Question

Consider the equations

$\mathbf{\left|}\begin{array}{c}\mathbf{y}\mathbf{+}\mathbf{2}\mathbf{k}\mathbf{z}\mathbf{=}\mathbf{0}\\ \mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{6}\mathbf{z}\mathbf{=}\mathbf{2}\\ \mathbf{k}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{z}\mathbf{=}\mathbf{1}\end{array}\mathbf{\right|}$

where$\mathit{k}$ is an arbitrary constant.

a. For which values of the constant $\mathit{k}$does this system have a unique solution?

b. When is there no solution?

c. When are there infinitely many solutions?

Short answer

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

(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, $k=\frac{1}{2}$.

Step-by-step solution

Transform the equations into a matrix form

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 operations

Perform the row Echelon operation to reduce the matrix.

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

Solve the equations

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.

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

Consider the conditions for finding the type of solutions

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.

Final answer

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

(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,$k=\frac{1}{2}$ .