91Ó°ÊÓ

In the system of equations, the variables can be eliminated by performing arithmetic operations on the equations.

Represent the system of equations in the form of matrix.

$\left|\begin{array}{ccccc}0& 1& 2k& |& 0\\ 1& 2& 6& |& 2\\ k& 0& 2& |& 1\end{array}\right|$

Perform the row Echelon operation to reduce the matrix.

$\begin{array}{c}\left(\begin{array}{cccc}0& 1& 2k& 0\\ 1& 2& 6& 2\\ k& 0& 2& 1\end{array}\right)=\left(\begin{array}{cccc}k& 0& 2& 1\\ 0& 2& \frac{6k−2}{k}& \frac{2k−1}{k}\\ 0& 0& \frac{2k^2−3k+1}{k}& −\frac{2k−1}{2k}\end{array}\right)\\ =\left(\begin{array}{cccc}1& 0& 0& \frac{1}{k−1}\\ 0& 1& 0& \frac{k}{k−1}\\ 0& 0& 1& −\frac{1}{2(k−1)}\end{array}\right)\end{array}$

Consider the third row of the original matrix.

$\left|\begin{array}{ccccc}0& 1& 2k& |& 0\\ 1& 2& 6& |& 2\\ k& 0& 2& |& 1\end{array}\right|$

Put$k=\frac{1}{2}$

$\left|\begin{array}{ccccc}0& 1& 1& |& 0\\ 1& 2& 6& |& 2\\ \frac{1}{2}& 0& 2& |& 1\end{array}\right|$

Perform the row Echelon operation to reduce the matrix.

$\begin{array}{c}\left(\begin{array}{cccc}0& 1& 1& 0\\ 1& 2& 6& 2\\ \frac{1}{2}& 0& 2& 1\end{array}\right)=\left(\begin{array}{cccc}1& 2& 6& 2\\ 0& 1& 1& 0\\ 0& 0& 0& 0\end{array}\right)\\ =\left(\begin{array}{cccc}1& 0& 4& 2\\ 0& 1& 1& 0\\ 0& 0& 0& 0\end{array}\right)\end{array}$

As the number of variables equals the number of equations, thus, the system has unique solution, except for the values,$k=\frac{1}{2},1$ .

When$k=1$, the third column equals to undefined value. Thus, for, $k=1$the system has no solution.

For,$k=\frac{1}{2}$ the system will have infinitely many solutions as the third row equals to zero.

(a)The system of equations will have a unique solution for the value, $k≠\frac{1}{2},1$.

(b)The system of equations will have no solution for the value,$k=1$ .

(c)The system of equations will have infinitely many solutions for the value,$k=\frac{1}{2}$ .