Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 119
Three.IV.4.23 With H a 2×3 matrix, in looking for a matrix G such that the combination HG acts
as the 2×2 identity we need G to be 3×2. Setting up the equation
( ) ⎛
1 0 1 ⎝ m n
⎞
( )
p q⎠ 1 0
=
0 1 0
0 1
r s
and solving the resulting linear system
m +r = 1
n +s = 0
p = 0
q = 1
gives infinitely many solutions.
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
m 1 −1 0
n
0
0
−1
{
p
⎜ q
=
0
⎟ ⎜1
+ r ·
0
⎟ ⎜ 0
+ s ·
0
∣ ⎟ ⎜ 0
r, s ∈ R}
⎟
⎝ r ⎠ ⎝0⎠
⎝ 1 ⎠ ⎝ 0 ⎠
s 0 0 1
Thus H has infinitely many right inverses.
As for left inverses, the equation
⎛
( ) ( )
a b 1 0 1
= ⎝ 1 0 0
⎞
0 1 0⎠
c d 0 1 0
0 0 1
gives rise to a linear system with nine equations and four unknowns.
a = 1
b = 0
a = 0
c = 0
d = 1
c = 0
e = 0
f = 0
e = 1
This system is inconsistent (the first equation conflicts with the third, as do the seventh and ninth)
and so there is no left inverse.
Three.IV.4.24 With respect to the standard bases we have
⎛
Rep E2 ,E 3
(η) = ⎝ 1 0
⎞
0 1⎠
0 0
and setting up the equation to find the matrix inverse
( ) ⎛
a b c
⎝ 1 0
⎞
( )
0 1⎠ 1 0
= = Rep
d e f
0 1
E2 ,E 2
(id)
0 0
gives rise to a linear system.
a = 1
b = 0
d = 0
e = 1
There are infinitely many solutions in a, . . . , f to this system because two of these variables are entirely
unrestricted
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
a 1 0 0
b
0
0
0
{
c
⎜d
=
0
⎟ ⎜0
+ c ·
1
⎟ ⎜0
+ f ·
0
∣ ⎟ ⎜0
c, f ∈ R}
⎟
⎝e⎠
⎝1⎠
⎝0⎠
⎝0⎠
f 0 0 1