if the system of equations :
`{:(2x+3y-z=0),(3x+2y+kz=0),(4x+y+z=0):}`
have a set of non-zero integral solutions then , find the smallest positive value of z.

Answer

To solve the given system of equations for non-zero integral solutions, we will express the system in matrix form and use the determinant condition for non-trivial solutions.

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

We can represent this system in matrix form as:
\[ \begin{bmatrix} 2 & 3 & -1 \\ 3 & 2 & k \\ 4 & 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 2: Calculate the determinant of the coefficient matrix
To find the condition for non-trivial solutions, we need to calculate the determinant of the coefficient matrix:
\[ \Delta = \begin{vmatrix} 2 & 3 & -1 \\ 3 & 2 & k \\ 4 & 1 & 1 \end{vmatrix} \]

Calculating the determinant using the rule of Sarrus or cofactor expansion:
\[ \Delta = 2 \begin{vmatrix} 2 & k \\ 1 & 1 \end{vmatrix} - 3 \begin{vmatrix} 3 & k \\ 4 & 1 \end{vmatrix} - 1 \begin{vmatrix} 3 & 2 \\ 4 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} 2 & k \\ 1 & 1 \end{vmatrix} = 2 \cdot 1 - k \cdot 1 = 2 - k\)

2. \(\begin{vmatrix} 3 & k \\ 4 & 1 \end{vmatrix} = 3 \cdot 1 - k \cdot 4 = 3 - 4k\)

3. \(\begin{vmatrix} 3 & 2 \\ 4 & 1 \end{vmatrix} = 3 \cdot 1 - 2 \cdot 4 = 3 - 8 = -5\)

Substituting these back into the determinant:
\[ \Delta = 2(2 - k) - 3(3 - 4k) - 1(-5) \]
\[ = 4 - 2k - 9 + 12k + 5 \]
\[ = (4 - 9 + 5) + (12k - 2k) \]
\[ = 0 + 10k = 10k \]

### Step 3: Set the determinant to zero for non-trivial solutions
For the system to have non-zero integral solutions, we need:
\[ 10k = 0 \implies k \neq 0 \]

### Step 4: Find the smallest positive value of \(z\)
Now, we need to find the smallest positive value of \(z\) when \(k\) is a non-zero integer. We can choose \(k = 1\) for simplicity.

Substituting \(k = 1\) back into the original equations:
1. \(2x + 3y - z = 0\)
2. \(3x + 2y + z = 0\)
3. \(4x + y + z = 0\)

From the first equation:
\[ z = 2x + 3y \]

From the second equation:
\[ z = -3x - 2y \]

Setting these equal:
\[ 2x + 3y = -3x - 2y \]
\[ 5x + 5y = 0 \implies x + y = 0 \implies y = -x \]

Substituting \(y = -x\) into \(z = 2x + 3y\):
\[ z = 2x + 3(-x) = 2x - 3x = -x \]

For \(x = 1\), \(y = -1\), we get:
\[ z = -1 \]

To find the smallest positive value of \(z\), we can try \(x = -1\):
\[ z = 1 \]

Thus, the smallest positive value of \(z\) is:
\[ \boxed{1} \]