The number of non-zero values of k for which the linear equations

4x + ky + z = 0

kx + 4y + z = 0

2x + 2y + z = 0

possess a non-zero solution is:

Concept:

Cramer's rule for Linear Equations of Three Variables:

a1x + b1y + c1z = d1

a2x + b2y + c2z = d2

a3x + b3y + c3z = d3

\(\rm D=\begin{vmatrix} \rm a_{1} & \rm a_{2} & \rm a_{3}\\ \rm b_{1} & \rm b_{2} & \rm b_{3}\\ \rm c_{1} & \rm c_{2} & \rm c_{3}\end{vmatrix}\)

\(\rm \rm D_{x}=\begin{vmatrix}\rm d_{1} & \rm d_{2} & \rm d_{3}\\ \rm b_{1} & \rm b_{2} & \rm b_{3}\\ \rm c_{1} & \rm c_{2} & \rm c_{3}\end{vmatrix}\quad \rm D_{y}=\begin{vmatrix}\rm a_{1} & \rm a_{2} & \rm a_{3}\\ \rm d_{1} & \rm d_{2} & \rm d_{3}\\ \rm c_{1} & \rm c_{2} & \rm c_{3}\end{vmatrix}\quad \rm D_{z}=\begin{vmatrix}\rm a_{1} & \rm a_{2} & \rm a_{3}\\ \rm b_{1} & \rm b_{2} & \rm b_{3}\\ \rm d_{1} & \rm d_{2} & \rm d_{3}\end{vmatrix}\)

If D ≠ 0: a unique solution (consistent).

The solution is: \(\rm x=\frac{D_x}{D}\quad y=\frac{D_y}{D}\quad z=\frac{D_z}{D}\).

If D = 0: either infinitely many solutions (consistent and dependent) or no solution (inconsistent). To find out if the system is dependent or inconsistent, another method, such as elimination or Rouché–Capelli theorem, will have to be used.

Calculation:

For the given set of equations:

4x + ky + z = 0

kx + 4y + z = 0

2x + 2y + z = 0

It can be observed that x = y = z = 0 is one solution of the system.

In order to have infinitely many (including non-zero) solutions, D must be zero.

⇒ \(\rm D=\begin{vmatrix} 4 & \rm k & 2 \\ \rm k & 4 & 2\\ 1 & 1 & 1\end{vmatrix}\) = 0

⇒ 4(4 - 2) + k(2 - k) + 2(k - 4) = 0

⇒ 8 + 2k - k² + 2k - 8 = 0

⇒ k² - 4k = 0

⇒ k(k - 4) = 0

⇒ k = 0 OR k - 4 = 0

⇒ k = 0 OR k = 4

Hence, there are 2 possible values of k but a non-zero solution is one.