# Find all values of K for which the system of linear equations..

• Feb 26th 2011, 05:59 PM
Oiler
Find all values of K for which the system of linear equations..
Three planes have the following equations...

$x+2y+z = 3$
$3x+4y+5z=k$
$2x-y+kz=1$

$$\emph{Augumented Matrix} = \left( {\begin{array}{cccc} 1 & 2 & 1 & 3 \\ 3 & 4 & 5 & k \\ 2 & -1 & k & 1 \end{array} } \right)$$

$$\emph{Reduced - Row echelon form} = \left( {\begin{array}{cccc} 1 & 0 & 0 & k+\frac{3}{2} \\ 0 & 1 & 0 & \frac{-k}{2}+2 \\ 0 & 0 & 1 & \frac{-5}{2} \end{array} } \right)$$

$\emph{Find all values of k for which the system of linear equations}$

$\emph{(i) has infinitely many solutions}$
$\emph{(ii) has a unique solution}$
$\emph{(iii) has no solutions}$

Not sure on how to approach this question..
• Feb 26th 2011, 06:12 PM
Ackbeet
I would work with determinants. Check this thread out. That might give you a few ideas. Make sure to read all the way to the end, if you're interested in a rigorous proof, as you can see that my first idea was invalid.
• Feb 27th 2011, 06:26 AM
HallsofIvy
Quote:

Originally Posted by Oiler
Three planes have the following equations...

$x+2y+z = 3$
$3x+4y+5z=k$
$2x-y+kz=1$

$$\emph{Augumented Matrix} = \left( {\begin{array}{cccc} 1 & 2 & 1 & 3 \\ 3 & 4 & 5 & k \\ 2 & -1 & k & 1 \end{array} } \right)$$

This can't be right. In reducing the first two columns, you will get a term involving k in the third column of the third row (I get "k- 6"), then have to divide by that term. You should have a "k" in the denominator.
Quote:

$$\emph{Reduced - Row echelon form} = \left( {\begin{array}{cccc} 1 & 0 & 0 & k+\frac{3}{2} \\ 0 & 1 & 0 & \frac{-k}{2}+2 \\ 0 & 0 & 1 & \frac{-5}{2} \end{array} } \right)$$

$\emph{Find all values of k for which the system of linear equations}$

$\emph{(i) has infinitely many solutions}$
$\emph{(ii) has a unique solution}$
$\emph{(iii) has no solutions}$

Not sure on how to approach this question..
• Feb 27th 2011, 10:11 AM
LoblawsLawBlog
I solved this using Maxima and its answer agrees with Oiler's reduced echelon form.

Just use that form to get expressions for your variables, Oiler. You can read them right off. Looking at the last row, what's z? Then find x and y similarly (these depend on k). Is there any k that won't give an answer, or will give more than one answer?
• Feb 27th 2011, 10:28 AM
Plato
• Feb 27th 2011, 12:11 PM
LoblawsLawBlog
Oops, I answered without thinking this through. It looks like those programs cancel factors in the denominators. If it's reduced manually then there are ways to get more than one solution.