m=

0 1 2

1 0 3

4 -3 8

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- Jun 1st 2009, 09:53 AMeunse16how can i find the inverse of matrix?
m=

0 1 2

1 0 3

4 -3 8 - Jun 1st 2009, 10:13 AMcraig
- Jun 1st 2009, 01:25 PMHallsofIvy
Using the determinant and cofactors is certainly one way to find an inverse matrix and, in fact, that was the first method I learned. But I think it is easier to use "row-reduction".

Write the given matrix and identity matrix next to each other:

$\displaystyle \begin{bmatrix}0 & 1 & 2 \\ 1 & 0 & 3 \\ 4 & -3 & 8\end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$

Now use row reduction to reduce your matrix to the identity matrix while applying those same row operations all the way across both matrices. The same row operations that change your matrix to the identity matrix will change the identity matrix to the inverse of the original matrix. - Jun 1st 2009, 01:30 PMcraig
- Jun 1st 2009, 01:39 PMcraig
Just been reading through some info on the Web, what ever you have to add or subtract to/from the first matrix to get the identity, do you place this value in the corresponding position in the resulting inverse matrix?

For example, would the inverse be:

$\displaystyle \begin{bmatrix}1 & -1 & -2 \\ -1 & 1 & -3 \\ -4 & 3 & -7\end{bmatrix}$

**Edit**- This does not make sense at all, ignore ;) - Jun 1st 2009, 05:32 PMSoroban
Hello, eunse16!

Quote:

$\displaystyle M \:=\:\begin{bmatrix}0&1&2 \\ 1&0&3 \\ 4&\text{-}3&8 \end{bmatrix}$

Find $\displaystyle M^{-1}$

We have: .$\displaystyle \left[\begin{array}{ccc|ccc}

0&1&2&1&0&0 \\ 1&0&3&0&1&0 \\ 4&\text{-}3&9 &0&0&1 \end{array}\right] $

. . $\displaystyle \begin{array}{c}\text{Switch}\\R_1,R_2 \\ \\ \end{array} \left[\begin{array}{ccc|ccc}

1&0&3&0&1&0 \\ 0&1&2&1&0&0 \\ 4&\text{-}3&8&0&0&1\end{array}\right]$

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

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

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

$\displaystyle \begin{array}{c}R_1-3R_3 \\ R_2-2R_3 \\ \end{array} \left[\begin{array}{ccc|ccc}1&0&0&\text{-}\frac{9}{2} & 7 &\text{-}\frac{3}{2} \\ 0&1&0 & \text{-}2 & 4 & \text{-}1 \\ 0&0&1 & \frac{3}{2} & \text{-}2 & \frac{1}{2} \end{array}\right] $

Therefore: .$\displaystyle M^{-1}\;=\;\begin{bmatrix}\text{-}\frac{9}{2} & 7 & \frac{3}{2} \\ \text{-}2 & 4 & \text{-}1 \\ \frac{3}{2}& \text{-}2 & \frac{1}{2}\end{bmatrix}$

- Jun 2nd 2009, 11:54 AMHallsofIvy
Aww, I wanted to do that!(Giggle)