(Feldman’s notes) Example III.19 1 1

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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.

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