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

Given the equations:
1. \(2x - y + 3z = 8\) (Equation 1)
2. \(-x + 2y + z = 4\) (Equation 2)
3. \(3x + y - 4z = 0\) (Equation 3)

### Step 1: Write the equations in matrix form
We can express the system of equations in the form of \(AX = B\), where:
\[ A = \begin{pmatrix} 2 & -1 & 3 \\ -1 & 2 & 1 \\ 3 & 1 & -4 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 8 \\ 4 \\ 0 \end{pmatrix} \]

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

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

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

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

### Step 3: Calculate \(D_x\), \(D_y\), and \(D_z\)
#### Calculate \(D_x\)
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. \( \begin{vmatrix} 2 & 1 \\ 1 & -4 \end{vmatrix} = -9 \) (calculated earlier)
2. \( \begin{vmatrix} 4 & 1 \\ 0 & -4 \end{vmatrix} = (4)(-4) - (1)(0) = -16 \)
3. \( \begin{vmatrix} 4 & 2 \\ 0 & 1 \end{vmatrix} = (4)(1) - (2)(0) = 4 \)

Substituting back into \(D_x\):
\[ D_x = 8(-9) + 1(-16) + 3(4) = -72 - 16 + 12 = -76 \]

#### Calculate \(D_y\)
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. \( \begin{vmatrix} -1 & 1 \\ 3 & -4 \end{vmatrix} = 1 \) (calculated earlier)
3. \( \begin{vmatrix} -1 & 4 \\ 3 & 0 \end{vmatrix} = -12 \)

Substituting back into \(D_y\):
\[ D_y = 2(-16) - 8(1) + 3(-12) = -32 - 8 - 36 = -76 \]

#### Calculate \(D_z\)
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. \( \begin{vmatrix} 2 & 4 \\ 1 & 0 \end{vmatrix} = (2)(0) - (4)(1) = -4 \)
2. \( \begin{vmatrix} -1 & 4 \\ 3 & 0 \end{vmatrix} = -12 \) (calculated earlier)
3. \( \begin{vmatrix} -1 & 2 \\ 3 & 1 \end{vmatrix} = -7 \) (calculated earlier)

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

### Step 4: Calculate the values of \(x\), \(y\), and \(z\)
Using Cramer's Rule:
\[ x = \frac{D_x}{D} = \frac{-76}{-38} = 2 \]
\[ y = \frac{D_y}{D} = \frac{-76}{-38} = 2 \]
\[ z = \frac{D_z}{D} = \frac{-52}{-38} = \frac{26}{19} \]

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