Analisi Matematica 2 Esercizi

A =

| λ 1 1 | | 2 λ 2 | | 3 2 3 |

ESERCIZI COMPITO

1. 2λx - y + λz = 1 λx + 2y - λz = λ 3z + y + z = -2λ

A₁ = | 2λ 1 λ | = (2λ)(-1)¹⁺¹(2-2λ)-(λ(-1)) = 2λ(2λ+λ)

A₂ = | 2λ -1 | = (-1)(-1)²⁺¹(λ-2λ)-(3(-λ)) = λ(2λ²+3λ)

A₃ = | λ 1 | = λ(-1)¹⁺³(λ x) - (3 - λ) = λ(λ - 6)

A_... | 2λ -1 | = 2λ(4λ + λ) + 1 (2λ²λ) + λ(λ - 6) = 8λ

A=

| λ 1 1 |

| 2λ 2 2 |

| 3λ 3 3 |

ESERCIZIO COMPLETO

1. {(2λ-1)x-y+λz=1
2. λx+2y-λz=λ
3. 3x+y+2( )z==-2λ

A₁=

= (2λ) (λ-1)⁺¹ (2-2λ)-(λ-(λ-λ))) = 2λ (4λ+λ)

A₂=

= λ (2-1) (λ)²⁺¹ (λ λ-2λ)-(3(λ-λ))) = λ (2λ² + 3λ)

A₃=

= (-λ) (λ) λ ⁺³ (λ-1) (λ - 1) (λ-1) (λ - λ) = λ (λ)

A_{( )}=

= λ (λ) λ⁺ (2λ λ-2λ) (2λ + λ ² ( )) + λ (λ - 6) λ₊ 1

Ora sostituisco i risultati delle equazioni al posto dei valori con x

¹⁄_λ( ¹⁄_λ(-2²) - λ( ¹⁄_λ(-2²)) - λ(2)) - ( ¹⁄_λ (-2²) + (2)))

= 2( ¹⁄_λ(2² - (2))) + ( ¹⁄_λ(2 - (2) + λ(1 - 2)))

+ λ( - 1 - 2) + ¹⁄_λ(-2²)) =

= 5λ² + 5λ

Ora sostituisco i risultati delle equazioni al posto dei valori con y

= -2λ + 3λ - ^{2 λ}⁄₃

= (-2 , - ^{5 λ3}⁄₃)

Ora sostituisco i risultati delle equazioni al posto dei valori con z

¹⁄_(-1)(λ( ⁴⁄_λ) + λ( ^-22⁄₃) + λ(1)⁶) =

= 8λ² + (6λ² - 1 + 6λ - ²⁄λ₂) =

= ( ^{-12 λ2}⁄λ₂ - 6)

Calcolo il dominio

(3λ³) (3) (-λ) = 5λ + 3

3) = ^λ⁄3 + 3

= λ ⁷⁄₁₃

PROVA

Si scelgono 3 valori, compresi nel range di:

λ = 1

• x = ^{5 · 1 +5}/_{13 + 3} = ^{5 + 5}/_{13 + 3} = ¹⁰/₁₆ = ⁵/₈
• y = ^{-2 · 1 + 3}/_{13 + 3} = ^{-5 + 3}/_{13 + 3} = ^-2/₁₆ = ^-1/₈
• z = ^{-12 · 1 + 12 + 6}/_{13 + 3} = ^{-12 + 12 + 6}/_{13 + 3} = ⁶/₁₆ = ³/₈

λ = 2

• x = ^{10 + 20}/_{52 + 6} = ³⁰/₅₈ = ¹⁵/₂₉
• y = ^{-16 - 20 + 6}/₄₆ = ^-30/₄₆ = ^-15/₂₃
• z = ^{-48 + 4 - 6}/₄₆ = ^-50/₄₆ = ^-25/₂₃

ORA TERMINO LA PROVA

1. 2x - y + λz = 1
2. λx + 2y - λz = 1
3. 3x + y + 2z = 2λ

• (2)(¹/₁) + (-1)(^-1/₁) + (1)(^-2/₁) = 1
• (-7)(⁰/₂) + (2)(⁰/₂) + (-7)(^-1/₂) = (-7)
• (3)(¹⁵/₂₃) + (2)(⁷/₂₃) + (2)(^-29/₂₃) = (2)(^z/_z)

• 1 = 1 ✓
• -1 = -1 ✓
• 4 = 4 ✓

1)

{_xx + y + z = 3λ

x - λy + z = 1

2λx - λy + z = 2(Λ + 1)

D_λ = | λ 1 1 |

| 1 -λ 1 |

| 2λ λ 1 | = λ(λ - λ) - λ(1-2λ) + 1(λ+2λ) = λ x -λx - λ x + 2λ² = 3λ + 1

A_x = | 3λ 1 1 |

| 1 -λ 1 |

| (2+2λ) λ 1 | = 3(λ - λ) - λ(1 - (2 + 2λ)) + (λ-λ)((2 + 2 λ)) = -3λ ⁷ - 3¹x² + 2λ + λ + 2λ + 2 = 6λ x + x +5 x² + 5λ + 1

A_y = | λ 3λ 1 |

| 1 1 1 |

| 2λ (3λ +2) 1 | = λ(1 - 2 − 2λ) ≠ 3 λ (1-2λ) + 1 (2 +2 λ) - 2λ = λ - 2λ - λ ² x (3 λ + 6 λ +2 + 2 λ - 2 λ = 4 λ ⁶ + 1 - 4λ + λ²

A_z =| λ 1 3 λ |

| 1 -λ 1 |

| 2λ λ (2+2λ) | = λ(λ − λ) (2 + 2 λ) - λ ) – λ (2 λ + 2 λ) + 3 λ (λ² + 1)

= -σ λ x³ - 2λ ¹ x ⁵ λ(λ − λ) − 3(λ + 2 λ) + 3λ (λ (2λ - 2) ₃ x + − 2 λ + λ x 3 λ + 6 λ³= 4 λ₇ - 2

x = -4^λ² + 5λ + 1 / 3λ - 1

y = 4λ²/3λ + 1 + 2

z = 4λ³ - 2 / 3λ - 1

Dominio

• 3λ - 1 ≠ 0
• 3λ ≠ 1
• λ ≠ 1/3

Prova

λ = 0

• x = 0 + 0 + 1 / 0 - 1 = -1
• y = 0 - 0 - 2 / 0 - 1 = -2
• z = 0 - 2 / 0 - 1 = 2

λ = 1

• x = -4 + 5 + 1 + 2 / 3 - 1 = 2
• y = 4 + 4 + 2 / 3 + 1 = 5/2
• z = 4 - 2 / 3 - 1 = 3/2

x + y + z = 3

x - λy + z = 1

2(x + λy + z) = 2λ

• x + y + z = 3λ
• 0 + z ≠ 0

1)

• x - 3y + λz = λ - z
• x + y - λz = λ
• x - y + z = 2

A = ^{|1 -3 λ|} _{|1 1 -λ|} ^{|1 -1 2|}

= λ ^{|λ-1 λ}| + 3 ^{|2 λ+1}| + λ ^|λ-1 | = 2λ - λ² + λ ^|3λ - λ + 3λ ^{|-λ - λ2}

= -3λ² + 5)

A_x = _{|λ -2 -3}| _{|λ 1 -λ| |2 -1 2}|

= λ ^{|-2 -1}| + 3 ^{|2 λ}| + ^|-1

= -2λ + λ

= 2 ^|6λ + λ

= 6λ² + 2λ)

A_y = ^{|1 λ -2}| _{|-2 1 -λ| |2 -1 2|}

= λ ^{|2 2}|

= -2 ^{|-3 |λ} -1

= 2 ^|+4 + 2λ ^|2

= 2λ² + ^|6λ

A_z = ^{|1 1 λ}| _{|λ 1 2| |λ -1} ¹

|

= λ ^|2 +1| + 3 ^{|2 2}

= 2λ

= -3λ² + 3λ + 8

x = ^{|6λ2 + 2λ|}/_{3λ² + 5}

y = ^{|2λ2 + 6λ|}/_{3λ² + 5}

z = ^{|-3λ2 + 8|}/_{3λ² + 5}

λ ≠ 0

λ(3λ + 5) ≠ 0

λ ≠ 5/3

x - y + z

x - 2

y

z

λ = -1

λ = 1

• x = ^{6 + 2}/_{3 + 5} ≡ ⁸/₈ = 1
• y = ^{2 + 6}/_{3 + 5} ≡ ⁸/₈ = 1
• z = -^{3 + 3 + 8}/_{3 + 5} ≡ ⁸/₈ = 1

λ = 2

• y = ^{24 + 4}/_{12 + 10} ≡ ²/₂₂
• y = ^{8 + 12}/_{12 + 10} = ²⁰/₂₂
• z = -^{12 + 6 + 8}/_{12 + 10} ≡ ²/₂₂

x x - 3y + λ z = λ z

x - y - λ z = λ

λ x - y + z λ z = z

{ ^{2 4}/_{6 + 1} = -1 = z

1 + 1/z - x = 1

^{2 8}/₂₂ = ⁴/₁₁ = z

-3 = -3

1 = 1

z = z

( λ x + y - z = 2 λ )

( x + 2 y + z = -1 )

( 3 x - λ y - 2 z = 0 )

A = |

• x 1 -1 |
• 12 λ 1
• 3 λ -2

A_x = |

• 2λ 1 -1 |
• -1 2λ 1
• 0 λ -2

A_y = |

• 1 2λ -1 |
• 1 λ 1
• 3 0 -2

A_z = |

• 1 x 2λ |
• 1 2λ -1
• 3 λ 0

x =

y =

z =

=

x = -7 - z / -3 + 7 + 5

y = -z - 7 / -3 + 7 + 5

z = -7 - 14 + 3 / -3 + z + 5

x = -7 - z / -9 / 9 = -1

y = -2 + 7 / -5 = -1

z = 1 - 14 + 3 / -5 = -10 / -5 = 2

( x + y - z = 2 )

( x - z ) y + z = 1 - )

3 x - y - 2z = 0

0 + 0 - 0 = 0 √

-1 - 2 2 = -1 X

3 x 1 - 4 = 0 x

2)

{ x + y + z = 4

{ x - y + z = 1

S = { ( 1 ) + λ ( 1 )}

( 0 ) ( 2 )

( 1 ) ( 1 )

{ y = 4 - x - 2z

{ x = 1 + y - z

{ y = 3 - 2z

{ x = 1 + y - z

{ y = 4 - ( 1 + y - z ) - 2z

{ x = 1 + y - z

{ y = 3 - z

{ x = 1 + y - z

{ y + y = 3 - z

{ x + x = 5 - 3z

{ y = 4 - ( 1 + y - z ) - 2z

{ x = 1 + ( 4 - x - 2z ) - z

{ 2y = 3 - z

{ 2x = 5 - 3z

{ y = 3/2 - z/2

{ x = 5/2 - 5/2 z

z = 0 z = 1

A = ( 5/3 2/3 ) B = ( 1 )

( -1/3 1/3 ) ( 1 )

( 0 )

B - A = ( 1 ) + ( 2/5 - 2/5 )

( 1 ) ( -3/5 1/5 )

( 1 ) ( 1/5 2/5 )

R = B + μ ( B - A ) = {( 1 ) + μ ( -3/5 )}

( 1 ) ( 1/5 )

( 1 ) ( 1/5 )

λ → 2 + 2b + c = 0

μ → 3b - b + 2c = 0

z = -2b + c

b = -3b + 2c

z = -2 ( 3b + 2c ) ≠ c

b = 3 ( 2b + c ) + 2c

{ z = 6z + 4c

{ b = 6b + 3c + 2c

{ z = 6z = -3c

{ b = 6b = 5c

{ 5z = -5c

{ -5b = 5c

{ -2z = -3/5

{ z = -c

b = -c

c = 1

P λ - Q μ = d ( -1 )

( -1 )

( 1 )

(0 1) = λ (1 2) + μ (3 -1) ; 2 (1 1)

1 -1 -2

d = (1 -1) + λ 3μ

-2 = (0 -1) +2 λ +μ

d = (-1 + μ) - 2 λ -1 2d = λ + 3μ

λ= -3μ + d

μ = 1 -2λ -2

d = 3μ + d - 2d

λ = 2

μ = 0

0 = 7 -2λ - 2

λ = 2

2 = -1 -2λ

(0 1) = 1/3 (1 2 1 -1 2 + 0) 3 1/2 -1

(0 1) 3/3 (1/3 2 2/3 3 2/3 1/3)

2)

R = { 3x - y + z = 1 x + y + 2z = 4}

S ={( _-1 1 1)+λ( ₁ -1 0)| λ ∈ ℝ}

$\begin{cases} y = 1 + z + 3x \\ x = -4 + y + 2z \end{cases}$

$\begin{cases} y = -1 + z + 3(-4 + y + 2z) \\ x = -4 + y + 2z \end{cases}$

$\begin{cases} y = -1 + z – 12 + 3y + 6z \\ x = -4 + y + 2z \end{cases}$

$\begin{cases} y - 3y = -13 + 7z \\ x = -4 + y + 2z \end{cases}$

$\begin{cases} -2y = -13 + 7z \\ x = -4 + y + 2z \end{cases}$

$\begin{cases} y = \frac{13}{2} - \frac{7}{2}z \\ x = -4 + \frac{13}{2} - \frac{7}{2}z + 2z \end{cases}$

$\begin{cases} y = \frac{13}{2} - \frac{7}{2}z \\ x = 5 + \frac{3}{2}z \end{cases}$

... (Text omitted) ...

R = B + μ(B-A) ={ ( ₁ 3 1 ) + μ ( _-3 -7 2 )}

• λ -> 2 - b - c = 0
• µ -> 3z - 7b - 7c = 0
• c = 10b

... (Text omitted) ...

b = 1 a = ( _-1 1 2)

P λ = Q µ = a( ₁ 2)

-2 + λ + 3μ - z = 2

2 - λ + 7μ = 2

λ - 2μ = 2λ

λ = -d - 3μ - z

d = -λ + 7μ - z

μ = ²/₂ d₁ - λ + z

λ - d - 3μ - z

d = -λ + 7μ - z

μ = -d + ¹/₂

(^x/_-1) = λ - 7μ + z - 3μ - z

x = λ + 3μ + 7μ + 7μ z

μ = : λ - 7μ + z + ^λ/₂

{

μ = 0

μ = 0

3^1/2λ = -8μ + z

}

{

μ = 0

3 = 0

λ = ¹⁶/₃μ = ⁴/₃

}

{

μ = 0

μ = 0

λ = -⁴/₃

}

(^-1/_y) (^-1/₃ ^x/_y) (^-1/_-3 ^y/₄) (⁰/₀)(^-/₃)

(^-8/₃) = 0 X

(^-10/₃)

(^-2/₃)

(^-3/_-1)

(^-7/_-3)

3)

U = span

• (
  • 1
  • 0
  • 0
  • 1
  )

V = span

• (
  • 1
  • 0
  • 0
  • 1
  )

• _U ∩ _V

α

• (
  • 1
  • 0
  • 0
  • 1
  )

• (
  • 0
  • 1
  • 1
  • 0
  )

• (
  • 0
  • 1
  • 1
  • 0
  )

• (
  • 1
  • 0
  • 0
  • 1
  )

• α = δ
• β = γ + δ

• β = γ δ = β
• α = β γ = δ

• β = γ β
• 2β = β
• β = 2β

• (
  • α
  • β
  • γ
  • δ
  )
= (2, 1, 1, 1)

• • 2
  • 1
  • 0
  • 1
  + _U
  • • (1, 0, 0, 1)
    • (0, 1, 1, 0)
    • (1, 0, 0, 1)

  V^⊥

  V₁ - <V₁, W₁> W₁

  W₁ =( 0 ) ( 3 ) ( 1 )( 0 ) -3 ( 7 ) = ( 4 ) = W₃( 0 ) ( 1 ) ( 3 )( 1 ) ( - 3 )( 7 ) +3 ( 4 )( 3 ) ( 1 )

  1 U + V

  < W₁, W₂ > W₃ = W₃ W₃ W₂ = W₂

  ( 4 ) ( -3 )( -3 ) ( 4 )( -3 ) ( 3 )( -3 ) ( - 3 + 4 + - 3 + 4 + - 3 + - 3 + 1 + 1 ( 49 ) ( 38 ) (98) 0 35+74 49 = 7 49 / W₄( 35 ) ( 49 ) (49)1 ( 7 ) ( -7 ) ( 2 ) = 7 ( 3 ) ( 1 ) ( 7 ) 1 (U+V)^⊥ ( a ) W₁ ( b ) ( c )

  W₁ → 3a + bc + z = 0

  W₂ → 3b + 3c + d = 0