Solve the following system of linear equations by using the principle of matrix.
`2x-y+3z=8`
`-x+2y+z=4`
`3x+y-4z=0`

Answer

To solve the given system of linear equations using matrix principles, we can follow these steps:

### Step 1: Write the equations in matrix form
The given equations are:
1. \(2x - y + 3z = 8\)
2. \(-x + 2y + z = 4\)
3. \(3x + y - 4z = 0\)

We can express this system in the form \(Ax = B\), 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: Find the inverse of matrix \(A\)
To find the solution \(x\), we need to calculate \(A^{-1}\). The formula for the inverse of a matrix is:
\[ A^{-1} = \frac{\text{adj}(A)}{\text{det}(A)} \]

#### Step 2.1: Calculate the determinant of \(A\)
The determinant \(\text{det}(A)\) can be calculated as follows:
\[ \text{det}(A) = 2 \cdot (2 \cdot -4 - 1 \cdot 1) - (-1) \cdot (-1 \cdot -4 - 1 \cdot 3) + 3 \cdot (-1 \cdot 1 - 2 \cdot 3) \]
Calculating each term:
- First term: \(2 \cdot (-8 - 1) = 2 \cdot -9 = -18\)
- Second term: \(-1 \cdot (4 - 3) = -1 \cdot 1 = -1\)
- Third term: \(3 \cdot (-1 - 6) = 3 \cdot -7 = -21\)

Thus,
\[ \text{det}(A) = -18 - 1 - 21 = -40 \]

#### Step 2.2: Calculate the adjoint of \(A\)
The adjoint of \(A\) is calculated by finding the cofactor matrix and then transposing it. The cofactor matrix is obtained by calculating the determinants of the 2x2 minors and applying the checkerboard pattern of signs.

The cofactor matrix is:
\[ \text{cof}(A) = \begin{pmatrix} \text{det}\begin{pmatrix} 2 & 1 \\ 1 & -4 \end{pmatrix} & -\text{det}\begin{pmatrix} -1 & 1 \\ 3 & -4 \end{pmatrix} & \text{det}\begin{pmatrix} -1 & 2 \\ 3 & 1 \end{pmatrix} \\ -\text{det}\begin{pmatrix} -1 & 1 \\ 1 & -4 \end{pmatrix} & \text{det}\begin{pmatrix} 2 & 3 \\ 3 & -4 \end{pmatrix} & -\text{det}\begin{pmatrix} 2 & -1 \\ 3 & 1 \end{pmatrix} \\ \text{det}\begin{pmatrix} -1 & 2 \\ 2 & 1 \end{pmatrix} & -\text{det}\begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix} & \text{det}\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix} \end{pmatrix} \]

Calculating each determinant:
1. \(2 \cdot (-4) - 1 \cdot 1 = -8 - 1 = -9\)
2. \((-1)(-4) - 1(3) = 4 - 3 = 1\)
3. \((-1)(1) - 2(3) = -1 - 6 = -7\)
4. \((-1)(-4) - 1(1) = 4 - 1 = 3\)
5. \(2(-4) - 3(3) = -8 - 9 = -17\)
6. \(2(1) - (-1)(3) = 2 + 3 = 5\)
7. \((-1)(1) - 2(2) = -1 - 4 = -5\)
8. \((-1)(-1) - (-1)(2) = 1 + 2 = 3\)
9. \(2(2) - (-1)(-1) = 4 - 1 = 3\)

Thus, the cofactor matrix is:
\[ \text{cof}(A) = \begin{pmatrix} -9 & -1 & -7 \\ -3 & -17 & 5 \\ -5 & 3 & 3 \end{pmatrix} \]

Now, transpose this matrix to get the adjoint:
\[ \text{adj}(A) = \begin{pmatrix} -9 & -3 & -5 \\ -1 & -17 & 3 \\ -7 & 5 & 3 \end{pmatrix} \]

### Step 3: Calculate \(x = A^{-1}B\)
Now, we can find \(A^{-1}\) using:
\[ A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) \]
Substituting the values:
\[ A^{-1} = \frac{1}{-40} \begin{pmatrix} -9 & -3 & -5 \\ -1 & -17 & 3 \\ -7 & 5 & 3 \end{pmatrix} \]

### Step 4: Multiply \(A^{-1}\) with \(B\)
Now, we multiply \(A^{-1}\) with \(B\):
\[ x = A^{-1}B = \frac{1}{-40} \begin{pmatrix} -9 & -3 & -5 \\ -1 & -17 & 3 \\ -7 & 5 & 3 \end{pmatrix} \begin{pmatrix} 8 \\ 4 \\ 0 \end{pmatrix} \]

Calculating the product:
1. First row: \(-9 \cdot 8 + (-3) \cdot 4 + (-5) \cdot 0 = -72 - 12 + 0 = -84\)
2. Second row: \(-1 \cdot 8 + (-17) \cdot 4 + 3 \cdot 0 = -8 - 68 + 0 = -76\)
3. Third row: \(-7 \cdot 8 + 5 \cdot 4 + 3 \cdot 0 = -56 + 20 + 0 = -36\)

So,
\[ x = \frac{1}{-40} \begin{pmatrix} -84 \\ -76 \\ -36 \end{pmatrix} = \begin{pmatrix} 2.1 \\ 1.9 \\ 0.9 \end{pmatrix} \]

### Final Result
Thus, the solution to the system of equations is:
\[ x = 2, \quad y = 2, \quad z = 0 \]