Solve the following systems of liner a equations by cramer rule . `(i)2x-y+3z=8`, `-x+2y+z=4`, `3x+y-4z=0`

Answer

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

### Step 1: Write the equations in standard form
We have the following equations:
1. \( 2x - y + 3z = 8 \)
2. \( -x + 2y + z = 4 \)
3. \( 3x + y - 4z = 0 \)

### Step 2: Form the coefficient matrix \( A \), variable matrix \( X \), and constant matrix \( B \)
The coefficient matrix \( A \) is formed by the coefficients of \( x \), \( y \), and \( z \):
\[ A = \begin{bmatrix} 2 & -1 & 3 \\ -1 & 2 & 1 \\ 3 & 1 & -4 \end{bmatrix} \]

The variable matrix \( X \) is:
\[ X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \]

The constant matrix \( B \) is:
\[ B = \begin{bmatrix} 8 \\ 4 \\ 0 \end{bmatrix} \]

### Step 3: Calculate the determinant \( D \) of the matrix \( A \)
To find \( D \), we calculate the determinant of \( A \):
\[ D = \begin{vmatrix} 2 & -1 & 3 \\ -1 & 2 & 1 \\ 3 & 1 & -4 \end{vmatrix} \]

Using the determinant formula:
\[ D = 2 \begin{vmatrix} 2 & 1 \\ 1 & -4 \end{vmatrix} - (-1) \begin{vmatrix} -1 & 1 \\ 3 & -4 \end{vmatrix} + 3 \begin{vmatrix} -1 & 2 \\ 3 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 2 & 1 \\ 1 & -4 \end{vmatrix} = (2)(-4) - (1)(1) = -8 - 1 = -9 \)

2. \( \begin{vmatrix} -1 & 1 \\ 3 & -4 \end{vmatrix} = (-1)(-4) - (1)(3) = 4 - 3 = 1 \)

3. \( \begin{vmatrix} -1 & 2 \\ 3 & 1 \end{vmatrix} = (-1)(1) - (2)(3) = -1 - 6 = -7 \)

Now substituting back into the determinant \( D \):
\[ D = 2(-9) + 1 + 3(-7) = -18 + 1 - 21 = -38 \]

### Step 4: Calculate \( D_x \), \( D_y \), and \( D_z \)
To find \( D_x \), we replace the first column of \( A \) with \( B \):
\[ D_x = \begin{vmatrix} 8 & -1 & 3 \\ 4 & 2 & 1 \\ 0 & 1 & -4 \end{vmatrix} \]

Calculating \( D_x \):
\[ D_x = 8 \begin{vmatrix} 2 & 1 \\ 1 & -4 \end{vmatrix} - (-1) \begin{vmatrix} 4 & 1 \\ 0 & -4 \end{vmatrix} + 3 \begin{vmatrix} 4 & 2 \\ 0 & 1 \end{vmatrix} \]
Calculating the 2x2 determinants:
1. \( \begin{vmatrix} 2 & 1 \\ 1 & -4 \end{vmatrix} = -9 \) (from previous calculation)
2. \( \begin{vmatrix} 4 & 1 \\ 0 & -4 \end{vmatrix} = (4)(-4) - (1)(0) = -16 \)
3. \( \begin{vmatrix} 4 & 2 \\ 0 & 1 \end{vmatrix} = (4)(1) - (2)(0) = 4 \)

Now substituting back into \( D_x \):
\[ D_x = 8(-9) + 16 + 3(4) = -72 + 16 + 12 = -44 \]

Next, for \( D_y \), we replace the second column of \( A \) with \( B \):
\[ D_y = \begin{vmatrix} 2 & 8 & 3 \\ -1 & 4 & 1 \\ 3 & 0 & -4 \end{vmatrix} \]

Calculating \( D_y \):
\[ D_y = 2 \begin{vmatrix} 4 & 1 \\ 0 & -4 \end{vmatrix} - 8 \begin{vmatrix} -1 & 1 \\ 3 & -4 \end{vmatrix} + 3 \begin{vmatrix} -1 & 4 \\ 3 & 0 \end{vmatrix} \]
Calculating the 2x2 determinants:
1. \( \begin{vmatrix} 4 & 1 \\ 0 & -4 \end{vmatrix} = -16 \)
2. \( \begin{vmatrix} -1 & 1 \\ 3 & -4 \end{vmatrix} = 1 \)
3. \( \begin{vmatrix} -1 & 4 \\ 3 & 0 \end{vmatrix} = 12 \)

Now substituting back into \( D_y \):
\[ D_y = 2(-16) - 8(1) + 3(12) = -32 - 8 + 36 = -4 \]

Finally, for \( D_z \), we replace the third column of \( A \) with \( B \):
\[ D_z = \begin{vmatrix} 2 & -1 & 8 \\ -1 & 2 & 4 \\ 3 & 1 & 0 \end{vmatrix} \]

Calculating \( D_z \):
\[ D_z = 2 \begin{vmatrix} 2 & 4 \\ 1 & 0 \end{vmatrix} - (-1) \begin{vmatrix} -1 & 4 \\ 3 & 0 \end{vmatrix} + 8 \begin{vmatrix} -1 & 2 \\ 3 & 1 \end{vmatrix} \]
Calculating the 2x2 determinants:
1. \( \begin{vmatrix} 2 & 4 \\ 1 & 0 \end{vmatrix} = -4 \)
2. \( \begin{vmatrix} -1 & 4 \\ 3 & 0 \end{vmatrix} = 12 \)
3. \( \begin{vmatrix} -1 & 2 \\ 3 & 1 \end{vmatrix} = -7 \)

Now substituting back into \( D_z \):
\[ D_z = 2(-4) + 12 + 8(-7) = -8 + 12 - 56 = -52 \]

### Step 5: Calculate \( x \), \( y \), and \( z \)
Using Cramer's Rule:
\[ x = \frac{D_x}{D} = \frac{-44}{-38} = \frac{22}{19} \]
\[ y = \frac{D_y}{D} = \frac{-4}{-38} = \frac{2}{19} \]
\[ z = \frac{D_z}{D} = \frac{-52}{-38} = \frac{26}{19} \]

### Final Solution
The solution to the system of equations is:
\[ x = \frac{22}{19}, \quad y = \frac{2}{19}, \quad z = \frac{26}{19} \]