Question

The number of values of k for which the system of linear equations, $(2k+1)x+5ky=k+2$ and $kx+(k+2)y=2$ has no solution, is:

• A. $Infinitelymany$
• B. $3$
• C. $1$
• D. $2$

Answer 1

The correct option is C $1$

A non-homogeneous system of linear equations has a unique non-trivial solution if and only if its determinant is non-zero. If this determinant is zero, then the system has either no nontrivial solutions or an infinite number of solutions.
$\left[\begin{array}{cc}(k+2)& 10\\ k& (k+3)\end{array}\right]$ $\left[\begin{array}{c}x\\ y\end{array}\right]$ $=\left[\begin{array}{c}k\\ k-1\end{array}\right]$
Now it is of the Form $Ax=B$
Now to for the system to have no Solution , determinant of $A$ must be $0$,as follows
$\Rightarrow |A|=(k+2)(k+3)-k\times 10=0\Rightarrow k^2-5k+6=(k-2)(k-3)=0$
Therefore for $k=2,3$ system will have no solution.
For $k=2$, we get infinitely many solutions, after substituting the value of $k=2$ in the equations.

Thus $k=3$

Thus, the number of solutions is $1$.