Matrix

[mickdano]

Here is the problem:
-.60s1 + .30s2 + .25s3 = 0
.30s1 - .45s2 + .30s3 = 0
.30s1 + .15s2 - .55s3 = 0
s1 + s2 + s3 = 1

s1 = ?
s2= ?
s3 =?

I need to know the next step to obtain the answers for S1, s2 and s3.
Directions inform me to use Matrix method or elimination to obtain the answer.
Any help would be greatly appreciated.
(The sample above was the last question, so steps to use for my next question is ideal, thanks!)

 

[Deleted member 4993]

Here is the problem:
-.60s1 + .30s2 + .25s3 = 0
.30s1 - .45s2 + .30s3 = 0
.30s1 + .15s2 - .55s3 = 0
s1 + s2 + s3 = 1

s1 = ?
s2= ?
s3 =?

I need to know the next step to obtain the answers for S1, s2 and s3.
Directions inform me to use Matrix method or elimination to obtain the answer.
Any help would be greatly appreciated.
(The sample above was the last question, so steps to use for my next question is ideal, thanks!)

Here you have three unknowns and 4 sets of homogeneous equations.

What method/s have you been taught to solve such system?

What do you get following the direction - elimination method?

Please show us your work, indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[mmm4444bot]

I need to know the next step

The next step is a function of what you've already learned.

:?: Which method are you trying to use?

 

[soroban]

Hello, mickdano!

This problem is unreasonably messy . . . decimals and subscripts ?!
. . (And the answers are not pleasant.)

I'll modify it for my convenience.

\(\displaystyle \begin{array}{ccccccc} x \;+\; y \;+\; z &=& 1 & (a) \\ \;0.30x + 0.15y - 0.55z &=& 0 & (b) \\ \;0.30x - 0.45y + 0.30z &=& 0 & (c) \\ \text{-}0.60x + 0.30y + 0.25z &=& 0 & (d) \end{array}\)

We are given four equations, but (b), (c) and (d) are not indenpendent.
. . We will use only (a), (b) and (c).

\(\displaystyle \begin{array}{ccccccc}\text{We have equation (a):} & x \;+\; y \;+\; z &=& 1 \\ \text{Multiply (b) by 20:} & 6x + 3y - 11z &=& 0 \\ \text{Multiply (c) by }\frac{20}{3}: & 2x - 3y + 2z &=& 0 \end{array}\)


\(\displaystyle \text{We have: }\;\left[\begin{array}{ccc|c}1&1&1&1 \\ 6&3&\text{-}11 & 0 \\ 2& \text{-}3 & 2 & 0 \end{array}\right]\)


\(\displaystyle \begin{array}{c} \\ R_2-6R_2 \\ R_3 - 2R_1\end{array} \left[\begin{array}{ccc|c} 1&1&1&1 \\ 0 & \text{-}3 & \text{-}17 & \text{-}6 \\ 0 & \text{-}5 & 0 & \text{-}2 \end{array}\right]\)


. . \(\displaystyle \begin{array}{c} \\ -\frac{1}{3}R_2 \\ -\frac{1}{5}R_3 \end{array} \left[\begin{array}{ccc|c}1 & 1 & 1 & 1 \\ 0 & 1 & \frac{17}{3} & 2 \\ 0 & 1 & 0 & \frac{2}{5} \end{array}\right]\)


\(\displaystyle \begin{array}{c}R_1-R_2 \\ \\ R_3-R_2\end{array} \left[\begin{array}{ccc|c}1 & 0 & \text{-}\frac{14}{3} & \text{-}1 \\ \\[-3mm] 0 & 1 & \frac{17}{3} & 2 \\ \\[-3mm] 0 & 0 & \text{-}\frac{17}{3} & \text{-}\frac{8}{5} \end{array}\right]\)


. . \(\displaystyle \begin{array}{c} \\ \\ \text{-}\frac{3}{17}R_3\end{array} \left[\begin{array}{ccc|c} 1 & 0 & \text{-}\frac{14}{3} & \text{-}1 \\ \\[-3mm] 0 &1 & \frac{17}{3} & 2 \\ 0 & 0 & 1 & \frac{24}{85}\end{array}\right]\)



\(\displaystyle \begin{array}{c}R_1 + \frac{14}{3}R_3 \\ R_2 - \frac{17}{3}R_3 \\ \\ \end{array} \left[\begin{array}{ccc|c} 1 & 0 & 0 & \frac{27}{85} \\ \\[-3mm] 0 & 1 & 0 & \frac{2}{5} \\ \\[-3mm]0 & 0 & 1 & \frac{24}{85} \end{array}\right]\)


\(\displaystyle \text{Therefore: }\;\begin{Bmatrix} x &=& \dfrac{27}{85} \\ \\[-2mm] y &=& \dfrac{2}{5} \\ \\[-2mm] z &=& \dfrac{24}{85} \end{Bmatrix}\)