Question

Задание приложено...

Answer 1

Ответ:

36.

а)

$\displaystyle \left \{ {{x + y = 1} \atop {x - y = 2}} \right.$

$\Delta = \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = -1 \cdot 1 - 1 \cdot 1 =-2$

$\Delta_{x} = \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} = -1 \cdot 1 - 1 \cdot 2 =-3$

$\Delta_{y} = \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} = 1 \cdot 2 - 1 \cdot 1 = 1$

$x = \dfrac{\Delta_{x} }{\Delta} = \dfrac{-3}{-2} = 1,5$

$y = \dfrac{\Delta_{y} }{\Delta} = \dfrac{1}{-2} = -0,5$

б)

$\left \{\begin{array}{l} x -3y + z= 2 \\ 2x + y + 3z = 3 \\ 2x - y - 2z = 8 \end{array} \right$

$\Delta = \begin{vmatrix} 1 & -3 & 1 \\ 2 & 1 & 3 \\ 2 & -1 & -2 \end{vmatrix} r_{3} + r_{2} = \begin{vmatrix} 1 & -3 & 1 \\ 2 & 1 & 3 \\ 4 & 0 & 1 \end{vmatrix} c_{1} - 4c_{3} =$

$= \begin{vmatrix} 1 - 4 \cdot 1& -3 & 1 \\ 2 - 4 \cdot3 & 1 & 3 \\ 4 - 4 \cdot1 & 0 & 1 \end{vmatrix} = \begin{vmatrix}-3& -3 & 1 \\ -10 & 1 & 3 \\ 0 & 0 & 1 \end{vmatrix} = 1 \cdot (-1)^{3 + 3} \begin{vmatrix} -3 & -3 \\ -10 & 1 \end{vmatrix} =$

$= -3 \cdot 1 - (-3) \cdot (-10) = -33$

$\Delta_{x} = \begin{vmatrix} 2 & -3 & 1 \\ 3 & 1 & 3 \\ 8 & -1 & -2 \end{vmatrix}r_{3} + r_{2} = \begin{vmatrix} 2 & -3 & 1 \\ 3 & 1 & 3 \\ 8 + 3 & -1 + 1 & -2 + 3 \end{vmatrix} = \begin{vmatrix} 2 & -3 & 1 \\ 3 & 1 & 3 \\ 11 & 0 & 1 \end{vmatrix} c_{1} - 11c_{3}$

$= \begin{vmatrix} 2 - 11 \cdot 1 & -3 & 1 \\ 3 - 11 \cdot3 & 1 & 3 \\ 11- 11 \cdot1 & 0 & 1 \end{vmatrix} = \begin{vmatrix} -9 & -3 & 1 \\ -30 & 1 & 3 \\ 0 & 0 & 1 \end{vmatrix} = 1 \cdot (-1)^{3 + 3} \begin{vmatrix} -9 & -3 \\ -30 & 1 \end{vmatrix} =$

$= -9 \cdot 1 - (-3) \cdot (-30) = -99$

$\Delta_{y} = \begin{vmatrix} 1 & 2 & 1 \\ 2 & 3 & 3 \\ 2 & 8 & -2 \end{vmatrix} c_{3} -c_{1} = \begin{vmatrix} 1 & 2 & 1 - 1 \\ 2 & 3 & 3 - 2 \\ 2 & 8 & -2 -2 \end{vmatrix} = \begin{vmatrix} 1 & 2 & 0 \\ 2 & 3 & 1 \\ 2 & 8 & -4 \end{vmatrix}c_{2} - 2c_{1} =$

$= \begin{vmatrix} 1 & 2 - 2 \cdot 1 & 0 \\ 2 & 3 - 2 \cdot 2 & 1 \\ 2 & 8 - 2 \cdot2 & -4 \end{vmatrix} = \begin{vmatrix} 1 & 0 & 0 \\ 2 & -1 & 1 \\ 2 & 4& -4 \end{vmatrix} = 1 \cdot (-1)^{1 + 1} \begin{vmatrix} -1 & 1 \\ 4 & -4 \end{vmatrix} =$

$= (-1) \cdot (-4) - 1 \cdot 4 =0$

$\Delta_{z} = \begin{vmatrix} 1 & -3 & 2 \\ 2 & 1 & 3 \\ 2 & -1 & 8 \end{vmatrix}r_{3} - r_{2};r_{2} - 2r_{1} = \begin{vmatrix} 1 & -3 & 2 \\ 2 - 2 \cdot 1 & 1 - 2 \cdot(-3) & 3 - 2 \cdot2\\ 2 - 2 & -1 - 1 & 8 -3\end{vmatrix} =$

$= \begin{vmatrix} 1 & -3 & 2 \\ 0 & 7 & -1 \\ 0 & -2 & 5\end{vmatrix} = 1 \cdot (-1)^{1 + 1} \begin{vmatrix} 7 & -1 \\ -2 & 5 \end{vmatrix} = 7 \cdot 5 - (-2) \cdot (-1) = 33$

$x = \dfrac{\Delta_{x} }{\Delta} = \dfrac{-99}{-33} = 3$

$y = \dfrac{\Delta_{y} }{\Delta} = \dfrac{0}{-33} = 0$

$z = \dfrac{\Delta_{z} }{\Delta} = \dfrac{33}{-33} = -1$

в)

$\left \{\begin{array}{l} 3x + y + 2z = -4 \\ x - 2y - z = -1 \\ 2x + 3y + 2z = 0 \end{array} \right$

$\Delta = \begin{vmatrix} 3 & 1 & 2 \\ 1 & -2 & -1 \\ 2 & 3 & 2 \end{vmatrix} r_{1} - 3r_{2};r_{3} - 2r_{2} = \begin{vmatrix} 3 - 3 \cdot 1 & 1 - 3 \cdot(-2) & 2 - 3 \cdot(-1) \\ 1 & -2 & -1 \\ 2 - 2 \cdot 1 & 3 - 2 \cdot(-2) & 2 - 2 \cdot(-1) \end{vmatrix} =$

$= \begin{vmatrix} 0 & 7 & 5 \\ 1 & -2 & -1 \\ 0 & 7 & 4 \end{vmatrix} = 1 \cdot (-1)^{2 + 1} \begin{vmatrix} 7 & 5 \\ 7 & 4 \end{vmatrix} = -(7 \cdot 4 - 7 \cdot 5) = 7$

$\Delta_{x} = \begin{vmatrix} -4 & 1 & 2 \\ -1 & -2 & -1 \\ 0 & 3 & 2 \end{vmatrix} = -\begin{vmatrix} -4 & 1 & 2 \\ 1 & 2 & 1 \\ 0 & 3 & 2 \end{vmatrix}r_{1} + 4r_{2} = -\begin{vmatrix} -4 + 4 \cdot 1 & 1 + 4 \cdot2 & 2 + 4 \cdot1 \\ 1 & 2 & 1 \\ 0 & 3 & 2 \end{vmatrix}=$

$= -\begin{vmatrix} 0 & 9 & 6 \\ 1 & 2 & 1 \\ 0 & 3 & 2 \end{vmatrix}= -1 \cdot 1 \cdot (-1)^{2 + 1} \begin{vmatrix} 9 & 6 \\ 3 & 2 \end{vmatrix} = 9 \cdot 2 - 6 \cdot 3 =0$

$\Delta_{y} = \begin{vmatrix} 3 & -4 & 2 \\ 1 & -1 & -1 \\ 2 & 0 & 2 \end{vmatrix} c_{3} - c_{1} = \begin{vmatrix} 3 & -4 & 2 - 3 \\ 1 & -1 & -1 - 1 \\ 2 & 0 & 2 - 2 \end{vmatrix} = \begin{vmatrix} 3 & -4 & -1 \\ 1 & -1 & -2 \\ 2 & 0 & 0 \end{vmatrix} =$

$= 2 \cdot (-1)^{3 + 1} \begin{vmatrix} -4 & -1 \\ -1 & -2 \end{vmatrix} = 2(4 \cdot 2 - 1 \cdot 1) = 14$

$\Delta_{z} = \begin{vmatrix} 3 & 1 & -4 \\ 1 & -2 & -1 \\ 2 & 3 & 0 \end{vmatrix} = -\begin{vmatrix} 3 & 1 & 4 \\ 1 & -2 & 1 \\ 2 & 3 & 0 \end{vmatrix}r_{1} - 4r_{2}=$

$= -\begin{vmatrix} 3 - 4 \cdot 1 & 1 - 4 \cdot (-2) & 4 - 4 \cdot 1\\ 1 & -2 & 1 \\ 2 & 3 & 0 \end{vmatrix} = -\begin{vmatrix} -1& 9 & 0\\ 1 & -2 & 1 \\ 2 & 3 & 0 \end{vmatrix} =$

$= -1 \cdot 1 \cdot (-1)^{2 + 3} \begin{vmatrix} -1 & 9 \\ 2 & 3 \end{vmatrix} = -1 \cdot 3 - 2 \cdot 9 =-21$

$x = \dfrac{\Delta_{x} }{\Delta} = \dfrac{0}{7} = 0$

$y = \dfrac{\Delta_{y} }{\Delta} = \dfrac{14}{7} =2$

$z = \dfrac{\Delta_{z} }{\Delta} = \dfrac{-21}{7} = -3$