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 December 1st, 2013, 01:24 AM #1 Newbie   Joined: Dec 2013 Posts: 1 Thanks: 0 Linear systems Hello. I have a question about linear systems. Ex: we have 2 equations with 3 unknown. 2x1+1x2+x3 = 5 1x1+ax2+x3 = 2 What to do with a if we put it in a matrix? Do we set it to 1 since x2 is behind or 1a? Like [ 2 1 1 | 5] [ 1 1 1 | 2] Or how Would you do? I am just curious
December 1st, 2013, 07:12 AM   #2
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
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Re: Linear systems

Hello, Predator!

What are the instructions?

Quote:
 We have 2 equations with 3 unknowns. [color=beige]. . [/color]$\begin{array}{ccc}2x\,+\,y\,+\,z=&5 \\ \\ \\ x\,+\,ay\,+\,z=&2 \end{array}=$ What to do with $a$ if we put it in a matrix?

I will assume that we are to solve the system.
And maybe answer the question:
[color=beige]. . [/color]"For what value(s) of $a$ will the system have no solutions?"

$\text{W\!e have: }\:\left|\begin{array}{ccc|c} 2&1&1&5 \\ \\ 1&a&1&2 \end{array}\right|$

$\begin{array}{c} R_1-R_2 \\ \\ \\ \\ \\ \\ \end{array}\;\left|\begin{array}{ccc|c} 1&1-a&0&3 \\ \\ \\ 1&a&1&2 \end{array}\right|$

$\begin{array}{c} \\ \\ \\ \\ \\ \\ \\ R_2-R_1\end{array}\;\left|\begin{array}{ccc|c}1&1-a&0&3 \\ \\ \\ 0&2a-1&1&-1 \end{array}\right|$

$\begin{array} \\ \\ \\ \\ \\ \\ \frac{1}{2a-1}R_2 \end{array}\;\left|\begin{array}{ccc|c} 1&1-a&0&3 \\ \\ \\
0&1&\frac{1}{2a-1} & \frac{-1}{2a-1}\end{array}\right|$

$\begin{array}{c}R_1-(1-a)R_2 \\ \\ \\ \\ \\ \\ \\ \\ \end{array}\;\left|\begin{array}{ccc|c}1&0&\frac{a-1}{2a-1} & \frac{5a-2}{2a-1} \\ \\ \\ 0&1&\frac{1}{2a-1} & \frac{-1}{2a-1} \end{array}\right|$

$\text{W\!e have: }\:\begin{Bmatrix}x\,+\,\frac{a-1}{2a-1}z=&\frac{5a-2}{2a-1} \\ \\ \\ y\,+\,\frac{1}{2a-1}z=&\frac{-1}{2a-1} \end{Bmatrix} \;\;\;\Rightarrow\;\;\;\begin{Bmatrix}x=&\frac{5a-2}{2a-1}\,-\,\frac{1-a}{2a-1}z \\ \\ \\ y=&\frac{-1}{2a-1}\,-\,\frac{1}{2a-1}z \\ \\ \\ \\ \\ \ z=&z \end{Bmatrix}=$

$\text{On the right, replace }z\text{ with a parameter }t.$

[color=beige]. . . . [/color]$\begin{Bmatrix}x=&\frac{5a-2}{2a-1} \,-\,\frac{a-1}{2a-1}t \\ \\ \\ y=&\frac{-1}{2a-1}\,-\,\frac{1}{2a-1}t \\ \\ \\ \\ \\ \\ z=&t \end{Bmatrix}=$

$\text{If }a= \frac{1}{2},\text{ the system has no solution.}$

$\text{Otherwise, there is a unique solution for every real number }t.$

 December 4th, 2013, 05:11 PM #3 Newbie   Joined: Dec 2013 Posts: 1 Thanks: 0 Re: Linear systems Hi, I have published a course in Mathematics on "Matrix Algebra" @ https://www.udemy.com/matrix-algebra. Check it out!!

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