Question

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

Answer 1

Hint: It is given that the system of equations has no solution. So, firstly, write the system of the equations in the form of a determinant and equate it to zero because as it has no solution then simplify the determinant to find the value of $k$.

Complete step by step solution:
Here we have three equations written as;
$kx + y + z = 1 \\ x + ky + z = k \\ x + y + zk = {k^2} \\ $
Now, take the determinant is equal to zero we get;
Here to calculate the determinant we avoid the right-hand side and assume it to be zero because if we take this, we are not able to find the solution.
Therefore, we have;
$\begin{vmatrix} k&1&1 \\ 1&k&1 \\ 1&1&k \end{vmatrix} = 0$
Here, we have the first equation in row 1, the second equation in row 2 and the third equation in row 3, and x, y, and z are in columns 1,2 and 3.
Further, by solving the above determinate we get;
$k({k^2} - 1) - 1(k - 1) + 1(1 - k) = 0 \\ \Rightarrow (k - 1)({k^2} + k - 1 - 1) = 0 \\ \Rightarrow (k - 1)({k^2} + k - 2) = 0 \\ \Rightarrow k - 1 = 0,{k^2} + k - 2 = 0 \\ $
Further, by solving the above equation we get;
$k - 1 = 0,{k^2} + k - 2 = 0 \\ \Rightarrow k = 1,{k^2} + 2k - k - 2 = 0 \\ \Rightarrow k = 1,k(k + 2) - 1(k + 2) = 0 \\ \Rightarrow k = 1,(k - 1)(k + 2) = 0 \\ $
Which gives the result as;
$k = 1,k = 1,k = - 2$

As a result, the given equation has no solutions at $k = 1$ or $k = -2$.

Note: Remember that the system of equations has a unique solution, when it has no solution and when it has infinitely many solutions it has the exact same solution.