Tiesiniu̧ lygčiu̧ sistemu̧ sprendimas Gauss’o metodu. Rimantas Grigutis

1. Išsprȩskite lygčiu̧ sistemas.

1) \{ \begin{matrix}2x_{1} − x_{2} + x_{3} − 3x_{4} = −4 \\ 3x_{1} + x_{2} − 2x_{3} + 5x_{4} = −2 \\ 4x_{1} + 2x_{2} − 3x_{3} + 7x_{4} = −2 \\ 2x_{1} + 3x_{2} − x_{3} + 2x_{4} = 3 \end{matrix} .

2) \{ \begin{matrix}2x_{1} + x_{2} − x_{3} − 3x_{4} = 1 \\ 3x_{1} − x_{2} + 3x_{3} + x_{4} = 2 \\ 5x_{1} + 2x_{3} − 2x_{4} = 3 \\ 4x_{1} + 2x_{2} − 2x_{3} − 6x_{4} = 5 \end{matrix} .

3) \{ \begin{matrix}x_{1} − 2x_{2} + x_{3} + x_{4} = 2 \\ 2x_{1} − x_{2} + 2x_{3} + x_{4} = 2 \\ x_{1} + 2x_{2} + x_{3} − 2x_{4} = 1 \\ 3x_{1} + x_{2} + 3x_{3} − x_{4} = 3 \end{matrix} .

2. Raskite n -ojo laipsnio polinoma̧, jeigu

1) n = 3, \begin{matrix}x & 1 & 2 & 3 & 4 \\ f (x) & -2 & -4 & -2 & 10 \end{matrix} .

2) n = 4, \begin{matrix}x & -2 & 1 & -3 & 2 & -1 \\ f (x) & 10 & 4 & 60 & -10 & -4 \end{matrix} .

3) n = 5, f (2) = 5, f′ (2) = 19, f^{(2)} (2) = 40, f^{(3)} (2) = 48, f^{(4)} (2) = 24.

4) n ≤ 4, f (1) = −3, f′ (1) = −3, f^{(2)} (1) = 12, f^{(3)} (1) = 42, f (−1) = 3.

5) n ≤ 5, f (1) = −2, f′ (1) = −7, f^{(2)} (1) = −14, f^{(3)} (1) = 24, f (2) = −4, f′ (2) = 25.

3.Kiek sprendiniu̧ priklausomai nuo parametro λ turi tiesiniu̧ lygčiu̧ sistema. Išsprȩskite sistema̧.

1) \{ \begin{matrix}λx + y + z = 1 \\ x + λy + z = 1 \\ x + y + λz = 1 \end{matrix} . 2) \{ \begin{matrix}x + 2y − 3x & = 4 \\ 3x − y + 5z & = 2 \\ 4x + y + (λ^{2} − 14) z & = λ + 2 \end{matrix} , λ ∈ Q.

3)\{ \begin{matrix}λx + y + z + t = 1 \\ x + λy + z + t = 1 \\ x + y + λz + t = 1 \\ x + y + z + λt = 1 \end{matrix} .4) \{ \begin{matrix}2x_{1} + 3x_{2} + x_{3} + 2x_{4} = 3 \\ 4x_{1} + 6x_{2} + 3x_{3} + 4x_{4} = 5 \\ 6x_{1} + 9x_{2} + 5x_{3} + 6x_{4} = 7 \\ 8x_{1} + 12x_{2} + 7x_{3} + λx_{4} = 9 \end{matrix} .

4. Išspȩskite sistema̧ \{ \begin{matrix}x_{1} + x_{2} & = 0 \\ x_{1} + x_{2} + x_{3} & = 0 \\ x_{2} + x_{3} + x_{4} & = 0 \\ ⋯ \\ x_{n−2} + x_{n−1} + x_{n} & = 0 \\ x_{n−1} + x_{n} & = 0 \end{matrix} .

5. Išspȩskite sistema̧ \{ \begin{matrix}x_{2} + x_{3} + ⋯ + x_{n−1} + x_{n} & = 0 \\ x_{1} + x_{3} + ⋯ + x_{n−1} + x_{n} & = 1 \\ x_{1} + x_{2} + x_{4} + ⋯ + x_{n−1} + x_{n} & = 0 \\ ⋯ \\ x_{1} + x_{2} + x_{3} + ⋯ + x_{n−1} & = 0 \end{matrix} .

6. Su kuriomis λ reikšmėmis lygtis

\{ \begin{matrix}−λx_{1} + x_{2} + ⋯ + x_{n} = 0 \\ x_{1} − λx_{2} + ⋯ + x_{n} = 0 \\ ⋯ \\ x_{1} + x_{2} + ⋯ − λx_{n} = 0 \end{matrix}

turi nenulini̧ sprendini̧ ?

7. Išspȩskite sistema̧ \{ \begin{matrix}(1 − n) x_{1} + x_{2} + ⋯ + x_{n} = 0 \\ x_{1} + (1 − n) x_{2} + ⋯ + x_{n} = 0 \\ ⋯ \\ x_{1} + x_{2} + ⋯ + (1 − n) x_{n} = 0 \end{matrix} .

8. Išspȩskite sistemas pritaikȩ Cramer’io formules.

1) \{ \begin{matrix}2x + y + z = 3 \\ x + 2y + z = 0 \\ x + y + 2z = 9 \end{matrix} . 2)\{ \begin{matrix}x + y + z = 6 \\ −x + y + z = 0 \\ x − y + z = 2 \end{matrix} . .