Hi everyone.
I need to consider the following system of equations, where m belong to R is an unknown constant:
x + 2z = 1
2x + y + 3z = m
-2x + my - 5z = -3
How to write down augmented matrix of the system and transform it into reduced row echelon form.????
Is it any chance to find the value of m which the system has:
no solution,
infinitly many solutions or
unique solution???
Thank you
If we wrote this system in matrix form, we would get $\displaystyle \left[\begin{array}{ccc}1&0&2\\2&1&3\\-2&m&-5\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}1\\m\\-3\end{array}\right]$, where $\displaystyle m\in\mathbb{R}$Originally Posted by Snowboarder
This matrix equation can be written in augmented form:
$\displaystyle \left[\begin{array}{ccc}1&0&2\\2&1&3\\-2&m&-5\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}1\\m\\-3\end{array}\right]\implies\left[\begin{array}{ccccr}1&0&2&:&1\\2&1&3&:&m\\-2&m&5&:&-3\end{array}\right]$
Your goal is this:
Get $\displaystyle \left[\begin{array}{ccccr}1&0&2&:&1\\2&1&3&:&m\\-2&m&5&:&-3\end{array}\right]$ into this form: $\displaystyle \left[\begin{array}{ccccr}1&0&0&:&x_0\\0&1&0&:&y_0\\0&0& 1&:&z_0\end{array}\right]$, where $\displaystyle x_0,~y_0,~\text{and }z_0$ are the solutions.
Try to take it from here. If you get stuck, let us know.
--Chris
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