Question

Find the value of k for which the equations kx-2y+z=1, x-2ky+z=0, x-2y+kz=0 have no solution

Answer 1

Answer:

k=1

Step-by-step explanation:

1) For values of k,which has no solution, determinant of coefficient matrix must be 0.

So,

$\begin{vmatrix}k&-2&1\\1&-2k&1\\1&-2&k\end{vmatrix}=0\\ \\ =>k(-2k^2+2)-2(1-k)+1(-2+2k)=0\\ \\=>-2k(k^2-1)+2(k-1)+2(k-1)=0\\ \\=>(k-1)[-2k(k+1)+2+2]=0\\ \\=>(k-1)(-2k^2-2k+4)=0\\ \\=>(k-1)(k+2)(k-1)=0$

k=1,2

When k=1, Equations are

x-2y+z=1

x-2y+z=0

x-2y+2z=0

First two equations has no solution .

Hence, k=1 gives no solution.

2) When k=2,equations are :

2x-2y+z=1

x-4y+z=0

x-2y+2z=0

This has solutions since none of the are parallel .

=> k=2 gives solution.

Finally,k=1 gives no solution .