The integral value of k for which the system of linear equations $kx + y + 2z = 0$, $ky = x - 3z$ and $2x + y + kz = 0$ has a non-zero solution is

For a system of homogeneous linear equations to have a non-zero (non-trivial) solution, the determinant of the coefficient matrix must be equal to zero.

Step-by-Step Solution:

1. Rearrange the equations into standard form ($Ax = 0$):

Equation 1: $kx + y + 2z = 0$

Equation 2: $x - ky - 3z = 0$

Equation 3: 2x + y + kz = 0

2. Set the coefficient determinant to zero:

$\begin{vmatrix} k & 1 & 2 \\ 1 & -k & -3 \\ 2 & 1 & k \end{vmatrix} = 0$

3. Expand the determinant along the first row:

$k[(-k)(k) - (-3)(1)] - 1[(1)(k) - (-3)(2)] + 2[(1)(1) - (-k)(2)] = 0$

$k(-k^2 + 3) - 1(k + 6) + 2(1 + 2k) = 0$

$-k^3 + 3k - k - 6 + 2 + 4k = 0$

$-k^3 + 6k - 4 = 0$

$k^3 - 6k + 4 = 0$

4. Find the roots of the cubic equation:

By inspection (testing integer values), let's check $k = 2$:

$2^3 - 6(2) + 4 = 8 - 12 + 4 = 0$

Since $k = 2$ satisfies the equation perfectly, $k = 2$ is an integral solution.

Dividing $(k^3 - 6k + 4)$ by $(k - 2)$ yields the remaining quadratic term:

$(k - 2)(k^2 + 2k - 2) = 0$

The roots for $k^2 + 2k - 2 = 0$ are non-integers ($k = -1 \pm \sqrt{3}$).

Answer:

The integral value of $k$ is 2.