Matrix Methods For Linear Equations

**luxdelux** · January 28th 2009, 10:25 PM

Good evening, everyone!

There are two problems that I'm having difficulty figuring out on my homework. I'd really appreciate the help!

(1) 2x - y = 2
-x + 3y + 2z = 1
x + 2y + 2z = -1

I created an augmented matrix from this and row reduced to the point where I have:

|1 -3 -2 | -1|
|0 1 4/5 | 0|
|0 5 4 | 0|

Not sure if I did incorrect calculations or what, but I have no idea where to go from here (provided I did the correct calculations up to this point). Please help me out!

(2) w + 2x - y - 2z = -1
x - y = -1
-w + 2x + z = -1

Here I also row reduced but only went through one round of reducing before coming to the conclusion that there's no solution...Yes or no?

Please, please help -- I really am quite lost with both of these!

-Lux

**Danny** · January 29th 2009, 05:51 AM

Originally Posted by luxdelux

Good evening, everyone!

There are two problems that I'm having difficulty figuring out on my homework. I'd really appreciate the help!

(1) 2x - y = 2
-x + 3y + 2z = 1
x + 2y + 2z = -1

I created an augmented matrix from this and row reduced to the point where I have:

|1 -3 -2 | -1|
|0 1 4/5 | 0|
|0 5 4 | 0|

Not sure if I did incorrect calculations or what, but I have no idea where to go from here (provided I did the correct calculations up to this point). Please help me out!

(2) w + 2x - y - 2z = -1
x - y = -1
-w + 2x + z = -1

Here I also row reduced but only went through one round of reducing before coming to the conclusion that there's no solution...Yes or no?

Please, please help -- I really am quite lost with both of these!

-Lux

Not sure how you got the middle row of the first augmented matrix. Here's what I got

$\left[ \begin{array}{ccc|c} 1 & -3 & -2 & -1\\ 2 & -1 & 0 & 2\\ 1 & 2 & 2 & -1 \end{array} \right]\;\;\; \begin{array}{c} 2R_1 - R_2 \rightarrow R_2\\ R_1 - R_3 \rightarrow R_3 \end{array} \left[ \begin{array}{ccc|c} 1 & -3 & -2 & -1\\ 0 & -5 & -4 & -4\\ 0 & -5 & -4 & 0 \end{array} \right] $

and subtracting the last two rows gives 0 = -4 so no solution. With your second problem, I got an infinite number of solutions

$x = t,\;\;\;y = t+1,\;\;\;z = 3t+1,\;\;\;w = 5t + 2$

**Soroban** · January 29th 2009, 06:12 AM

Hello, Lux!

$(1)\;\;\begin{array}{ccc}2x -y \qquad&=& 2 \\ \text{-}x + 3y + 2z &=& 1 \\ x + 2y + 2z &=& \text{-}1\end{array}$

Multiply the second equation by -1 and list it first: . $\begin{array}{ccc}x - 3y - 2x &=& \text{-}1 \\ 2x - y \qquad\;\; &=& 2 \\ x + 3y + 2x &=& \text{-}1 \end{array}$

We have: . $\left[\begin{array}{ccc|c}1 & \text{-}3 & \text{-}2 & \text{-}1 \\ 2&\text{-}1&0&2 \\ 1&2&2&\text{-}1 \end{array}\right]$

$\begin{array}{c}\\ R_2-2R_1 \\ R_3-R_1\end{array}\left[\begin{array}{ccc|c} 1 & \text{-}3&\text{-}2&\text{-}1 \\ 0&5&4&4\\0&5&4&0 \end{array}\right]$

$\begin{array}{c}\\ \\ R_3-R_2\end{array}\left[\begin{array}{ccc|c}1&\text{-}3&\text{-}2&\text{-}1 \\ 0&5&4&4 \\ 0&0&0&4 \end{array}\right]$

The bottom row gives us: . $0 \,=\,4$ . . . The system is inconsistent (no solution).

$(2)\;\;\begin{array}{ccc}w+2x-y-2z &=& \text{-}1 \\ x-y &= &\text{-}1 \\ \text{-}w + 2x \qquad + z\;\; &=& \text{-}1 \end{array}$

We have: . $\left[\begin{array}{cccc|c}1&2&\text{-}1&\text{-}2&\text{-}1 \\ 0&1&\text{-}1&0&\text{-}1 \\ \text{-}1&2&0&1&\text{-}1 \end{array}\right]$

$\begin{array}{c}\\ \\ R_3+R_1\end{array} \left[\begin{array}{cccc|c} 1&2&\text{-}1&\text{-}2&\text{-}1 \\ 0&1&\text{-}1&0&\text{-}1 \\ 0&4&\text{-}1&\text{-}1&\text{-}2 \end{array}\right]$

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

. . . $\begin{array}{c}\\ \\ \frac{1}{3}R_3\end{array} \left[\begin{array}{cccc|c}1&0&1&\text{-}2 & 1 \\ 0&1&\text{-}1&0&1 \\ 0&0&1&\text{-}\frac{1}{3} & \frac{2}{3} \end{array}\right]$

$\begin{array}{c}R_1-R_3 \\ R_2+R_3 \\ \\ [4mm] \end{array} \left[\begin{array}{cccc|c} 1&0&0&\text{-}\frac{5}{3} & \frac{2}{3} \\ \\[-4mm] 0&1&0&\text{-}\frac{1}{3} & \text{-}\frac{1}{3} \\ \\[-4mm] 0&0&1&\text{-}\frac{1}{3} & \frac{2}{3}\end{array}\right]$

The system has an infinite number of solutions.

We have: . $\begin{array}{ccc}w - \frac{5}{3}z &=&\frac{1}{3} \\ \\[-4mm]x - \frac{1}{3}x &=& \text{-}\frac{1}{3} \\ \\[-4mm] y - \frac{1}{3}z &=& \frac{2}{3} \end{array}$ . $\Rightarrow\quad \begin{array}{ccc}w &=& \frac{5}{3}z + \frac{1}{3} \\ x &=& \frac{1}{3}z - \frac{1}{3} \\ y &=& \frac{1}{3}z + \frac{2}{3} \\ z &=& z\quad \end{array}$

On the right, replace $z$ with a parameter $t.$

. . . . $\begin{array}{ccc}w &=& \dfrac{5t+1}{3} \\ \\[-4mm] x &=& \dfrac{t-1}{3} \\ \\[-4mm] y &=& \dfrac{t+2}{3} \\ \\[-4mm] z &=& t \quad \end{array}$

This represents all the solutions to the system,
. . one for each value of $t.$

Edit: Too fast for me, Danny . . .
. . . . And I like your solutions!
.

**luxdelux** · January 29th 2009, 07:42 PM

Thanks guys! I really am having a lot of troubles with Gauss-Jordan.
I'm so sorry, but I accidently typed the last problem incorrectly (2) --- the first line of that equation should read 2x-y = -2. I did the problem again with the correct equation and row reduced to the point where I have:

1 -3 -2 | -1
0 1 4/5 | 0
0 5 4 | 0

I know that I need to change the -3 and the 5 in the second column to zero, but I don't know what to do from here. And I have a feeling I may have miscalculated again.

Next, I'm dealing with a bit of a monsterous problem (it is a word problem, and we need to create a linear equation and subsequently a matrix from the numbers given). I got help from a math tutor at school setting up the equations involved, but again, I know that I am miscalculating something while I'm row reducing because I'm not getting the right answer. Gauss-Jordan is particularly frustrating, because if you mess up even the tiniest little thing, everything is affected. Here are my two equations:

100,000x + 70,000y +40,000z = 1,600,000
100x +50y = 1000
20,000x + 5,000y = 150,000

100,000x + 70,000y +40,000z = 1,140,000
100x +50y = 700
20,000x + 5,000y = 110,000

I have pages of scratch-work for this, nothing makes sense anymore.

Thirdly, when a problem wants to know if the answers are inconsistent or not, what does this mean? How do you know if something is inconsistent? Do I have to keep row-reducing until I can't reduce anymore...and that means it's inconsistent? Please help, I'm way lost.

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