# Inverse of a matrix

• Nov 18th 2011, 12:37 AM
terence
Inverse of a matrix
I am trying to do an inverse of the following matrix but somehow I can't work it out.

The given matrix is this:
Code:

1 1 0 1 1 1 2 1 1
I added the I matrix to that :
Code:

1 1 0 | 1 0 0 1 1 1 | 0 1 0 2 1 1 | 0 0 1
and started to reducing to upper triangular form:

I tried
Row2 = Row2 - Row1
Row3 = Row3 - 2Row1

but this was the result

Code:

1 1 0 | 1 0 0 0 0 1 | -1 1 0 0 -1 1| -2 0 1
and I got stuck there.

Isit perhaps because I am doing something wrong?
• Nov 18th 2011, 01:10 AM
alexmahone
Re: Inverse of a matrix
Quote:

Originally Posted by terence
but this was the result

Code:

1 1 0 | 1 0 0 0 0 1 | -1 1 0 0 -1 1| -2 0 1
and I got stuck there.

Isit perhaps because I am doing something wrong?

You now need to exchange rows 2 and 3.
• Nov 18th 2011, 08:20 AM
terence
Re: Inverse of a matrix
Quote:

You now need to exchange rows 2 and 3.
Well, That's what I actually thought because it becomes perfect then, but this is just a matrix not a system of equations.
eg
Code:

1 2 3 | 5 4 5 6 | 7 7 8 9 | 8
(just invented for an example)

I think that is possible in only system of equations or am i wrong?
• Nov 18th 2011, 08:43 AM
Soroban
Re: Inverse of a matrix
Hello, terence!

alexmahone's suggestion is excellent!
Or you can solve it head-on.

Quote:

$\text{Find the inverse of }\:A \;=\;\begin{vmatrix}1&1&0 \\ 1&1&1 \\ 2&1&1\end{vmatrix}$

$\text{We have: }\:\begin{array}{|ccc|ccc|}1&1&0&1&0&0 \\ 1&1&1&0&1&0 \\ 2&1&1&0&0&1 \end{array}$

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

$\begin{array}{c} R_1+R_3 \\ R_2-R_3 \\ \ \end{array}\begin{array}{|ccc|ccc|} 1&0&1 & \text{-}1&0&1 \\ 0&1&0 & 1&1&\text{-}1 \\ 0&\text{-}1 &1 & \text{-}2&0&1 \end{array}$

$\begin{array}{c} \\ \\ R_3 + R_2 \end{array}\begin{array}{|ccc|ccc|} 1&0&1 & \text{-}1&0&1 \\ 0&1&0 & 1&1&\text{-}1 \\ 0&0&1 & \text{-}1&1&0 \end{array}$

$\begin{array}{c}R-R_3 \\ \\ \\ \end{array} \begin{array}{|ccc|ccc|} 1&0&0 & 0&\text{-}1&1 \\ 0&1&0 & 1&1&\text{-}1 \\ 0&0&1 & \text{-}1&1&0 \end{array}$

Therefore: . $A^{-1} \;=\;\begin{array}{|ccc|} 0&\text{-}1 & 1 \\ 1 & 1 & \text{-}1 \\ \text{-}1 & 1 & 0 \end{array}$

• Nov 18th 2011, 08:49 AM
HallsofIvy
Re: Inverse of a matrix
I am confused as to what you are asking. There are three types of "row operation": multiply a single row by a number, add a multiple of one row to another (the multiple can be negative), and interchange two rows. You can use them to solve a system of equations by writing the equations as matrices but that is not the only use of row operations. They can be used to find inverse matrices, as you do here, or to find the determinant of a matrix.

What you give would be written as
$\begin{bmatrix}1 & 2 & 3 & 5 \\ 4 & 5 & 6 & 7\\ 7 & 8 & 9 & 8\end{bmatrix}$
(I removed the vertical line because that is not part of a matrix- that would be used to identify the right hand side of a system of equations, which you say you are not talking about.)

Yes, you can apply all row operations to that matrix- but to what purpose? What do you want to do with this matrix?