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What use is the inverse matrix? I would not use it to solve linear systems but there must be some concrete or real life applications where it is used.

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Why wouldn't you use it to solve linear systems? I imagine it's certainly the best way for a computer to solve a linear system, even in cases where a human could "just see" the answer faster than compute the inverse of a matrix. –  Matt Pressland Jun 7 '12 at 13:22
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@RahulNarain That is surprising and interesting, thanks! –  Matt Pressland Jun 7 '12 at 13:45
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If you have a matrix $A$ of a linear transformation $f$ in a certain basis, and then you want to work with a different basis, you would use some matrix $X$ and its inverse to compute the new matrix representation of $f$ in the new basis. –  rschwieb Jun 7 '12 at 13:54
    
@Matt, read this (page 3) for another eloquent of putting that gaussian elimination was superseded in numerics over 50 years ago, and more!, by the informative Nick Trefethen. –  Peter Sheldrick Jun 7 '12 at 14:18

1 Answer 1

Lots of reasons :

  • You CAN use it.

  • In certain cases, it is the fastest way (Diagonal Matrices).

  • As preconditioners to linear systems.

Consider a system $Ax=b$ which is being solved numerically, if it is possible to find a matrix K which approximates the inverse of A, it is possible to solve $KAx=Kb \implies A'x=b' $ for $x$ much faster.

The theory of preconditioners is very rich and active. The hunt for such Approximate inverses is fairly field specific and is crucial in solving linear systems using modern algorithms such as the Krylov Subspace Methods. In such algorithms, the crucial parameter is the condition number which is the ratio of the maximum and minimum eigenvalues. Smaller the ratio (min. being 1), FASTER is the convergence (Identity matrix will converge in 1 step). So, preconditioners essentially reduce the condition number and in the process reduce the number of iterations (& hence time) required for the algorithms to converge.

Example : $$A=\begin{bmatrix} 0.6551 & 0.4984 & 0.5853\\ 0.1626 & 0.9597 & .2238\\ 0.1190 & 0.3404 & 0.7513\\\end{bmatrix} eig(A) = \begin{bmatrix} 1.3665\\ 0.4732\\ 0.5264\\\end{bmatrix}$$ Use a preconditioner $K=\begin{bmatrix} 1.9000 & -0.5000 & -1.3000\\ -0.3000 & 1.2000 & -0.2000\\ -0.2000 & -0.5000 & 1.6000\\ \end{bmatrix}$

For the system $KAx=Kb \Longleftrightarrow A'x = b'$, $$A' = \begin{bmatrix} 1.0087 & 0.0245 & 0.0235\\ -0.0252 & 0.9341 & -0.0573\\ -0.0219 & -0.0349 & 0.9731\\ \end{bmatrix}; eig(A') = \begin{bmatrix} 0.9169\\ 0.9995 + 0.0035i\\ 0.9995 - 0.0035i\\ \end{bmatrix} $$ How K was obtained is a question that involves delving into the source of the matrix but that's not essential for the motivation of the topic. All that is important is that $K$ attempts to approximate the inverse of the matrix A.

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