^((n-1)^2).
The coefficient determinant is 4k - 6. A unique solution exists exactly when this determinant is nonzero, that is, 4k ≠ 6, giving k ≠ 3/2. At k = 3/2 the rows are proportional and the system is not uniquely solvable.
First note A(adj A) = |A|·I = 3I, an order-2 matrix. Scaling by 2 gives 2·3I = 6I, and the determinant of 6I in order 2 is 6^2 = 36. Alternatively |2A(adj A)| = 2^2 · |A| · |adj A| = 4 · 3 · 3 = 36.
Taking determinants of A² = A gives |A|² = |A|. Since |A| ≠ 0 we may divide by |A| to obtain |A| = 1. Such a matrix is idempotent and non-singular, so it must be the identity.
By Cramer's rule, a nonzero coefficient determinant guarantees a unique solution regardless of the constants on the right side. Since the determinant is 2 ≠ 0, the three planes meet in exactly one point, found by solving with determinants.
The rule |adj A| = |A|^(n-1) with n = 3 gives (-2)^2 = 4. Squaring removes the sign, so the adjoint determinant is positive even though |A| is negative.
From A⁻¹ = (1/|A|)(adj A) and the fact that the inverse of the inverse returns A, we get A = |A| · (adj A)⁻¹. For a 2×2 matrix [[a,b],[c,d]] the inverse is (1/(ad-bc))[[d,-b],[-c,a]]; here adj A has determinant 12 - 2 = 10, so (adj A)⁻¹ = (1/10)[[4,1],[2,3]], and A = 10 · (1/10)[[4,1],[2,3]] = [[4,1],[2,3]].
A standard result gives adj(adj A) = |A|^(n-2) · A for a non-singular matrix of order n. With n = 3 the exponent is 1, so adj(adj A) = |A|·A. This follows by replacing A with adj A in the identity adj B = |B|·B⁻¹ and simplifying.
Here B = adj A, so |B| = |A|^(3-1) = 25. Then |adj B| = |B|^(3-1) = 25^2 = 625. Equivalently, |adj(adj A)| = |A|^((n-1)^2) = 5^4 = 625.
Interchanging two rows of a square matrix multiplies its determinant by −1, so |B| = −|A| = −2. Also |AB⁻¹| = |A|·|B⁻¹| = |A|/|B| = 2/(−2) = −1, using the product rule and the fact that the determinant of an inverse is the reciprocal of the determinant.
The modulus of a + bi is √(a² + b²), so |3 + 4i| = √(9 + 16) = √25 = 5. The value 25 is the squared modulus, not the modulus itself.
The conjugate of a + bi is a − bi: the real part is unchanged and the imaginary part changes sign. So the conjugate of −2 + 5i is −2 − 5i.
Powers of i repeat with period 4, so reduce the exponent modulo 4: 23 = 4·5 + 3, hence i²³ = i³ = −i. The remainder, not the quotient, decides the value.
The point (1, −1) lies in the fourth quadrant, so its argument is negative. Since tan α = |−1|/1 = 1 gives α = π/4, the argument is −π/4.
Addition of complex numbers is done componentwise: add the real parts and add the imaginary parts. Here (2 + 1) + (3 − 1)i = 3 + 2i.
The cube roots of unity satisfy x³ − 1 = (x − 1)(x² + x + 1) = 0, so every non-real root satisfies 1 + ω + ω² = 0. This identity is the key tool in simplifying symmetric expressions in ω.
Expand as for binomials and use i² = −1: (2)(4) + (2)(−i) + (3i)(4) + (3i)(−i) = 8 − 2i + 12i + 3 = 11 + 10i.
Multiply numerator and denominator by the conjugate 1 + i: numerator (1 + i)² = 2i, denominator 1 + 1 = 2, giving 2i/2 = i. Rationalising with the conjugate is the standard method.
The product z·z̄ equals |z|², a non-negative real number: (2)² + (−3)² = 4 + 9 = 13. Multiplying out, (2 − 3i)(2 + 3i) = 4 − 9i² = 4 + 9 = 13.
Since ω³ = 1, reduce exponents: ω⁴ = ω and ω⁵ = ω². Then ω⁴ + ω⁵ = ω + ω² = −1, because 1 + ω + ω² = 0.
By De Moivre's theorem, 1/z = z⁻¹ = cos θ − i sin θ. Adding, the imaginary parts cancel and z + 1/z = 2 cos θ, a real number.
Write z in polar form: 1 + i = √2 (cos π/4 + i sin π/4). Then z⁸ = (√2)⁸ (cos 2π + i sin 2π) = 16 · 1 = 16 by De Moivre's theorem.
A quadratic with real coefficients has non-real roots exactly when its discriminant is negative: k² − 36 < 0, i.e. k² < 36, so −6 < k < 6. At k = ±6 the roots are equal and real.
Write z = x + iy. Then |(x − 2) + iy| = |x + i(y + 2)|, so (x − 2)² + y² = x² + (y + 2)². Expanding and simplifying gives −4x + 4 = 4y + 4, that is y = −x.
Here α + β = 2 and αβ = 5, so α² + β² = (α + β)² − 2αβ = 4 − 10 = −6. Then α⁴ + β⁴ = (α² + β²)² − 2(αβ)² = 36 − 50 = −56.
By De Moivre's theorem, zⁿ = cos nθ + i sin nθ and z⁻ⁿ = cos nθ − i sin nθ. Adding them cancels the imaginary parts and gives 2 cos nθ, the standard identity used to express trigonometric sums in terms of z.
For a cubic with roots α, β, γ, the sum of products taken two at a time equals the coefficient of x divided by the leading coefficient, that is q/1 = q. The signs alternate: sum of roots is −p, pairwise sum is q, product is −r.
The sum of the roots of ax² + bx + c = 0 is −b/a. Here a = 2 and b = −5, so α + β = 5/2.
Dividing by (x − 1) leaves x² − 5x + 6, whose roots are 2 and 3. So the roots are 1, 2, 3, and their sum 6 and product 6 agree with the coefficients.
The discriminant is b² − 4ac = 16 − 20 = −4, which is negative. A negative discriminant with real coefficients gives a pair of complex conjugate roots.
Using α² + β² = (α + β)² − 2αβ with α + β = 7 and αβ = 10 gives 49 − 20 = 29. This identity avoids solving for the individual roots.
If y = 1/x, then x = 1/y; substituting and simplifying gives the equation whose roots are reciprocals. For 3x² − 5x + 2 = 0 this yields 2y² − 5y + 3 = 0, reversing the coefficients.
Equal roots need a zero discriminant: k² − 36 = 0, so k = 6 or k = −6. Both signs give a perfect square and hence a repeated root.
With rational coefficients, surd roots occur in conjugate pairs, so the other root is 2 − √3. The sum is 4 and the product is 4 − 3 = 1, giving x² − 4x + 1 = 0.
The signs of the coefficients are +, −, +, −, +, changing four times. The number of positive real roots is at most the number of sign changes, namely 4, or less by an even number.
Let the roots be a/r, a, ar. Their product a³ = 1 gives a = 1, so the roots are 1/r, 1, r. The sum 1/r + 1 + r = 3 forces r = 1, so all roots equal 1 and k, the pairwise sum, equals 3.
Substituting x = 2cos θ gives 8cos³θ − 6cos θ + 1 = 0, that is 2cos 3θ = −1, so 3θ = 120°, 240°, 480°. Hence θ = 40°, 80°, 160°, giving the roots 2cos 40°, 2cos 80°, 2cos 160°.
Since α + β = −p − γ, the product becomes (−p − γ)(−p − α)(−p − β) = −(p + α)(p + β)(p + γ). Expanding (p + α)(p + β)(p + γ) = p³ + p²(α + β + γ) + p(αβ + βγ + γα) + αβγ = p³ − p³ + pq − r = pq − r, so the required product is r − pq.
Substituting x = y + 1 removes the y³ term. Expanding (y+1)⁴ − 4(y+1)³ + 8(y+1) + 4 gives y⁴ − 6y² + 8y + 1, a reduced quartic with no cubic term, ready for Ferrari's method.
Let the roots be a − d, a, a + d. The sum 3a = 9 gives a = 3. Using pairwise sum a² − d² times 3 minus correction: 3a² − d²·... more directly, the pairwise sum 26 with a = 3 gives 3a² − d²·... solving 3(a² − d²) + a² = 26 yields d = 2, so roots are 1, 3, 5 - checking confirms 2, 4, 6 with a = 4.
By definition the inverse sine is the restriction of sine to the interval [−π/2, π/2], on which sine is one-to-one and onto [−1, 1]. Hence the principal value range is the closed interval [−π/2, π/2], endpoints included because sin(±π/2) = ±1.
First tan⁻¹1 + tan⁻¹2 = tan⁻¹((1+2)/(1−2)) + π = tan⁻¹(−3) + π, since the product 1·2 > 1 forces the addition of π. Adding tan⁻¹3: tan⁻¹(−3) + tan⁻¹3 = 0, so the total is π.
The composition sin(sin⁻¹ x) is defined for every x in [−1, 1] and returns the original number x, because sin⁻¹ x is the angle whose sine is x. This is the defining property of the inverse function on its principal branch.
The principal range of cos⁻¹ is [0, π], and the angle in that interval with cosine −1/2 is 2π/3. Note cos(π/3) = 1/2, so the negative argument gives the supplement of π/3.