Show that the homogenous system of equations
`x - 2y + z = 0, x + y - z = 0, 3 x + 6y - 5z = 0 `
has a non-trivial solution. Also find the solution

Answer

To show that the homogeneous system of equations:

1. \( x - 2y + z = 0 \)
2. \( x + y - z = 0 \)
3. \( 3x + 6y - 5z = 0 \)

has a non-trivial solution, we can use the determinant method. If the determinant of the coefficients of the variables is zero, then there exists a non-trivial solution.

### Step 1: Write the coefficient matrix and calculate the determinant

The coefficient matrix \( A \) for the system of equations is:

\[ A = \begin{pmatrix} 1 & -2 & 1 \\ 1 & 1 & -1 \\ 3 & 6 & -5 \end{pmatrix} \]

Now, we calculate the determinant \( \Delta \) of matrix \( A \):

\[ \Delta = \begin{vmatrix} 1 & -2 & 1 \\ 1 & 1 & -1 \\ 3 & 6 & -5 \end{vmatrix} \]

Using the formula for the determinant of a 3x3 matrix:

\[ \Delta = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Where:
- \( a = 1, b = -2, c = 1 \)
- \( d = 1, e = 1, f = -1 \)
- \( g = 3, h = 6, i = -5 \)

Calculating \( \Delta \):

\[ \Delta = 1 \cdot (1 \cdot -5 - (-1) \cdot 6) - (-2) \cdot (1 \cdot -5 - (-1) \cdot 3) + 1 \cdot (1 \cdot 6 - 1 \cdot 3) \]

Calculating each term:

1. \( 1 \cdot (-5 + 6) = 1 \)
2. \( 2 \cdot (-5 + 3) = 2 \cdot -2 = -4 \)
3. \( 1 \cdot (6 - 3) = 3 \)

So,

\[ \Delta = 1 + 4 + 3 = 8 \]

### Step 2: Check if the determinant is zero

Since \( \Delta \neq 0 \), we made a mistake in our calculations. Let's recalculate the determinant correctly.

Calculating it step by step:

\[ \Delta = 1 \cdot (-5 - (-6)) - (-2) \cdot (-5 - (-3)) + 1 \cdot (6 - 3) \]

1. \( 1 \cdot (-5 + 6) = 1 \)
2. \( -(-2) \cdot (-5 + 3) = 2 \cdot -2 = -4 \)
3. \( 1 \cdot (6 - 3) = 3 \)

So we have:

\[ \Delta = 1 - 4 + 3 = 0 \]

### Conclusion: Non-trivial solution exists

Since \( \Delta = 0 \), the system has a non-trivial solution.

### Step 3: Finding the non-trivial solution

Now, we will express \( z \) in terms of \( x \) and \( y \).

Let \( z = k \) (where \( k \) is a parameter). We can express \( x \) and \( y \) in terms of \( k \).

From the second equation:

\[ x + y - k = 0 \implies y = k - x \]

Substituting \( y \) in the first equation:

\[ x - 2(k - x) + k = 0 \]

Expanding and simplifying:

\[ x - 2k + 2x + k = 0 \implies 3x - k = 0 \implies x = \frac{k}{3} \]

Now substituting \( x \) back to find \( y \):

\[ y = k - \frac{k}{3} = \frac{2k}{3} \]

Thus, we have:

\[ x = \frac{k}{3}, \quad y = \frac{2k}{3}, \quad z = k \]

### Final Non-Trivial Solution

The non-trivial solution can be expressed as:

\[ (x, y, z) = \left(\frac{k}{3}, \frac{2k}{3}, k\right) \]

where \( k \) is any non-zero real number.