Linear Algebra Exercises-n-Answers.pdf

28 Linear Algebra, by Hefferon
(a) The first gives
while the second gives
( )
−4ρ 1 +ρ 2 1 2
−→
0 0
( ) ( )
ρ 1 ↔ρ 2 1 2 −2ρ 2 +ρ 1 1 0
−→ −→
0 1
0 1
The two reduced echelon form matrices are not identical, and so the original matrices are not row
equivalent.
(b) The first is this.
⎛
−3ρ 1 +ρ 2
−→ ⎝ 1 0 2 ⎞ ⎛
0 −1 −5⎠ −ρ 2+ρ 3
−→ ⎝ 1 0 2 ⎞ ⎛
0 −1 −5⎠ −ρ 2
−→ ⎝ 1 0 2 ⎞
0 1 5⎠
−5ρ 1+ρ 3
0 −1 −5
0 0 0 0 0 0
The second is this.
−2ρ 1 +ρ 3
−→
⎛
⎝ 1 0 2 ⎞
0 2 10
0 0 0
⎠ (1/2)ρ 2
−→
⎛
⎝ 1 0 2 ⎞
0 1 5⎠
0 0 0
These two are row equivalent.
(c) These two are not row equivalent because they have different sizes.
(d) The first,
( ) ( ) ( )
ρ 1 +ρ 2 1 1 1 (1/3)ρ 2 1 1 1 −ρ 2 +ρ 1 1 0 0
−→ −→ −→
0 3 3 0 1 1 0 1 1
and the second.
These are not row equivalent.
(e) Here the first is
while this is the second.
These are not row equivalent.
( ) ( ) ( )
ρ 1↔ρ 2 2 2 5 (1/2)ρ 1 1 1 5/2 −ρ 2+ρ 1 1 0 17/6
−→ −→
−→
0 3 −1 (1/3)ρ 2
0 1 −1/3 0 1 −1/3
( ) ( )
(1/3)ρ 2 1 1 1 −ρ 2 +ρ 1 1 1 0
−→ −→
0 0 1 0 0 1
( ) ( )
ρ 1 ↔ρ 2 1 −1 1 ρ 2 +ρ 1 1 0 3
−→ −→
0 1 2 0 1 2
One.III.2.12 First, the only matrix row equivalent to the matrix of all 0’s is itself (since row operations
have no effect).
Second, the matrices that reduce to (
1
)
a
0 0
have the form (
b
)
ba
c ca
(where a, b, c ∈ R).
Next, the matrices that reduce to (
0
)
1
0 0
have the form (
0
)
a
0 b
(where a, b ∈ R).
Finally, the matrices that reduce to (
1
)
0
0 1
are the nonsingular matrices. That’s because a linear system for which this is the matrix of coefficients
will have a unique solution, and that is the definition of nonsingular. (Another way to say the same
thing is to say that they fall into none of the above classes.)