Solve the following system of homogenous linear equations :
`x+y-z=0`, `x-2y+z=0`, `3x+6y-5z=0`

Answer

To solve the system of homogeneous linear equations given by:

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

we can follow these steps:

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

The coefficient matrix \( A \) is:
\[ A = \begin{pmatrix} 1 & 1 & -1 \\ 1 & -2 & 1 \\ 3 & 6 & -5 \end{pmatrix} \]

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

### Step 2: Calculate the determinant of the coefficient matrix \( A \)
To find the determinant of \( A \), we can use the formula for the determinant of a 3x3 matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \( A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \).

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

Calculating the determinant:
\[ \text{det}(A) = 1((-2)(-5) - (1)(6)) - 1((1)(-5) - (1)(3)) + (-1)((1)(6) - (-2)(3)) \]
\[ = 1(10 - 6) - 1(-5 - 3) - (6 + 6) \]
\[ = 1(4) - 1(-8) - 12 \]
\[ = 4 + 8 - 12 = 0 \]

### Step 3: Analyze the determinant
Since \( \text{det}(A) = 0 \), this indicates that the system has non-trivial solutions (infinite solutions).

### Step 4: Express one variable in terms of another
We can express \( z \) in terms of \( x \) and \( y \). From the first equation:
\[ z = x + y \]

### Step 5: Substitute \( z \) into the other equations
Substituting \( z \) into the second equation:
\[ x - 2y + (x + y) = 0 \implies 2x - y = 0 \implies y = 2x \]

Now substituting \( y \) into the expression for \( z \):
\[ z = x + y = x + 2x = 3x \]

### Step 6: Write the solution in terms of a parameter
Let \( x = k \) where \( k \) is any real number. Then:
\[ y = 2k, \quad z = 3k \]

### Final Solution
Thus, the solution to the system of equations can be expressed as:
\[ (x, y, z) = (k, 2k, 3k) \quad \text{for any } k \in \mathbb{R} \]