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New answer posted
11 months agoContributor-Level 10
x + 2y + z = 2
αx + 3y – z = α
–αx + y + 2z = –α
Δ = | (1, 2, 1), (α, 3, -1), (-α, 1, 2) | = 1 (6+1) – 2 (2α–α) + 1 (α+3α) = 7+2α
α = –7/2
New answer posted
11 months agoContributor-Level 10
Given the matrix P = [2, -1], [5, -3].
The characteristic equation is det (P - λI) = 0, which is (2-λ) (-3-λ) - (-1) (5) = 0.
This simplifies to λ² + λ - 1 = 0.
By the Cayley-Hamilton theorem, the matrix P satisfies this equation: P² + P - I = 0, so P² = I - P.
To find P³: P³ = P * P² = P (I-P) = P - P² = P - (I-P) = 2P - I.
The problem asks for N=6, likely related to a higher power P? Continuing the pattern:
P? = 2P² - P = 2 (I-P) - P = 2I - 3P.
P? = 2P - 3P² = 2P - 3 (I-P) = 5P - 3I.
P? = 5P² - 3P = 5 (I-P) - 3P = 5I - 8P.
The solution N=6 must relate to a different question not fully transcribed, for example, if P^N = 5I - 8P.
New answer posted
11 months ago
Contributor-Level 10
1 = (2-1)¹ (The n is likely 1).
3? = (7-4)³ (This seems to be a pattern matching (a-b)^c).
4²? = (12-8)? ! = 4²?
The blank space must be (5-3)² = 2² = 4.
New answer posted
11 months agoContributor-Level 10
Evaluate the integral:
∫ (2x-1)cos (√ (4x²-4x+6) / √ (4x²-4x+6) dx
∫ (2x-1)cos (√ (2x-1)²+5) / √ (2x-1)²+5) dx
Let (2x-1)² + 5 = t².
Differentiating both sides:
2 (2x-1)*2 dx = 2t dt
2 (2x-1) dx = t dt
(2x-1) dx = (t/2) dt
Substitute into the integral:
∫ cos (t)/t * (t/2) dt
= 1/2 ∫ cos (t) dt
= 1/2 sin (t) + C
= 1/2 sin (√ (2x-1)²+5) + C
= 1/2 sin (√ (4x²-4x+6) + C
New answer posted
11 months agoContributor-Level 10
Given the equations:
t? (A + 2B) = -1 which expands to t? (A) + 2t? (B) = -1 . (I)
t? (2A - B) = 3 which expands to 2t? (A) - t? (B) = 3 . (II)
Solving equations (I) and (II) simultaneously, we get:
t? (A) = 1
t? (B) = -1
Therefore, t? (A) - t? (B) = 1 - (-1) = 2.
New answer posted
11 months agoContributor-Level 10
The system of equations has no solution if the determinant of the coefficient matrix is zero.
Δ = |k 1|
|1 k 1|
|1 k|
Δ = k (k² - 1) - 1 (k - 1) + 1 (1 - k) = 0
Δ = k³ - k - k + 1 + 1 - k = 0
⇒ k³ - 3k + 2 = 0 ⇒ (k - 1)² (k + 2) = 0
∴ k = -2, 1
If k = 1 then all the equations are identical (infinite solutions). Hence k = -2 for no solution.
New answer posted
11 months agoContributor-Level 10
A = [ x 1 ]
[ 1 0 ]
A² = [ x 1 ] [ x 1 ] = [ x²+1 x ]
[ 1 0 ] [ 1 0 ] [ x 1 ]
A? = [ x²+1 x ] [ x²+1 x ]
[ x 1 ] [ x 1 ]
= [ (x²+1)²+x² x (x²+1)+x ]
[ x (x²+1)+x²+1 ]
a? = (x² + 1)² + x² = 109
⇒ x = ±3
a? = x² + 1 = 10
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