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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT 256 23 "The inverse of a matrix
" }}{PARA 0 "" 0 "" {TEXT -1 245 "Here is one way how Maple can find t
he inverse of a matrix. Using MAple I will show you the algorithm of f
inding the inverse, using at the same time the elementary matrices tha
t acieve at each step in the algorithm the elementary row operations.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "with(LinearAlgebra);" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 190 "Example 1. We look at  exercise
 7(a) in Anton&Rorres. We will define the elementary matrices that wil
l perform the row operations on the matrix A to reduceit to its reduce
d-row-echelon form." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "A:=M
atrix([[3,4,-1],[1,0,3],[2,5,-4]]);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 17 "MatrixInverse(A);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 200 "The first step is to get an 1 on the first row, first column. \+
We will interchange the first and second row. At the same time we defi
ne the elementary matrix E1 that does just that. We also check that " 
}{XPPEDIT 18 0 "E1*A;" "6#*&%#E1G\"\"\"%\"AGF%" }{TEXT -1 28 " achieve
s exactly that step." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "E1:
=Matrix([[0,1,0],[1,0,0],[0,0,1]]);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 19 "A1:=Multiply(E1,A);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 244 "Now we will get zeros on the first column in the secnd and thi
rd row by subtracting three times the first row from the second row an
d subtracting twice the first row from the third row, respectively. Ag
ain we define two new elementary matrices:" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 78 "E2:=Matrix([[1,0,0],[-3,1,0],[0,0,1]]);E3:=Matrix([
[1,0,0],[0,1,0],[-2,0,1]]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 109 "A
gain, by multiplying the matrix A1 to the left by the matrix E2 and th
en by E3 we obtain the desired result." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "A2:=Multiply(E2,A1);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "A3:=Multiply(E3,A2);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 183 "Now we want to get 1 on the second row. We just divide b
y 4 the entire row (Maple can handle that). So here is the matrix that
 achieves that row operation and the transformed matrix: " }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "E4:=Matrix([[1,0,0],[0,1/4,0],[0,0,
1]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "A4:=Multiply(E4,A3
);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 110 "Next we get zero in the se
cond column, row three by subtracting fives times the second row from \+
the third row:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "E5:=Matri
x([[1,0,0],[0,1,0],[0,-5,1]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 20 "A5:=Multiply(E5,A4);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 46 "N
ext step is to multiply the third row by 2/5:" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 40 "E6:=Matrix([[1,0,0],[0,1,0],[0,0,2/5]]);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "A6:=Multiply(E6,A5);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 187 "Now we have to more stps to get z
ero in the third column, above the pivot. We first add 5/2 the third r
ow to the second row and then subtract three times the third row from \+
the first row:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "E7:=Matri
x([[1,0,0],[0,1,5/2],[0,0,1]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 20 "A7:=Multiply(E7,A6);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 39 "E8:=Matrix([[1,0,-3],[0,1,0],[0,0,1]]);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 20 "A8:=Multiply(E8,A7);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 16 "So we have that " }{XPPEDIT 18 0 "E8*E7*E6*E5*E4*E3*E2*E1
*A = I;" "6#/*4%#E8G\"\"\"%#E7GF&%#E6GF&%#E5GF&%#E4GF&%#E3GF&%#E2GF&%#
E1GF&%\"AGF&%\"IG" }{TEXT -1 259 " and therefore the inverse of the ma
trix A (think that you multiply by the inverse of A to the right of th
e equation above) equals the product of the eight elementary matrices.
 Let us check that. Unfortunately, we have to multiply only two matric
es at a time:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "B1:=Multip
ly(E8,E7);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "B2:=Multiply(
B1,E6);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "B3:=Multiply(B2,
E5);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "B4:=Multiply(B3,E4)
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "B5:=Multiply(B4,E3);" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "B6:=Multiply(B5,E2);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "B7:=Multiply(B6,E1);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 161 "And this last matrix, B7 should b
e the inverse of A, and as you can verify, it is the same matrix given
 by the Matrixinverse command at the top of the worksheet." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "37" 0 }{VIEWOPTS 1 1 0 
1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }