Solve the following systems of linear homogenous equations :
`2x+y-3z=0`, `x+3y+z=0` and `3x-2y+z=0`

Answer

To solve the system of linear homogeneous equations given by:

1. \( 2x + y - 3z = 0 \)
2. \( x + 3y + z = 0 \)
3. \( 3x - 2y + z = 0 \)

we can use the matrix method. Here are the steps to find the solution:

### Step 1: Write the equations in matrix form
We can express the system of equations in the matrix form \( A \mathbf{x} = 0 \), where \( A \) is the coefficient matrix and \( \mathbf{x} \) is the variable matrix.

The coefficient matrix \( A \) is:
\[ A = \begin{bmatrix} 2 & 1 & -3 \\ 1 & 3 & 1 \\ 3 & -2 & 1 \end{bmatrix} \]

And the variable matrix \( \mathbf{x} \) is:
\[ \mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \]

### Step 2: Calculate the determinant of the coefficient matrix
To determine the nature of the solution, we need to calculate the determinant of matrix \( A \).

\[ \text{det}(A) = \begin{vmatrix} 2 & 1 & -3 \\ 1 & 3 & 1 \\ 3 & -2 & 1 \end{vmatrix} \]

Using the determinant formula for a \( 3 \times 3 \) matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \( a, b, c \) are the elements of the first row, and \( d, e, f, g, h, i \) are the elements of the second and third rows respectively.

Calculating the determinant:
\[ = 2 \begin{vmatrix} 3 & 1 \\ -2 & 1 \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 3 & 1 \end{vmatrix} - 3 \begin{vmatrix} 1 & 3 \\ 3 & -2 \end{vmatrix} \]

Calculating the minors:
\[ = 2(3 \cdot 1 - (-2) \cdot 1) - 1(1 \cdot 1 - 3 \cdot 1) - 3(1 \cdot -2 - 3 \cdot 3) \]
\[ = 2(3 + 2) - 1(1 - 3) - 3(-2 - 9) \]
\[ = 2(5) - 1(-2) - 3(-11) \]
\[ = 10 + 2 + 33 \]
\[ = 45 \]

### Step 3: Analyze the determinant
Since \( \text{det}(A) = 45 \neq 0 \), we conclude that the system of equations has only the trivial solution.

### Step 4: State the solution
The only solution to the system of equations is:
\[ x = 0, \quad y = 0, \quad z = 0 \]