# Math Help - Matrices Again

1. ## Matrices Again

For the following system of linear equations determine the values of k for which it has (i) no solutions, (ii) infinitely many solutions, and (iii) a unique solution. For those values of k where solutions exist, find what those solutions are.

I am trying to do this with Gaussian or Gauss-Jordan elimination

2x1 + 6x2 - 4x3 = 0
3x1 + 9x2 - 6x3 = 0
-4x1 - 12x2 + kx3 = 0

I'm familiar with how to do Gaussian elimination, but I don't know how to go about solving each of the situations they ask for.

Thanks

For the following system of linear equations determine the values of k for which it has (i) no solutions, (ii) infinitely many solutions, and (iii) a unique solution. For those values of k where solutions exist, find what those solutions are.

I am trying to do this with Gaussian or Gauss-Jordan elimination

2x1 + 6x2 - 4x3 = 0
3x1 + 9x2 - 6x3 = 0
-4x1 - 12x2 + kx3 = 0

I'm familiar with how to do Gaussian elimination, but I don't know how to go about solving each of the situations they ask for.

Thanks
create the augmented matrix (from the coefficients):

$\left( \begin{array}{ccc|c} 2 & 6 & 4 & 0\\ 3 & 9 & -6 & 0\\ -4 & -12 & k & 0\end{array} \right)$

since you are familar with Gaus-Jordan elimination, perform it on the system above, then read of your solutions at the end. do you know how to do that?

For the following system of linear equations, determine the values of $k$
for which it has (i) no solutions, (ii) infinitely many solutions, and (iii) a unique solution.
For those values of $k$ where solutions exist, find what those solutions are.

. . $\begin{array}{ccc}2x + 6y - 4z& = &0\\
3x + 9y - 6z & = & 0\\ \text{-}4x - 12y + kz & = & 0\end{array}$

We have: . $\begin{vmatrix}\:2 & 6 & \text{-}4 & | & 0\: \\ \:3 & 9 & \text{-}6 & | & 0\: \\ \:\text{-}4 & \text{-}12 & k & | & 0\:\end{vmatrix}$

. . . . $\begin{array}{c}\frac{1}{2}R_1 \\ \frac{1}{3}R_2 \\ \text{-}\frac{1}{4}R_3\end{array} \;\begin{vmatrix}\:1 & 3 & \text{-}2 &|& 0\: \\ \:1 & 3 & \text{-}2 &|& 0\: \\ \:1 & 3 & \text{-}\frac{k}{4} & | & 0\: \end{vmatrix}$

$\begin{array}{c} \\ R_2-R_1\\R_3-R_1\end{array}\;
\begin{vmatrix}\;1 & 3 & \text{-}2 & | & 0\: \\ \:0 & 0 & 0 & | & 0\: \\ \:0 & 0 & 2\,\text{-}\,\frac{k}{4} & | & 0\; \end{vmatrix}$

It turns out that the three equations are almost identical.

If $k = 8$, the three equation are identical.
We have one equation: . $x + 3y - 2z\:=\:0$
. . which has infinitely many solutions.

If $k \neq 8$, we have: . $x + 3y \:=\:0,\;\;z \:=\:0$
. . which also has infinitely many solutions.

I see no way that the system can have a unique solution or no solutions.

4. The system is homogeneous and it has always at least a solution: $(0,0,0)$.

An homogeneous system has an unique solution ( (0,0,0) ) if the determinant of the matrix of system is not equal to 0.

In this problem, the determinant is 0 for all real k. So, the system has more than one solution.

5. Thanks a lot, that must've been why I was having trouble trying to find unique and no solutions.

And sorry for the double post, I had no idea I did that.