MA 125 HW-6 (Solutions) Due: March 4, 2023
1. Consider the plane in R3 given by equation x − 2y + 3z = 0, which corresponds to the set of
x
y
vectors U = z | x − 2y + 3z = 0
(a) Verify that U is a subspace of R3 .
Solution:
0
• 0
is in U since 0 − 2(0) + 3(0) = 0.
0 = 0
x x2
1
y1 y2
• Suppose u1 = z1 and u2 = z2 are in U . Then
0 = 0 + 0 = (x1 − 2y1 + 3z1 ) + (x2 − 2y2 + 3z2 ) = (x1 + x2 ) − 2(y1 + y2 ) + 3(z1 + z2 )
x1 + x2
y1 + y2
is in U .
and so u1 + u2 = z1 + y3
x
y
is in U , then for any scalar k ,
• Similarly, if u
= z
0 = k · 0 = k(x − 2y + 3z) = kx − 2(ky) + 3(kz )
and so ku is in U .
(b) Find a basis for U .
Solution: Any correct basis with justification is acceptable. For example, find a
spanning set of U by writing
2y − 3z 2 −3
y 1 0 ,
U= z | y, z in R = y 0 + z 1 | y, z in R
2 −3
1 0 . Then conclude the set
showing that U = Span 0 , 1
2 −3
1 0 is linearly independent as there are only two vectors which are not
0 , 1
2 −3
1 0
into row echelon and seeing both columns
scalars multiples, or putting 0 1
have pivots.
(c) Find a basis for the subspace of R3 corresponding to the intersection of the plane given
by x − 2y + 3z = 0 and the xy -plane.
1. Consider the plane in R3 given by equation x − 2y + 3z = 0, which corresponds to the set of
x
y
vectors U = z | x − 2y + 3z = 0
(a) Verify that U is a subspace of R3 .
Solution:
0
• 0
is in U since 0 − 2(0) + 3(0) = 0.
0 = 0
x x2
1
y1 y2
• Suppose u1 = z1 and u2 = z2 are in U . Then
0 = 0 + 0 = (x1 − 2y1 + 3z1 ) + (x2 − 2y2 + 3z2 ) = (x1 + x2 ) − 2(y1 + y2 ) + 3(z1 + z2 )
x1 + x2
y1 + y2
is in U .
and so u1 + u2 = z1 + y3
x
y
is in U , then for any scalar k ,
• Similarly, if u
= z
0 = k · 0 = k(x − 2y + 3z) = kx − 2(ky) + 3(kz )
and so ku is in U .
(b) Find a basis for U .
Solution: Any correct basis with justification is acceptable. For example, find a
spanning set of U by writing
2y − 3z 2 −3
y 1 0 ,
U= z | y, z in R = y 0 + z 1 | y, z in R
2 −3
1 0 . Then conclude the set
showing that U = Span 0 , 1
2 −3
1 0 is linearly independent as there are only two vectors which are not
0 , 1
2 −3
1 0
into row echelon and seeing both columns
scalars multiples, or putting 0 1
have pivots.
(c) Find a basis for the subspace of R3 corresponding to the intersection of the plane given
by x − 2y + 3z = 0 and the xy -plane.
