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Found in: Page 22

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Consider the equations$\mathbf{|}\begin{array}{c}\mathbf{y}\mathbf{+}\mathbf{2}\mathbf{k}\mathbf{z}\mathbf{=}\mathbf{0}\\ \mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{6}\mathbf{z}\mathbf{=}\mathbf{2}\\ \mathbf{k}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{z}\mathbf{=}\mathbf{1}\end{array}\mathbf{|}$where${\mathbit{k}}$ is an arbitrary constant.a. For which values of the constant ${\mathbit{k}}$does this system have a unique solution?b. When is there no solution?c. When are there infinitely many solutions?

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

See the step by step solution

## Step 1: 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.

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

## Step 2: Perform the row operations

Perform the row Echelon operation to reduce the matrix.

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

## Step 3: Solve the equations

Consider the third row of the original matrix.

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

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

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

Perform the row Echelon operation to reduce the matrix.

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

## Step 4: 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.

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