# Matrices Again

• Oct 1st 2007, 09:31 PM
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
• Oct 2nd 2007, 05:10 AM
Jhevon
Quote:

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?
• Oct 2nd 2007, 05:58 AM
Soroban

Quote:

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.

• Oct 2nd 2007, 08:22 AM
red_dog
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.
• Oct 2nd 2007, 06:39 PM