# Augmented matrix

• Sep 9th 2008, 06:36 PM
Snowboarder
Augmented matrix
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
• Sep 9th 2008, 08:05 PM
Chris L T521
Quote:

Originally Posted by Snowboarder
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 $\left[\begin{array}{ccc}1&0&2\\2&1&3\\-2&m&-5\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}1\\m\\-3\end{array}\right]$, where $m\in\mathbb{R}$

This matrix equation can be written in augmented form:

$\left[\begin{array}{ccc}1&0&2\\2&1&3\\-2&m&-5\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}1\\m\\-3\end{array}\right]\implies\left[\begin{array}{ccccr}1&0&2&:&1\\2&1&3&:&m\\-2&m&5&:&-3\end{array}\right]$

Get $\left[\begin{array}{ccccr}1&0&2&:&1\\2&1&3&:&m\\-2&m&5&:&-3\end{array}\right]$ into this form: $\left[\begin{array}{ccccr}1&0&0&:&x_0\\0&1&0&:&y_0\\0&0& 1&:&z_0\end{array}\right]$, where $x_0,~y_0,~\text{and }z_0$ are the solutions.

Try to take it from here. If you get stuck, let us know.

--Chris
• Sep 9th 2008, 09:17 PM
Snowboarder
ok so i've got:

{{1,0,0,2m-1},{0,1,0,-1},{0,0,1,1-m}}

how to find values of m for which the system has:

no solution
infinitly many solutions
a unique solution
???

thx for help