Question

The system of equations $kx+y+z=1$, $x+ky+z=k$ and $x+y+zk=k^2$ has no solution if $k$ is equal to

• A. $-2$
• B. $-1$
• C. $1$
• D. $0$

Answer 1

The correct option is A

$-2$

Explanation for the correct option:

Given equations are, $kx+y+z=1$, $x+ky+z=k$ and $x+y+zk=k^2$ .

For equations $a_{1}x+b_{1}y+c_{1}z=d_1$, $a_{2}x+b_{2}y+c_{2}z=d_2$ and $a_{3}x+b_{3}y+c_{3}z=d_3$, we difine

$\mathrm{\Delta }=\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|,{\mathrm{\Delta }}_1=\left|\begin{array}{ccc}{d}_1& b_1& c_1\\ d_2& b_2& c_2\\ d_3& b_3& c_3\end{array}\right|,{\mathrm{\Delta }}_2=\left|\begin{array}{ccc}{a}_1& d_1& c_1\\ a_2& d_2& c_2\\ a_3& d_3& c_3\end{array}\right|,{\mathrm{\Delta }}_3=\left|\begin{array}{ccc}{a}_1& b_1& d_1\\ a_2& b_2& d_2\\ a_3& b_3& d_3\end{array}\right|$

If, $\mathrm{\Delta }=0$ and ${\mathrm{\Delta }}_1\ne 0$ or ${\mathrm{\Delta }}_2\ne 0$ or ${\mathrm{\Delta }}_3\ne 0$, then the system of equations are inconsistent and have no solutions.

Here,

$$\begin{gathered} \left(a_1,b_1,c_1,d_1\right)=\left(k,1,1,1\right) \\ \left(a_2,b_2,c_2,d_2\right)=\left(1,k,1,k\right) \\ \left(a_3,b_3,c_3,d_3\right)=\left(1,1,k,k^2\right) \end{gathered}$$

Thus, if

$\begin{array}{cc}\mathrm{\Delta }=& \left|\begin{array}{ccc}\mathrm{k}& 1& 1\\ 1& \mathrm{k}& 1\\ 1& 1& \mathrm{k}\end{array}\right|=0\\ \Rightarrow & \mathrm{k}\left({\mathrm{k}}^2-1\right)-\left(\mathrm{k}-1\right)+\left(1-\mathrm{k}\right)=0\\ \Rightarrow & \mathrm{k}\left(\mathrm{k}+1\right)\left(\mathrm{k}-1\right)-\left(\mathrm{k}-1\right)+\left(1-\mathrm{k}\right)=0\\ \Rightarrow & \left(\mathrm{k}-1\right)\left({\mathrm{k}}^2+\mathrm{k}-1-1\right)=0\\ \Rightarrow & \left(\mathrm{k}-1\right)\left({\mathrm{k}}^2+\mathrm{k}-2\right)=0\\ \Rightarrow & \left(\mathrm{k}-1\right)\left({\mathrm{k}}^2-\mathrm{k}+2\mathrm{k}-2\right)=0\\ \Rightarrow & \left(\mathrm{k}-1\right)\left(\mathrm{k}-1\right)\left(\mathrm{k}+2\right)=0\end{array}$

$\Rightarrow k=1$ or $k=-2$

And, if

$\begin{array}{cc}{\mathrm{\Delta }}_1=& \left|\begin{array}{ccc}1& 1& 1\\ k& \mathrm{k}& 1\\ k^2& 1& \mathrm{k}\end{array}\right|\ne 0\\ \Rightarrow & 1\left(k^2-1\right)-1\left(k^2-k^2\right)+1\left(k-k^3\right)\ne 0\\ \Rightarrow & \left(k+1\right)\left(k-1\right)-0+k\left(1-k^2\right)\ne 0\\ \Rightarrow & \left(k+1\right)\left(k-1\right)-0+k\left(1+k\right)\left(1-k\right)\ne 0\\ \Rightarrow & \left(k+1\right)\left(k-1+k-k^2\right)\ne 0\\ \Rightarrow & \left(k+1\right)\left(-k^2+2k-1\right)\ne 0\\ \Rightarrow & \left(k+1\right)\left(k^2-2k+1\right)\ne 0\\ \Rightarrow & \left(k+1\right){\left(k-1\right)}^2\ne 0\end{array}$

$\Rightarrow k\ne \pm 1$

From the above two results, if $k=-2$, the set of equations has no solutions.

Hence, $\mathrm{Option}\left(\mathrm{A}\right)$ is correct.