Using Cramer's Rule , solve the following system of equations :
`{:(3x-4y+5z=-6),(x+y-2z=-1),(2x+3y+z=5):}`

Answer

To solve the system of equations using Cramer's Rule, we will follow these steps:

Given equations:
1. \(3x - 4y + 5z = -6\) (Equation 1)
2. \(x + y - 2z = -1\) (Equation 2)
3. \(2x + 3y + z = 5\) (Equation 3)

### Step 1: Write the system in standard form
We already have the equations in the standard form \(Ax + By + Cz = D\).

### Step 2: Identify coefficients
From the equations, we can identify the coefficients:
- For Equation 1: \(A_1 = 3\), \(B_1 = -4\), \(C_1 = 5\), \(D_1 = -6\)
- For Equation 2: \(A_2 = 1\), \(B_2 = 1\), \(C_2 = -2\), \(D_2 = -1\)
- For Equation 3: \(A_3 = 2\), \(B_3 = 3\), \(C_3 = 1\), \(D_3 = 5\)

### Step 3: Calculate the determinant \(\Delta\)
\[ \Delta = \begin{vmatrix} A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3 \end{vmatrix} = \begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix} \]

Calculating the determinant:
\[ \Delta = 3 \begin{vmatrix} 1 & -2 \\ 3 & 1 \end{vmatrix} - (-4) \begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} + 5 \begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} \]

Calculating the smaller determinants:
1. \(\begin{vmatrix} 1 & -2 \\ 3 & 1 \end{vmatrix} = (1)(1) - (-2)(3) = 1 + 6 = 7\)

2. \(\begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} = (1)(1) - (-2)(2) = 1 + 4 = 5\)

3. \(\begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} = (1)(3) - (1)(2) = 3 - 2 = 1\)

Now substituting back:
\[ \Delta = 3(7) + 4(5) + 5(1) = 21 + 20 + 5 = 46 \]

### Step 4: Calculate \(\Delta_x\)
\[ \Delta_x = \begin{vmatrix} D_1 & B_1 & C_1 \\ D_2 & B_2 & C_2 \\ D_3 & B_3 & C_3 \end{vmatrix} = \begin{vmatrix} -6 & -4 & 5 \\ -1 & 1 & -2 \\ 5 & 3 & 1 \end{vmatrix} \]

Calculating \(\Delta_x\):
\[ \Delta_x = -6 \begin{vmatrix} 1 & -2 \\ 3 & 1 \end{vmatrix} - (-4) \begin{vmatrix} -1 & -2 \\ 5 & 1 \end{vmatrix} + 5 \begin{vmatrix} -1 & 1 \\ 5 & 3 \end{vmatrix} \]

Calculating the smaller determinants:
1. \(\begin{vmatrix} 1 & -2 \\ 3 & 1 \end{vmatrix} = 7\) (as calculated before)

2. \(\begin{vmatrix} -1 & -2 \\ 5 & 1 \end{vmatrix} = (-1)(1) - (-2)(5) = -1 + 10 = 9\)

3. \(\begin{vmatrix} -1 & 1 \\ 5 & 3 \end{vmatrix} = (-1)(3) - (1)(5) = -3 - 5 = -8\)

Now substituting back:
\[ \Delta_x = -6(7) + 4(9) + 5(-8) = -42 + 36 - 40 = -46 \]

### Step 5: Calculate \(\Delta_y\)
\[ \Delta_y = \begin{vmatrix} A_1 & D_1 & C_1 \\ A_2 & D_2 & C_2 \\ A_3 & D_3 & C_3 \end{vmatrix} = \begin{vmatrix} 3 & -6 & 5 \\ 1 & -1 & -2 \\ 2 & 5 & 1 \end{vmatrix} \]

Calculating \(\Delta_y\):
\[ \Delta_y = 3 \begin{vmatrix} -1 & -2 \\ 5 & 1 \end{vmatrix} - (-6) \begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} + 5 \begin{vmatrix} 1 & -1 \\ 2 & 5 \end{vmatrix} \]

Calculating the smaller determinants:
1. \(\begin{vmatrix} -1 & -2 \\ 5 & 1 \end{vmatrix} = 9\) (as calculated before)

2. \(\begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} = 5\) (as calculated before)

3. \(\begin{vmatrix} 1 & -1 \\ 2 & 5 \end{vmatrix} = (1)(5) - (-1)(2) = 5 + 2 = 7\)

Now substituting back:
\[ \Delta_y = 3(9) + 6(5) + 5(7) = 27 + 30 + 35 = 92 \]

### Step 6: Calculate \(\Delta_z\)
\[ \Delta_z = \begin{vmatrix} A_1 & B_1 & D_1 \\ A_2 & B_2 & D_2 \\ A_3 & B_3 & D_3 \end{vmatrix} = \begin{vmatrix} 3 & -4 & -6 \\ 1 & 1 & -1 \\ 2 & 3 & 5 \end{vmatrix} \]

Calculating \(\Delta_z\):
\[ \Delta_z = 3 \begin{vmatrix} 1 & -1 \\ 3 & 5 \end{vmatrix} - (-4) \begin{vmatrix} 1 & -1 \\ 2 & 5 \end{vmatrix} - 6 \begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} \]

Calculating the smaller determinants:
1. \(\begin{vmatrix} 1 & -1 \\ 3 & 5 \end{vmatrix} = (1)(5) - (-1)(3) = 5 + 3 = 8\)

2. \(\begin{vmatrix} 1 & -1 \\ 2 & 5 \end{vmatrix} = (1)(5) - (-1)(2) = 5 + 2 = 7\)

3. \(\begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} = (1)(3) - (1)(2) = 3 - 2 = 1\)

Now substituting back:
\[ \Delta_z = 3(8) + 4(7) - 6(1) = 24 + 28 - 6 = 46 \]

### Step 7: Calculate values of \(x\), \(y\), and \(z\)
Using Cramer's Rule:
\[ x = \frac{\Delta_x}{\Delta} = \frac{-46}{46} = -1 \]
\[ y = \frac{\Delta_y}{\Delta} = \frac{92}{46} = 2 \]
\[ z = \frac{\Delta_z}{\Delta} = \frac{46}{46} = 1 \]

### Final Solution
Thus, the solution to the system of equations is:
\[ x = -1, \quad y = 2, \quad z = 1 \]