esercizi svolti metodi matematici

Es

A = ^{K - 1    0      1}

         _{K   2    1}

         _{2   K    1}

   b   ²

        ₀

|A| = ^{|K - 1    0    1|} = - |K - 1       | + |K - 1    |

           _{2        1}

          _K    1

           _{0    2 |}

                               |2      |

                               |K    1|

                               |K    2|

= -K² + 2K + K²

= - K² + 3K + 2

ρ(A) = 3 ≡ ^{i,1,2}

ν_E ∈ ^{i,2}

|K + 1    0      2| → | - 1    0       2|

|K    2        1|            → |  2       1      0|

|2    K       1|             |K       1       2|

K ∈ ^{i,1,2} ρ(A) = ρ(A|b) = 2

K ∈ ^{i,2} ρ(A) ≢ ρ(A|b) incompatible

K ∈ ^{i,1,2}

x₁ = ⇒ ^{2     0    2}

             _{2     0    1}

             _{2     K    1}

= ^{2       2     0} ⁄ (K-1)(K-2)

        _{2       0     1}

        _{2      K     1}

= ^{2     0  K} ⁄ (K - 1)(K - 2)

        _{2     1    2}

K₂ ⇒ + ^{K   0  0}

         _{2   1   0}

         _{2   2   1}

= ^{k     2  0} ⁄ (K-1)(K-2)

        _{2     1  2}

        _{2     2  K}

= ^{-2K + 2    -2K    +1} = ^{-4K + 1}

                     _{(K-1)(K-2)      (K-1)(K-2)}

x₃ ⇒ |K-1| ⇒ ^{|-1    2   0|} →

        |K| → ^{|K     2    0|}

        |2| → ^{|2      K    2|}

= ^{2      0      2} ⁄ (K-1)(K-2)

= ^{2     K    1} ⁄ (K-1)(K-2)

= ^{2       K    2} ⁄ (K-1)(K-2)

= ^{2K2 - 8 + (K+1) 2+1} ⁄ (K-1)(K-2)

= ^{2K2 + 6K - 8}

         _(K-1)(K-2)

V[ℝ]_{A}

(^{k 1 0} _{k 2 1} ^{2 k 1})

→ (^{1 -4 0} _{1 2 1})

→ |^{x₁ x₂ x₃}|

|^{1 1 0} _{1 2 1}|

x_A = |^{-1 0 2} _{2 1 1}|

Det = -1

x₂ = |^{2 + 0 |} _{-1 + 1}| = -2t

x₃ = - |^{-1 2} _{0 1}| = |^{(-1 0} ₂₎| = (-3t + 4)

|^x₁| = |⁰ _{x2 -2} ^x₃| = |⁴| + t |⁹ ₄ ³|

K ∈ {ℝ}

(₂ ₀ ₁ | ₀)(₁ ₀ ₁ | ₂)(₁ ₁ ₀ | ₂)

^X_{n, X1X3}(₂ ₀ ₁ | t)(₁ ₀ ₁ | t)(₁ ₁ ₀ | 2)

(₂ ₀ ₁ | t)(₁ ₁ ₀ | 2)

|I| = 2

X₃ =

X_h = ₁ + ₀₁² ₁

oppure

Δ_X2 = ₁ − ^t₂

X₁ + X₂ = 2 =

X₂ = 2 - X₁ =

X₂ = 2

Δ - ^t₂

X₂−4−1^t₂

X₂ = ^3+t₂

⎛ ⎞ ⎛ ⎞

⎜X₁⎟ ⎜1^t₂⎟

⎜X₂⎟ = ⎜3^t₂⎟ + t⎛^t₂⎞

⎜X₃⎟ ⎜0⎟ ⎜1⎝

Es.

g(x) = ^x+2_√x²+x−3

x²+2x−30

Δ=4+12=16

x_1,2 = ^−2±4₂ =¹₋₃

∆ = ]−∞3 ]∪ ]1;+∞ [

lim_x→−3⁻^x+2_√x²+x−3 = [¹₊ → −∞

lim_x→1⁺^x+2_√x²+x−3 = [³₊ ] → +∞

lim_x→+∞^x+2_√x²+x−3 = ^x(1+2/x)_√x(^1+3/x

X

−(1+²_x)

= 1

X = −3

AVV. SINISTRO

compensate

rg(A) = rg(A|b)

|A| = |₁ K 1|        | K K 1|        | 2 1 1| = -1| 1 K 1| - K| K K 1| + K| K 1 1|                  |₂ 1 1| = K² - K² + K + 2K - K = -K + 3K

|A| = 0 <=> -K² + 3K - 2 = 0 <=> K ∈ {1; 2}

Δ = 9 - 8 = 1

K = \(\frac{3 \pm 1}{2}\)

K ∈ {1; 2, 3 } ⇒ rg(A) = rg(A|b) = 3soluzione unica

|₁ K 1 0|         | K - 1 K |         | 2 1 0| → | K 1 0| -> | - K K 1|                   | 1 K | = -K + 2K

K ∈ {1, 2} ⇒ rg(A) ≠ rg(A|b) = 3