# Thread: Inverse of a 3 by 3 matrix

1. ## Inverse of a 3 by 3 matrix

i am having trouble with this problem. its been almost an hour and i still only get part of the answer. i dont know what i am doing wrong.

the question is find the inverse of this matrix.

5 5 5

15 25 20

15 30 25

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this is what i ended up with, but i know its wrong because i used the calculator. i'm trying to figure out step by step.

1 -1 .-2

-.6 .4 -.2

1.8 .-6 .4

so far no luck. no matter what i do i still get wrong answers. i tried interchanging rows and all that good stuff but nothing. i must be doing something wrong.

2. Show your working as to what you did and we'll try and see where you're making your mistake. These brutes are messy.

3. Originally Posted by Matt Westwood
Show your working as to what you did and we'll try and see where you're making your mistake. These brutes are messy.
you know how long thats gonna take for me to plug my work in.....? forget it. this was no help what so ever. wasted my time.. thanks anyways.

4. Hello, flexus!

Find the inverse of this matrix: .$\displaystyle \begin{pmatrix}5& 5& 5 \\ 15 &25& 20 \\ 15 &30 & 25 \end{pmatrix}$
First, note that all elements have a factor of 5.
We can factor it out . . . and restore it later.

$\displaystyle A \;=\;5\begin{pmatrix}1&1&1 \\ 3&5&4 \\ 3&6&5\end{pmatrix}$

We have: . $\displaystyle \left|\begin{array}{ccc|ccc} 1&1&1 & 1&0&0 \\ 3&5&4 & 0&1&0 \\ 3&6&5 & 0&0&1\end{array}\right|$

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

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

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

. . $\displaystyle \begin{array}{c}\\ \\ \text{-}1\!\cdot\!R_3\end{array} \left| \begin{array}{ccc|ccc}1&0&0 & 1&1&\text{-}1 \\ 0&1&1 & 0&\text{-}1&1 \\ 0&0&1 & 3&\text{-}3&2 \end{array}\right|$

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

Therefore: .$\displaystyle A^{-1} \;=\;\frac{1}{5}\begin{pmatrix}1&1&\text{-}1 \\ \text{-}3&2&\text{-}1 \\ 3&\text{-}3&2\end{pmatrix} \;=\;\begin{pmatrix}\frac{1}{5} & \frac{1}{5} & \text{-}\frac{1}{5} \\ \\[-4mm] \text{-}\frac{3}{5} & \frac{2}{5} & \text{-}\frac{1}{5} \\ \\[-4mm] \frac{3}{5} & \text{-}\frac{3}{5} & \frac{2}{5} \end{pmatrix}$

5. Originally Posted by flexus
you know how long thats gonna take for me to plug my work in.....? forget it. this was no help what so ever. wasted my time.. thanks anyways.
And how long do you think it will take someone to type a guide for the solution (I assume that's what you expect to get). Frankly, I find this attitude quite insulting.