Solve the following systems of three linear (c) \(i_{1}+i_{2}-i_{3}=1\) equations using Cramer's rule: (a) \(2 x-y-2 z=1\) \(x+y+3 z=6\) \(-2 x+3 z=-1\) (b) \(-2 x+3 z=-1\) \(\alpha+3 \beta+\gamma=2\) \(3 \alpha+2 \gamma=3\) \(v_{1}+\frac{1}{2} v_{2}-v_{3}=0.7\) \(-3 v_{1}+2 v_{3}=-3.1\) \(\theta_{1}+\theta_{2}+4 \theta_{3}=0\) \(2 \theta_{1}-\theta_{2}+\theta_{3}=5.4\) \(\theta_{1}+2 \theta_{2}-\theta_{3}=9.2\)

We are given the following system of linear equations for system (a): $$2x - y - 2z = 1$$ $$x + y + 3z = 6$$ $$-2x + 3z = -1$$

Create a coefficient matrix \(A\) from the given system of linear equations and compute its determinant: $$ A = \begin{bmatrix} 2 & -1 & -2 \\ 1 & 1 & 3 \\ -2 & 0 & 3 \end{bmatrix} $$ Now, find the determinant of matrix \(A\) (denoted as \(|A|\)): $$ |A| = 2\begin{vmatrix}1 & 3\\0 & 3\end{vmatrix} - (-1)\begin{vmatrix}1 & 3\\-2 & 3\end{vmatrix} -2\begin{vmatrix}1 & 1\\-2 & 0\end{vmatrix} $$ Now, calculate the determinants inside: $$ |A| = 2(1\cdot 3 - 0) -(-1)(1\cdot 3 - (-2)\cdot 3) - 2(1\cdot 0 - (-2)\cdot1) $$ $$ |A| = 6 + 9 - 4 = 11 $$

Replace the first column of matrix \(A\) with the constants from the equations, then find its determinant. Repeat this process for the second and third columns: $$ A_{x} = \begin{bmatrix} 1 & -1 & -2 \\ 6 & 1 & 3 \\ -1 & 0 & 3 \end{bmatrix} $$ $$ |A_{x}| = 1\begin{vmatrix}1 & 3\\0 & 3\end{vmatrix} - (-1)\begin{vmatrix}6 & 3\\-1 & 3\end{vmatrix} -2\begin{vmatrix}6 & 1\\-1 & 0\end{vmatrix} $$ $$ |A_{x}| = 3 + 21 - 2 = 22 $$ $$ A_{y} = \begin{bmatrix} 2 & 1 & -2 \\ 1 & 6 & 3 \\ -2 & -1 & 3 \end{bmatrix} $$ $$ |A_{y}| = 2\begin{vmatrix}6 & 3\\-1 & 3\end{vmatrix} - 1\begin{vmatrix}1 & 3\\-2 & 3\end{vmatrix} -2\begin{vmatrix}1 & 6\\-2 & -1\end{vmatrix} $$ $$ |A_{y}| = 30 - 1 + 8 = 37 $$ $$ A_{z} = \begin{bmatrix} 2 & -1 & 1 \\ 1 & 1 & 6 \\ -2 & 0 & -1 \end{bmatrix} $$ $$ |A_{z}| = 2\begin{vmatrix}1 & 6\\0 & -1\end{vmatrix} - (-1)\begin{vmatrix}1 & 6\\-2 & -1\end{vmatrix} - 1\begin{vmatrix}1 & 1\\-2 & 0\end{vmatrix} $$ $$ |A_{z}| = -2 + 13 + 2 = 13 $$

Divide the determinants \(|A_x|\), \(|A_y|\), and \(|A_z|\) by \(|A|\) to find the values of \(x\), \(y\), and \(z\): $$ x = \frac{|A_x|}{|A|} = \frac{22}{11} = 2 $$ $$ y = \frac{|A_y|}{|A|} = \frac{37}{11} = 3\frac{4}{11} $$ $$ z = \frac{|A_z|}{|A|} = \frac{13}{11} = 1\frac{2}{11} $$ So, the solution for system (a) is: $$ x=2, y=3\frac{4}{11}, z=1\frac{2}{11} $$