Exact parallel matrix inversion using para-Hensel codes with systolic processors

V. K. Murthy

doi:10.1364/AO.27.002022

Applied Optics
Vol. 27,
Issue 10,
pp. 2022-2024
(1988)
•https://doi.org/10.1364/AO.27.002022

Exact parallel matrix inversion using para-Hensel codes with systolic processors

V. K. Murthy

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V. K. Murthy, "Exact parallel matrix inversion using para-Hensel codes with systolic processors," Appl. Opt. 27, 2022-2024 (1988)

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Abstract

The application of para-Hensel codes for exact inversion or generalized inversion of a matrix using massively parallel processors is described.

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PHC EXPRESSION	p₁ = 7	p₂ = 11	p₃ = 13

$A_{1}^{+} = a_{1}^{T} {(a_{1}^{T} a_{1})}^{- 1}$	(3 0 6)	(9 0 7)	(8 0 3)
$d_{z} = A_{1}^{+} a_{2}$	(6)	(7)	(3)
c₂ = a₂ − A₁d₂	$[\begin{matrix} 1 \\ 1 \\ 3 \end{matrix}]$	$[\begin{matrix} 4 \\ 1 \\ 9 \end{matrix}]$	$[\begin{array}{r} 10 \\ 1 \\ 8 \end{array}]$
$b_{2} = c_{2}^{T} {(c_{2}^{T} c_{2})}^{- 1}$	(2 2 6)	(7 10 2)	(4 3 11)
$A_{2}^{+} = [\begin{matrix} A_{1}^{+} - d_{2} b_{2} \\ b_{2} \end{matrix}]$	$[\begin{matrix} 5 & 2 & 5 \\ 2 & 2 & 6 \end{matrix}]$	$[\begin{array}{r} 4 & 7 & 4 \\ 7 & 10 & 2 \end{array}]$	$[\begin{matrix} 9 & 4 & 9 \\ 4 & 3 & 11 \end{matrix}]$
$d_{3} = A_{2}^{+} a_{3}$	$[\begin{matrix} 3 \\ 0 \end{matrix}]$	$[\begin{matrix} 8 \\ 8 \end{matrix}]$	$[\begin{matrix} 5 \\ 1 \end{matrix}]$
c₃ = a₃ − A₂d₃	$[\begin{matrix} 5 \\ 6 \\ 1 \end{matrix}]$	$[\begin{matrix} 4 \\ 2 \\ 9 \end{matrix}]$	$[\begin{matrix} 9 \\ 11 \\ 2 \end{matrix}]$
$b_{3} = c_{3}^{T} {(c_{3}^{T} c_{3})}^{- 1}$	(2 1 6)	(2 1 10)	(2 1 12)
$A_{3}^{+} = [\begin{matrix} A_{2}^{+} - d_{3} b_{3} \\ b_{3} \end{matrix}]$	$[\begin{matrix} 6 & 6 & 1 \\ 2 & 2 & 6 \\ 2 & 1 & 6 \end{matrix}]$	$[\begin{array}{r} 10 & 10 & 1 \\ 2 & 2 & 10 \\ 2 & 1 & 10 \end{array}]$	$[\begin{array}{r} 12 & 12 & 1 \\ 2 & 2 & 12 \\ 2 & 1 & 12 \end{array}]$
$d_{4} = A_{3}^{+} a_{4}$	$[\begin{matrix} 0 \\ 1 \\ 1 \end{matrix}]$	$[\begin{matrix} 0 \\ 1 \\ 1 \end{matrix}]$	$[\begin{matrix} 0 \\ 1 \\ 1 \end{matrix}]$
c₄ = a₄ − A₃d₄	0=null vector	0	0
$b_{4} = {dIA}_{3}^{+} {(1 + {dId}_{4})}^{- 1}$	(6 1 4)	(5 1 3)	(10 1 8)
$A_{4}^{+} = [\begin{matrix} A_{3}^{+} - d_{4} b_{4} \\ b_{4} \end{matrix}]$	$[\begin{matrix} 6 & 6 & 1 \\ 3 & 1 & 2 \\ 3 & 0 & 2 \\ 6 & 1 & 4 \end{matrix}]$	$[\begin{array}{r} 10 & 10 & 1 \\ 8 & 1 & 7 \\ 8 & 0 & 7 \\ 5 & 1 & 3 \end{array}]$	$[\begin{array}{r} 12 & 12 & 1 \\ 5 & 1 & 4 \\ 5 & 0 & 4 \\ 10 & 1 & 8 \end{array}]$

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Applied Optics