To have a non-zero solution for the homogeneous system [ \begin{cases} 4x+ky+2z=0 \ kx+4y+z=0 \ 2x+2y+z=0 \end{cases} ] the coefficient matrix must be singular, i.e. its determinant must be zero.

1. Form the coefficient matrix

[ A=\begin{pmatrix} 4 & k & 2 \ k & 4 & 1 \ 2 & 2 & 1 \end{pmatrix} ]

For a homogeneous system, a non-trivial solution exists iff [ \det(A)=0. ]

2. Compute the determinant

[ \det(A)= \begin{vmatrix} 4 & k & 2 \ k & 4 & 1 \ 2 & 2 & 1 \end{vmatrix} ]

Expanding along the first row: [ \det(A)=4\begin{vmatrix}4&1\2&1\end{vmatrix} -k\begin{vmatrix}k&1\2&1\end{vmatrix} +2\begin{vmatrix}k&4\2&2\end{vmatrix} ]

Now evaluate each minor:

[ \begin{vmatrix}4&1\2&1\end{vmatrix}=4\cdot 1-1\cdot 2=2 ]

[ \begin{vmatrix}k&1\2&1\end{vmatrix}=k\cdot 1-1\cdot 2=k-2 ]

[ \begin{vmatrix}k&4\2&2\end{vmatrix}=k\cdot 2-4\cdot 2=2k-8 ]

So, [ \det(A)=4(2)-k(k-2)+2(2k-8) ] [ =8-k^2+2k+4k-16 ] [ =-k^2+6k-8 ]

Hence, [ \det(A)=0 \iff -k^2+6k-8=0 ]

i.e. [ k^2-6k+8=0 ] [ (k-2)(k-4)=0 ] So, [ k=2 \text{ or } k=4. ]

3. Number of values of (k)

There are 2 values of (k).

4. Compare with stored answer

Derived answer: 2

Stored correct answer: A = 2

They match.