Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 41
(d) Yes.
{b
( ) ( )
−1 0 0 1 ∣∣
+ c b, c ∈ R}
0 1 0 0
Two.I.2.21 No, it is not closed. In particular, it is not closed under scalar multiplication because it
does not contain the zero polynomial.
Two.I.2.22
(a) Yes, solving the linear system arising from
⎛
r 1
⎝ 1 ⎞ ⎛
0⎠ + r 2
⎝ 0 ⎞ ⎛
0⎠ = ⎝ 2 ⎞
0⎠
0 1 1
gives r 1 = 2 and r 2 = 1.
(b) Yes; the linear system arising from r 1 (x 2 ) + r 2 (2x + x 2 ) + r 3 (x + x 3 ) = x − x 3
2r 2 + r 3 = 1
r 1 + r 2 = 0
r 3 = −1
gives that −1(x 2 ) + 1(2x + x 2 ) − 1(x + x 3 ) = x − x 3 .
(c) No; any combination of the two given matrices has a zero in the upper right.
Two.I.2.23 (a) Yes; it is in that span since 1 · cos 2 x + 1 · sin 2 x = f(x).
(b) No, since r 1 cos 2 x + r 2 sin 2 x = 3 + x 2 has no scalar solutions that work for all x. For instance,
setting x to be 0 and π gives the two equations r 1 · 1 + r 2 · 0 = 3 and r 1 · 1 + r 2 · 0 = 3 + π 2 , which
are not consistent with each other.
(c) No; consider what happens on setting x to be π/2 and 3π/2.
(d) Yes, cos(2x) = 1 · cos 2 (x) − 1 · sin 2 (x).
Two.I.2.24
(a) Yes, for any x, y, z ∈ R this equation
⎛
r 1
⎝ 1 ⎞ ⎛
0⎠ + r 2
⎝ 0 ⎞ ⎛
2⎠ + r 3
⎝ 0 ⎞ ⎛
0⎠ = ⎝ x ⎞
y⎠
0 0 3 z
has the solution r 1 = x, r 2 = y/2, and r 3 = z/3.
(b) Yes, the equation
⎛
r 1
⎝ 2 ⎞ ⎛
0⎠ + r 2
⎝ 1 ⎞ ⎛
1⎠ + r 3
⎝ 0 ⎞ ⎛
0⎠ = ⎝ x ⎞
y⎠
1 0 1 z
gives rise to this
2r 1 + r 2
r 1
r 2
= x
= y
+ r 3 = z
−(1/2)ρ 1+ρ 3
−→
(1/2)ρ 2+ρ 3
−→ r 2 = y
2r 1 + r 2 = x
r 3 = −(1/2)x + (1/2)y + z
so that, given any x, y, and z, we can compute that r 3 = (−1/2)x + (1/2)y + z, r 2 = y, and
r 1 = (1/2)x − (1/2)y.
(c) No. In particular, the vector
⎛ ⎞
0
⎝0⎠
1
cannot be gotten as a linear combination since the two given vectors both have a third component
of zero.
(d) Yes. The equation
⎛
r 1
⎝ 1 ⎞ ⎛
0⎠ + r 2
⎝ 3 ⎞ ⎛
1⎠ + r 3
⎝ −1
⎞ ⎛
0 ⎠ + r 4
⎝ 2 ⎞ ⎛
1⎠ = ⎝ x ⎞
y⎠
1 0 0 5 z
leads to this reduction.
⎛
⎝ 1 3 −1 2 x
⎞
0 1 0 1 y
1 0 0 5 z
⎠ −ρ1+ρ3
−→
3ρ2+ρ3
−→
⎛
⎝ 1 3 −1 2 x
0 1 0 1 y
0 0 1 6 −x + 3y + z
We have infinitely many solutions. We can, for example, set r 4 to be zero and solve for r 3 , r 2 , and
r 1 in terms of x, y, and z by the usual methods of back-substitution.
⎞
⎠