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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 Senior Member     Joined: Sep 2015 From: USA Posts: 2,299 Thanks: 1218 $\begin{pmatrix} 1 &1 &1 &| &1 &0 &0\\ 0 &1 &1 &| &0 &1 &0\\ 0 &0 &1 &| &0 &0 &1 \end{pmatrix}$ $R1 \Rightarrow R1 - R2$ $\begin{pmatrix} 1 &0 &0 &| &1 &-1 &0\\ 0 &1 &1 &| &0 &1 &0\\ 0 &0 &1 &| &0 &0 &1 \end{pmatrix}$ $R2 \Rightarrow R2 - R3$ $\begin{pmatrix} 1 &0 &0 &| &1 &-1 &0\\ 0 &1 &0 &| &0 &1 &-1\\ 0 &0 &1 &| &0 &0 &1 \end{pmatrix}$ $A^{-1} =\begin{pmatrix} 1 &-1 &0\\ 0 &1 &-1\\ 0 &0 &1 \end{pmatrix}$ Thanks from v8archie and Leonardox
 September 22nd, 2018, 09:17 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,599 Thanks: 2587 Math Focus: Mainly analysis and algebra On a $\LaTeX$ note, the array environment is your friend. $\left(\begin{array}{@{}ccc|ccc@{}} 1 &1 &1 &1 &0 &0\\ 0 &1 &1 &0 &1 &0\\ 0 &0 &1 &0 &0 &1 \end{array}\right)$ Thanks from Leonardox
 September 22nd, 2018, 09:35 PM #4 Senior Member   Joined: Sep 2016 From: USA Posts: 556 Thanks: 321 Math Focus: Dynamical systems, analytic function theory, numerics 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. Thanks from Leonardox
September 22nd, 2018, 09:46 PM   #5
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Quote:
 Originally Posted by v8archie On a $\LaTeX$ note, the array environment is your friend. \begin{array}{@{}ccc|ccc@{}}
what's the @{} about?

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 Senior Member   Joined: Apr 2017 From: New York Posts: 119 Thanks: 6 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.
 September 22nd, 2018, 10:13 PM #8 Senior Member   Joined: Apr 2017 From: New York Posts: 119 Thanks: 6 after finding A^-1 are we going to find x or b or both?
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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