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