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*3

lAfip}-

Obtaining the inverse of the partition

x2y3 - x3y2 y2 - y3 x3 - x2 x$y\-x\yi — Ji x,-x3 xi y2 - x2yt y, - y2 x2 - x, xiy3-x3y2 x3yt y2 - y3

xi y3 xi y2-x2yx-3*3 - 3*1 y\ ~ J2 X| — x3 x2 — x]

and det [A„tt] = (x2>'3 - x^y2) - Ui y3 - X3V1) + (x, y2 - x2y 1 ) = *i(y2 - - >'i)+JC3(3'I - yi)

— 2x area of element = 2a, (see following derivation)

Then

adj[Aaa] _ det[Aaa] 2 a x2y3 - x3y2 jc3yi - X! >>3 X] y2 - x2yi yi - yi y3 - yi yi - yi

Hence, [A]"1 = — 2 a x2y3-x}y2 x3y[-x\n x\y2-x2y\ 0

Area of element

With reference to Fig. 9.42, area of triangular element, a = area of enclosing rectangle — (area of triangles b, c and d)

= (X2 -xt)(y3 - yi) - (1/2)0*2 - X{){y2 - yi) - (l/2)(x2 -*3)(j>3 - yi)

= x2y3 -x2y 1 -xiy3 +xiy{ - {\/2)[x2y2 - x2y\ - xxy2 + x\yx) + (x2yi - x2y2 - x3y3 + x3y2) + (x3y3 - X3.V1 - Xi y3 + x\ yt )] = (\/2)(x2y3 -x2y\ -xxy3+x{y2 - x3y2 + x3y\) = (l/2)[x,(y2 - y3)+x2(y3 - ji)+j:3(yi - y2)]

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