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September 22nd, 2018, 07:55 PM   #1
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Find A^-1 matrix

How can I do this ?
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September 22nd, 2018, 08:05 PM   #2
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$\begin{pmatrix}
1 &1 &1 &| &1 &0 &0\\
0 &1 &1 &| &0 &1 &0\\
0 &0 &1 &| &0 &0 &1
\end{pmatrix}$

$R1 \Rightarrow R1 - R2$

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

$R2 \Rightarrow R2 - R3$

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

$A^{-1} =\begin{pmatrix}
1 &-1 &0\\
0 &1 &-1\\
0 &0 &1
\end{pmatrix}$
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September 22nd, 2018, 09:17 PM   #3
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On a $\LaTeX$ note, the array environment is your friend.
$\left(\begin{array}{@{}ccc|ccc@{}}
1 &1 &1 &1 &0 &0\\
0 &1 &1 &0 &1 &0\\
0 &0 &1 &0 &0 &1
\end{array}\right)$
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September 22nd, 2018, 09:35 PM   #4
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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.
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September 22nd, 2018, 09:46 PM   #5
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Quote:
Originally Posted by v8archie View Post
On a $\LaTeX$ note, the array environment is your friend.
\begin{array}{@{}ccc|ccc@{}}
what's the @{} about?
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September 22nd, 2018, 09:55 PM   #6
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This is what I did to find A^-1
And then? Is that the answer of all question?
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September 22nd, 2018, 09:58 PM   #7
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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.
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September 22nd, 2018, 10:13 PM   #8
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after finding A^-1 are we going to find x or b or both?
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September 22nd, 2018, 10:19 PM   #9
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This is what I did. Did I continue correctly ?
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September 22nd, 2018, 10:21 PM   #10
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Oops sorry for the orientation. I corrected.
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