Solve Simultaneous Equations: No/1/Inf Solns?

[Guest2]

$x+2y-z=2 \\
x-y+z =5 \\
3x+3y-z=\mu$

The question is for what value of the parameter has the equation (a) no solutions, (b) one solution, and (c) infinitely many solutions.

The trouble when I solve this system using guassian elimination I get $

(x, y, z) = ( -19-\frac{1}{2}\mu, 7, -7-\frac{1}{2}\mu)$

 

[Prove It]

$x+2y-z=2 \\
x-y+z =5 \\
3x+3y-z=\mu$

The question is for what value of the parameter has the equation (a) no solutions, (b) one solution, and (c) infinitely many solutions.

The trouble when I solve this system using guassian elimination I get $

(x, y, z) = ( -19-\frac{1}{2}\mu, 7, -7-\frac{1}{2}\mu)$

Write the system in matrix form as

$\displaystyle \begin{align*} \left[ \begin{matrix} 1 & \phantom{-}2 & -1 \\ 1 & -1 & \phantom{-}1 \\ 3 & \phantom{-}3 & -1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] = \left[ \begin{matrix} 2 \\ 5 \\ \mu \end{matrix} \right] \end{align*}$

The system will have unique solution if $\displaystyle \begin{align*} \left| \begin{matrix} 1 & \phantom{-}2 & -1 \\ 1 & -1 & \phantom{-}1 \\ 3 & \phantom{-}3 & -1 \end{matrix} \right| \neq 0 \end{align*}$ and either no solution or infinite solutions if $\displaystyle \begin{align*} \left| \begin{matrix} 1 & \phantom{-}2 & -1 \\ 1 & -1 & \phantom{-}1 \\ 3 & \phantom{-}3 & -1 \end{matrix}\right| = 0 \end{align*}$.

 

[Guest2]

But those determinants won't give me a condition on $\mu$.

Is there a way to read this off from the reduced row echelon form?

 

[topsquark]

But those determinants won't give me a condition on $\mu$.

Is there a way to read this off from the reduced row echelon form?

As the question is stated there is no condition on \(\displaystyle \mu\).

-Dan

 

[Guest2]

As the question is stated there is no condition on \(\displaystyle \mu\).

-Dan

I found that $\mu = 9$ - and wolfram says that's the only value for which the system has solution.

 

[topsquark]

I found that $\mu = 9$ - and wolfram says that's the only value for which the system has solution.

(Doh) That just shows that thinking the problem backward (as I did) is no substitute for thinking of it by simply solving the problem.

My apologies and thanks for the catch.

-Dan

 

[Guest2]

I reduced the matrix to row reduced echelon form

$A =\begin{pmatrix} 1& 0& 1/3& 0\\ 0 &1 &-2/3 &0\\ 0& 0& 0& 1\end{pmatrix}$

 

[Guest2]

If I compute the determinant I get det(A) = 0.

Could someone please explain what the answer to the problem is.