Equations & Matrix

[DarkSun]

As I am new to the subhect, I am having a problems with the following question...

"For which values of ? does the system of equations have a single solution,
For which values of ? does it have more than a single solution?"
2y - ?z = 0
2x - ?y + z = 0
-?x + y + 3z = 0

Using a matrix, i can get to -
(2 -? 1
0 2 -?
-2? 2 6)

But what next?

Thanks!

 

[stapel]

But what next?

Keep going until you get reduced-row form. Then use what you know about system types and their matrices to create and solve equations in terms of lambda.

Eliz.

 

[soroban]

Hello, DarkSun!

Is there a typo?
There doesn't seem to be an elementary solution . . .

For which values of \(\displaystyle \lambda\) does the system of equations have a single solution,
For which values of \(\displaystyle \lambda\) does it have more than a single solution?

\(\displaystyle \begin{array}{ccc} \qquad2y - \lambda z &=& 0 \\ 2x - \lanbda y + z &=& 0 \\ \yrcy{-}\lambda x + y + 3x &=& 0 \end{array}\)

Solving the augmented matrix leads nowhere.
. . A little Thought would have clued you in. .**

In case you're not convinced, here are the (useless) steps . . .


\(\displaystyle \text{Re-arrange the equations: }\;\begin{array}{ccc}2x - \lambda y + z &=& 0 \\ \qquad 2y - \lambda z &=& 0 \\ \text{-}\lambda x + y + 3x &=&0 \end{array}\)


. . . . \(\displaystyle \text{We have: }\;\left|\begin{array}{ccc|c} 2 & \text{-}\lambda & 1 & 0 \\ 0 & 2 & \text{-}\lambda & 0 \\ \text{-}\lambda & 1 & 3 & 0 \end{array}\right|\)


. . . . . . . .\(\displaystyle \begin{array}{c}\frac{1}{2}R_1 \\ \\[-3mm] \frac{1}{2}R_2 \\ \\ \end{array}\left|\begin{array}{ccc|c}1 & \text{-}\frac{\lambda}{2} & \frac{1}{2} & 0 \\ \\[-3mm] 0 & 1 & \text{-}\frac{1}{2} & 0 \\ \\[-3mm] \text{-}\lambda & 1 & 3 & 0 \end{array}\right|\)

. . . .\(\displaystyle \begin{array}{c}R_1+\frac{\lambda}{2}R_2 \\ \\ R_3 + \lambda R_1 \end{array} \left|\begin{array}{ccc|c}1 & 0 & \frac{2-\lambda}{4} & 0 \\ 0 & 1 & \text{-}\frac{\lambda}{2} & 0 \\ 0 & \frac{2-\lambda^2}{2} & \frac{6+\lambda}{2} & 0 \end{array}\right|\)

\(\displaystyle \begin{array}{c}\\ \\ R_3 - (\frac{2-\lambda^2}{2})R_2 \end{array} \left|\begin{array}{ccc|c}1 & 0 & \frac{2-\lambda}{4} & 0 \\ 0 & 1 & \text{-}\frac{\lambda}{2} & 0 \\ 0 & 0 & \frac{12+4x-x^3}{4} & 0 \end{array}\right|\)

. . .\(\displaystyle \begin{array}{c}\\ \\ \frac{4}{12+4x-x^3}R_3 \end{array} \left|\begin{array}{ccc|c}1 &0 & \frac{2-x}{4} & 0 \\ 0 & 1 & \text{-}\frac{\lambda}{2} & 0 \\ 0 & 0 & 1 & 0 \end{array}\right|\)

\(\displaystyle \begin{array}{c}R_1- (\frac{2-\lambda}{4})R_3 \\ R_2 + \frac{\lambda}{2}R_3 \\ \end{array} \left|\begin{array}{ccc|c}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}\right| \quad\hdots\quad\text{See? }\:\text{a complete waste of time!}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

The three given equations represent three planes which pass through the origin.

What point do they have in common? . . . . Duh!