augmented matricies

[turkey16]

i have only gotten this far i had a doctors appointment so i was only in class for 10 min.
i am not shure how to work it step by step and get [1,0,0] and how to use that in my problem it harder than cramers rule and i am really worried and confused.
[0,1,0]
[0,0,1]


1. 3x -2y + z = 9
x + 4y = -2
-y + 4z = 5
this is my problem and if you could show me how to put the 1,0,0,.... matrix thing into it and explain as you go i would be a grateful person! thanx!!!

 

[fasteddie65]

Let's look at the system:
3x - 2y + z = 9
x + 4y = -2
-y + 4z = 5

The matrix would be [ 3 -2 1 9 ]
[ 1 4 0 -2 ]
[ 0 -1 4 5 ]

The quickest way to get a 1 in the upper left corner is to switch the first and second rows.

[ 1 4 0 -2 ]
[ 3 -2 1 9 ]
[ 0 -1 4 5 ]

Now we would like to get a 0 directly below the 1 in the first column. Using a row operation, we can multiply the first row by -3 and add it to the second row.

1 • (-3) = -3, and -3 + 3 = 0.
4 • (-3) = -12, and -12 + (-2) = -14
0 • (-3) = 0, and 0 + 1 = 1
(-2) • (-3) = 6, and 6 + 9 = 15.

[ 1 4 0 -2 ]
[ 0 -14 1 15 ]
[ 0 -1 4 5 ]

Row operations include: multiplying a row by a factor, adding a row to another row, multiplying a row and then adding it to another row. Using these row operations, we want to get a 1 where the -14 is. Why not try multiplying it by -1/14?
Continue doing this until you have the 1 0 0 etc.

 

[soroban]

Hello, turkey16!

\(\displaystyle 1)\;\;\begin{array}{ccc}3x -2y + z &=& 9 \\ x + 4y \qquad &=& \text{-}2 \\ \qquad \text{-}y + 4z & =& 5 \end{array}\)

\(\displaystyle \text{I would switch the order of the equations: }\;\begin{array}{ccc}x + 4y \qquad &=& \text{-}2 \\ \qquad\text{-}y + 4z &=& 5 \\ 3x - 2y + z &=& 9 \end{array}\)

\(\displaystyle \text{We have: }\;\left[\begin{array}{ccc|c}1 & 4 & 0 & \text{-}2 \\ 0 & \text{-}1 & 4 & 5 \\ 3 & \text{-}2 & 1 & 9 \end{array}\right]\)


\(\displaystyle \begin{array}{c}\\ \text{-}1\!\cdot\!R_2 \\ R_3-3R_1\end{array} \left[\begin{array}{ccc|c}1 & 4 & 0 & \text{-}2 \\ 0 & 1 & \text{-}4 & \text{-}5 \\ 0 & \text{-}14 & 1 & 15 \end{array}\right]\)


\(\displaystyle \begin{array}{c}R_1-4R_2 \\ \\ R_3 + 14R_2\end{array} \left[\begin{array}{ccc|c}1 & 0 & 16 & 18 \\ 0 & 1 & \text{-}4 & \text{-}5 \\ 0 & 0 & \text{-}55 & \text{-}55 \end{array}\right]\)


. . . \(\displaystyle \begin{array}{c} \\ \\ \text{-}\frac{1}{55}\!\cdot\!R_3 \end{array} \left[\begin{array}{ccc|c}1 & 0 & 16 & 18 \\ 0 & 1 & \text{-}4 & \text{-}5 \\ 0 & 0 & 1 & 1 \end{array}\right]\)


\(\displaystyle \begin{array}{c}R_1-16R_3 \\ R_2 + 4R_3 \\ \\ \end{array} \left[\begin{array}{ccc|c} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & \text{-}1 \\ 0 & 0 & 1& 1 \end{array}\right]\)


\(\displaystyle \text{Therefore: }\;\begin{Bmatrix} x &=&2 \\ y &=&\text{-}1 \\ z &=& 1 \end{Bmatrix}\)

 

[turkey16]

thanx for all the help i understand how to use it now it just takes a long time!