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I just did a past paper and came across this matrix that I need to get an inverse for. Obviously it has a lot of deliberate zeros to help but I do not know how to solve it?

enter image description here

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You should use matrix block multiplication.

$A=\begin{pmatrix}2&1\\1&1\end{pmatrix}$

$B=\begin{pmatrix}2&3\\3&5\end{pmatrix}$

$C=\begin{pmatrix}1&0\\0&\cfrac{1}{2}\end{pmatrix}$

$M=\begin{pmatrix}A&0&0\\0&B&0\\0&0&C\end{pmatrix}$

So $M^{-1}=\begin{pmatrix}A^{-1}&0&0\\0&B^{-1}&0\\0&0&C^{-1}\end{pmatrix}$

And to inverse those matrix, you can use $X^{-1}=\cfrac{Adj(X)}{det(X)}$ where $Adj(X)=\,^t C$ with $C$ the matrix of cofactors. (http://en.wikipedia.org/wiki/Adjugate_matrix)

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  • $\begingroup$ Awesome thanks. WHat does $^tCom(X)$ mean I have never seen this before? $\endgroup$ – Kane Blackburn May 12 '13 at 11:33
  • $\begingroup$ @Kane Blackburn: I edited. $\endgroup$ – xavierm02 May 12 '13 at 11:36

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