proof of matrix inverse calculation by Gaussian elimination

Let $A$ be an invertible matrix, and $A^{-1}$ its inverse, whosecolumns are $A^{-1}_{1},\\cdots,A^{-1}_{n}$ .Then, by definition of matrix inverse, $AA^{-1}=I_{n}$ .But this implies $AA^{-1}_{1}=e_{1},\\ldots,AA^{-1}_{n}=e_{n}$ ,with $e_{1},\\cdots,e_{n}$ being the first, $\\ldots$ , $n$ -th column of $I_{n}$ respectively.

$A$ being non singular (or invertible), for all $k\\leq n$ , $AA^{-1}_{k}=e_{k}$ has a solution for $A^{-1}_{k}$ , which canbe found by Gaussian elimination of $[A\\mid e_{k}]$ .

The only part that changes between the augmented matricesconstructed is the last column, and these last columns, once theGaussian elimination has been performed, correspond to the columnsof $A^{-1}$ . Because of this, the steps we need to take for theGaussian elimination are the same for each augmented matrix.

Therefore, we can solve the matrix equation by performing Gaussianelimination on $[A\\mid e_{1}\\cdots e_{n}]$ , or $[A\\mid I_{n}]$ .

单词	ProofOfMatrixInverseCalculationByGaussianElimination
释义	proof of matrix inverse calculation by Gaussian elimination Let $A$ be an invertible matrix, and $A^{-1}$ its inverse, whosecolumns are $A^{-1}_{1},\\cdots,A^{-1}_{n}$ .Then, by definition of matrix inverse, $AA^{-1}=I_{n}$ .But this implies $AA^{-1}_{1}=e_{1},\\ldots,AA^{-1}_{n}=e_{n}$ ,with $e_{1},\\cdots,e_{n}$ being the first, $\\ldots$ , $n$ -th column of $I_{n}$ respectively. $A$ being non singular (or invertible), for all $k\\leq n$ , $AA^{-1}_{k}=e_{k}$ has a solution for $A^{-1}_{k}$ , which canbe found by Gaussian elimination of $[A\\mid e_{k}]$ . The only part that changes between the augmented matricesconstructed is the last column, and these last columns, once theGaussian elimination has been performed, correspond to the columnsof $A^{-1}$ . Because of this, the steps we need to take for theGaussian elimination are the same for each augmented matrix. Therefore, we can solve the matrix equation by performing Gaussianelimination on $[A\\mid e_{1}\\cdots e_{n}]$ , or $[A\\mid I_{n}]$ .
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