Texas Instruments TI-89 Tip List page 208

and the inverse is
1 1
4 −
2
− 3 4
1 − 1 4
− 3 3
4 2
1
0
2
X −1 =
1
−1
2
9
−3
4
− 3 2
0
− 3 2 − 1
2
1 0
and the coefficient vector solution a is found simply by
a = X −1 $ z
-1
Again X is a constant matrix, so the coefficients for the interpolating polynomial are found with a single
matrix multiply. Another advantage to this method is that all of the elements of the inverted matrix can
be represented exactly with 89/92+ BCD arithmetic, so they do not contribute to round-off error in the
final result
.Once we have the polynomial coefficients, it is straight-forward to interpolate for z with
z = a $ u 2 $ v 2 + b $ u 2 $ v + c $ u $ v 2 + d $ u 2 + e $ v 2 + f $ u $ v + g $ u + h $ v + i
This function, intrp9z(), implements these ideas.
intrpz9(xl,yl,zmat,x,y)
Func
©({xlist},{ylist},[zmatrix],x,y) 9-point z-interpolation
©Uses matrix math\im1a
©1apr01/dburkett@infinet.com
local u,v
when(x<xl[2],(x-xl[1])/(xl[2]-xl[1]),(x-xl[2])/(xl[3]-xl[2])+1)→u
when(y<yl[2],(y-yl[1])/(yl[2]-yl[1]),(y-yl[2])/(yl[3]-yl[2])+1)→v
sum(mat▶list(math\im1a*(augment(augment(zmat[1],zmat[2]),zmat[3])))*{u^2*v^2,u^
2*v,u*v^2,u^2,v^2,u*v,u,v,1})
EndFunc
Note that the matrix im1a (from equation [9]) must be present and stored in the \math folder.
The input arguments are
1 1 1 − 1 2 1
4 − 4 −
2
3 3
1
2 −2 2 −
4
− 3 1 − 1
1 −2
4 4
1
0 −1
0 0
2
1
0 0 0 0
2
3 3
−3 4 −1
4 4
0 − 1 2
0 2 0
0 0 0 0
2
0 0 0 0 0
1 1
2 4
1 − 1 4
1 − 1
2 4
0 0
0 0
1
−1
4
0 0
0 0
0 0
[9]
[10]
[11]
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