# How can I solve this system of linear equations?

• Feb 8th 2013, 03:16 PM
cellae
How can I solve this system of linear equations?
Using either Gaussian or Gauss-Jordan elimination, find the value(s) of k, if any, for which the following system will have
(i) no solution
(ii) a unique solution, and
(iii) infinitely many solutions

x − 3z = −3
−2x − λy + z = 2
x + 2y + λz = 1

Thanks for any help
• Feb 8th 2013, 05:07 PM
zhandele
Re: How can I solve this system of linear equations?
There is no "k" in your equations. I assume you're looking for values of lambda, not "k", right?

Can you set this up as a matrix? Can you row-reduce that matrix?
• Feb 8th 2013, 06:37 PM
jakncoke
Re: How can I solve this system of linear equations?
if $\displaystyle \lambda$ = -5 (no solution) or 2 (infinitely many soln), for every other value, a unique solution exists. . I cheated and used determinant.
• Feb 8th 2013, 06:41 PM
cellae
Re: How can I solve this system of linear equations?
Sorry, do you mind briefly explaining how you got these answers, just I get what I'm asked to do, just confused by the process..
• Feb 8th 2013, 06:50 PM
jakncoke
Re: How can I solve this system of linear equations?
I took the determinant of
$\displaystyle A = \begin{bmatrix} 1 & 0 & -3 \\ -2 & -\lambda & 0 \\ 1 & 2 & \lambda \end{bmatrix}$
to get Det(A) = $\displaystyle - \lambda^{2} - 3 \lambda + 10$ and i used the quadratic formula to get the roots, $\displaystyle \lambda = -5, 2$. A system has zero or infinitely many solution only when Det(A) = 0. so Det(A) = 0, when $\displaystyle \lambda = -5, 2$. I then found that for $\displaystyle \lambda = -5$ using row reducing, that the system didint have a solution, and again for $\displaystyle \lambda = 2$, i found out by row reducing that it had infinite soln.
• Feb 8th 2013, 06:54 PM
cellae
Re: How can I solve this system of linear equations?
Makes so much sense, thank you !
• Feb 12th 2013, 11:05 PM
cellae
Re: How can I solve this system of linear equations?
Sorry Jakncoke,
I was just confirming, would the values for λ be 5, -2 ?

because of the negative in front of -λ^2 - 3λ + 10
-(λ - 5)(λ + 2)
• Feb 12th 2013, 11:14 PM
jakncoke
Re: How can I solve this system of linear equations?
Quote:

Originally Posted by cellae
Sorry Jakncoke,
I was just confirming, would the values for λ be 5, -2 ?

because of the negative in front of -λ^2 - 3λ + 10
-(λ - 5)(λ + 2)

$\displaystyle (\lambda - 5)(\lambda + 2) = \lambda^{2} + 2 \lambda - 5 \lambda - 10 = \lambda^{2} + -3 \lambda - 10$ negative of that is $\displaystyle -\lambda^{2} + 3 \lambda + 10$ which is not the determinant (the 3 has to be -3) so root ($\displaystyle \lambda$ is as stated -5, and 2.
• Feb 12th 2013, 11:20 PM
cellae
Re: How can I solve this system of linear equations?
Thanks again !!