no solution, a unique solution, infinitely many solutions?

[maeveoneill]

For what value(S) of k, if any, wil the systems ahve (a) no solutions, (b) a unique solution, and (c) infinitely many solutions?

x-2y +3z =2
x+y +z =k
2x -y +4z = k^2.

If someone could explain how I would figure this out.
Thank you very much!

 

[Deleted member 4993]

For what value(S) of k, if any, wil the systems ahve (a) no solutions, (b) a unique solution, and (c) infinitely many solutions?

x-2y +3z =2
x+y +z =k
2x -y +4z = k^2.

If someone could explain how I would figure this out.
Thank you very much!

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.
Hint:

What are the conditions for:

(a) no solutions,

(b) a unique solution, and

(c) infinitely many solutions?

Check the equations for independency.

 

[soroban]

Hello, maeveoneill!

For what value(s) of $\displaystyle k$, if any, will the system have (a) no solutions,
(b) a unique solution, and (c) infinitely many solutions?

. . $\displaystyle \begin{array}{cccc}(1) & x-2y +3z &=& 2 \\ (2) & x+y +z &=&k \\ (3) & 2x -y +4z &=& k^2 \end{array}$

Using an Augmented Matrix, try to solve the system.

$\displaystyle \text{Switch equations (2) and (3): }\:\begin{array}{ccc}x - 2y + 3z &=& 2 \\ 2x - y + z &=& k^2 \\ x + y + z &=& k \end{array}$


$\displaystyle \text{We have: }\;\left[\begin{array}{ccc|c} 1 & \text{-}2 & 3 & 2 \\ 2 & \text{-}1 & 4 & k^2 \\ 1 & 1 & 1 & k \end{array}\right]$


$\displaystyle \begin{array}{c} \\ R_2-2R_1 \\ R_3-R_1\end{array} \left[\begin{array}{ccc|c}1 & \text{-}2 & 3 & 2 \\ 0 & 3 & \text{-}2 & k^2-4 \\ 0 & 3 & \text{-}2 & k-2 \end{array}\right]$


. . . $\displaystyle \begin{array}{c}\\ \frac{1}{3}R_2 \\ \\ \end{array} \left[\begin{array}{ccc|c}1 & \text{-}2 & 3 & 2 \\ 0 & 1 & \text{-}\frac{2}{3} & \frac{k^2-4}{3} \\ 0 & 3 & \text{-}2 & k-2 \end{array}\right]$


$\displaystyle \begin{array}{c}R_1+2R_2 \\ \\ R_3-3R_2 \end{array} \left[\begin{array}{ccc|c}1 & 0 & \frac{5}{3} & \frac{2k^2-2}{3} \\ 0 & 1 & \text{-}\frac{2}{3} & \frac{k^2-4}{3} \\ 0 & 0 & 0 & \text{-}(k+1)(k-2) \end{array}\right]$



$\displaystyle \text{(b) Since the left side of the bottom row is "all zeros", there is }no\text{ unique solution.}$


$\displaystyle \text{(c) If the right side of the bottom row is also zero, there is an infinite number of solutions.}$
. . .$\displaystyle \text{This happens when: }\;-(k+1)(k-2) \:=\:0 \quad\Rightarrow\quad k \:=\:-1,\:2$


\(\displaystyle \text{(a) For any other value of }k\k \neq -1,2)\text{, there are }no\text{ solutions.}\)