(Feldman’s notes) Example III.19 1 1

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Example III.19 (Feldman’s notes)
1 1
Let A =
. We look for a matrix B that obeys AB = I. The
1 2 1
0
columns of I are b
e1 =
and b
e2 =
. Denote the columns of B by
0
1
~ 1 and B
~ 2 . That is
B
′
X X′
X
~1 B
~2
~1 =
~ 2 = X′
= B
B=
B
B
′
Y Y
Y
Y
We are to find a B obeying
X +Y
X′ + Y ′
~
~
b
b
=
=
A
B
A
B
AB =
e
e
1
2
1 2
X + 2Y X ′ + 2Y ′
~1 = b
~2 = b
~ 1 and
or, equivalently, AB
e1 , AB
e2 . We wish to solve for B
~ 2 . The augmented matrices for these two systems of equations are
B
Ab
e1 and A b
e2 . For efficiency, we combine them into a single larger
agumented matrix
A b
e1 b
e2 =
1
1
1 1
2 0
0
1
We can apply row operations to bring this to reduced row echelon form
1 1 1 0
1 0 2 −1 (1) − (2)
0 1 −1 1 (2) − (1)
0 1 −1 1
The augmented
on the right
matrix
represents the two systems of equa~ 1 = 2 and I B
~ 2 = −1 . Thus
tions I B
−1
1
2
−1
2
−1
~1 =
~2 =
~1 B
~2 =
B
B
B= B
−1
1
−1 1
c Joel Feldman. 2011. All rights reserved.
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