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 Leonardox September 22nd, 2018 06:55 PM

Find A^-1 matrix

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How can I do this ? :(

 romsek September 22nd, 2018 07:05 PM

$\begin{pmatrix} 1 &1 &1 &| &1 &0 &0\\ 0 &1 &1 &| &0 &1 &0\\ 0 &0 &1 &| &0 &0 &1 \end{pmatrix}$

$R1 \Rightarrow R1 - R2$

$\begin{pmatrix} 1 &0 &0 &| &1 &-1 &0\\ 0 &1 &1 &| &0 &1 &0\\ 0 &0 &1 &| &0 &0 &1 \end{pmatrix}$

$R2 \Rightarrow R2 - R3$

$\begin{pmatrix} 1 &0 &0 &| &1 &-1 &0\\ 0 &1 &0 &| &0 &1 &-1\\ 0 &0 &1 &| &0 &0 &1 \end{pmatrix}$

$A^{-1} =\begin{pmatrix} 1 &-1 &0\\ 0 &1 &-1\\ 0 &0 &1 \end{pmatrix}$

 v8archie September 22nd, 2018 08:17 PM

On a $\LaTeX$ note, the array environment is your friend.
$\left(\begin{array}{@{}ccc|ccc@{}} 1 &1 &1 &1 &0 &0\\ 0 &1 &1 &0 &1 &0\\ 0 &0 &1 &0 &0 &1 \end{array}\right)$

 SDK September 22nd, 2018 08:35 PM

While technically Romsek's answer is correct, its a bit of overkill for a problem like this. You already have the matrix in upper triangular form which means it is trivial to solve $Ax = b$ for any $b$ you like.

Choose $b$ to be a basis vector and back solve to determine each column of $A^{-1}$ instantly. In general, this works anytime you have already obtained an LU decomposition. In this case, the $L$ is just the identity matrix.

 romsek September 22nd, 2018 08:46 PM

Quote:
 Originally Posted by v8archie (Post 599506) On a $\LaTeX$ note, the array environment is your friend. \begin{array}{@{}ccc|ccc@{}}
what's the @{} about?

 Leonardox September 22nd, 2018 08:55 PM

1 Attachment(s)
This is what I did to find A^-1
And then? Is that the answer of all question?

 Leonardox September 22nd, 2018 08:58 PM

I guess Romsek found the A^-1 more practically then I did.:) so how should I continuity. I learned how to find inverse of any 3x3 matrix so far now.

 Leonardox September 22nd, 2018 09:13 PM

after finding A^-1 are we going to find x or b or both?

 Leonardox September 22nd, 2018 09:19 PM

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This is what I did. Did I continue correctly ?

 Leonardox September 22nd, 2018 09:21 PM

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Oops sorry for the orientation. I corrected.

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