For what values of $k$ in this set of linear equations $Ax = b$ has no solutions, an infinite number of solutions and an unique solution?

Question

For what values of $k$ in this set of linear equations $Ax = b$ has no solutions, an infinite number of solutions and a unique solutions?

I know I want to be using Gaussian Elimination here, I've augmented the matrix and I'm perfectly familiar with ERO's and back-solving for systems without unknown constants but this is new to me. \begin{array}{ccc|c} 2 & 2 & 0 & 2\\ 0 & k & 1 & 1\\ 1 & 2 & k & 2 \end{array}

Would I try to be putting this into Row-Echelon form? I have an inkling by playing with it that $k = -1$ for no solutions and $k = 1$ for an infinite number of solutions. I can't do the Gaussian steps properly with a $k$ involved to produce some decent working though.

Thank you in advance for any help, solutions or tips. :)

Answer 1

Hint :

In order to have unique solutions, the determinant should be nonzero :

$$ \det(A) = 0 \Leftrightarrow \begin{array}{|ccc|c} 2 & 2 & 0 & \\ 0 & k & 1 \\ 1 & 2 & k \end{array} =0 \Leftrightarrow 2(k^2-1) = 0 \Leftrightarrow k = +-1$$

Now, by plugging $k=1$ to our matrix and doing a Reduced Echelon Form Transformation :

$$\left( \begin{array}{cccc} 1 & 0 & -1 & 0 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)$$

and by plugging $k=-1$, executing again a Reduced Echelon Form Transformation :

$$\left( \begin{array}{cccc} 1 & 0 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)$$

Can you now derive conclusions for inconsistent, unique solution and infinite solutions?

Answer 2

As first comment already said you can use determinantats to find which answerz will dire t to a determinant equals to zero, which means a matrix which no aolution or with infinite solution. Using determinant you fimd that 1 and -1 are k values for determinant to be equals 0. Now if you try to use those two values and reduce the matrix to find the solution then you will find that k=1 lead to infinite solutions, since it leads to a system of three variables and two rows. On the other hand with k=-1 you will find at some point of resuction that two rows show contradictory information, for example one auggesting -x1 + x2 = 1 and the other one suggesting that -x1+x2=-2. In some way its like there are three rows but just two variables, since you have a row that you dont need in order to have a system of two rows two variables.