Question

The number of values of k for which the linear equations 4x+ky+2z=0 kx+4y+z=0 2x+2y+z=0 possesses a non zero solution is 0 or 3 or 2 or 1

Answer 1

Answer:

Step-by-step explanation:

4x + ky+ 2z = 0;

kx + 4y + z = 0;

2x + 2y + z = 0.

Now, for the system of equations to possess a non -zero solution, the value of the determinant should be zero.

Hence, D = 0 gives, [4(4-2) – k(k-2) + 2(2k-8)]

Hence, we have (16 – 8 – k2 + 2k + 4k – 16) = 0

This gives -k2 + 6k -8 = 0

Hence, k = 4, 2.

hope it would be helpful..................

Answer 2

Answer:

The number of values of k is 2.

Step-by-step explanation:

To find the non-zero solution we first need to find the determinant value and equate it to zero.

i.e., determinant ( $\Delta$ ) = 0.

Entering the coefficient values into the determinant, we get

                                        $\left|\begin{array}{ccc}4&k&2\\k&4&1\\2&2&1\end{array}\right|=0$

Expanding the determinant we obtain

$4 (4-2)-k (k-2)+2 (2k-8)=0$ ,

$8-k^2+2k+4k-16=0$,

$k^2-6k+8=0$,

$(k-2)(k-4)=0$,

$k=2, k=4$

Hence the number of values obtained for k is 2.