1 3 and let S be the set of vectors x E R³ such that Ax e Span { 2 (3) Let A = Find a basis for S 1 which contains the vector -3 1 Solution 1. Note that Ax e Span{} if and only if the first coordinate of Ax is 0, in other words if [1 1 3] x = [0], thus S = nul([1 1 3]). We find a basis for S by solving for the general solution in vector form; here the matrix is already in reduced echelon form, and x2, X3 are free variables; we solve for -x2 – 3x3 -31 xi to get x1 = -x2 3x3; thus the general solution is x = x2 = x2 1 + x3 To find X3 a basis containing note that it is contained in S (since [1 1 3|-3 0) so we can apply the 1 -3 1 procedure for finding a basis of the span of -3] 0 0 0 1 3 1 0 -1 1 1 -3 1 R2+=3R3 1 rearrange 1 3 1 1 1 1 0 0
1 3 and let S be the set of vectors x E R³ such that Ax e Span { 2 (3) Let A = Find a basis for S 1 which contains the vector -3 1 Solution 1. Note that Ax e Span{} if and only if the first coordinate of Ax is 0, in other words if [1 1 3] x = [0], thus S = nul([1 1 3]). We find a basis for S by solving for the general solution in vector form; here the matrix is already in reduced echelon form, and x2, X3 are free variables; we solve for -x2 – 3x3 -31 xi to get x1 = -x2 3x3; thus the general solution is x = x2 = x2 1 + x3 To find X3 a basis containing note that it is contained in S (since [1 1 3|-3 0) so we can apply the 1 -3 1 procedure for finding a basis of the span of -3] 0 0 0 1 3 1 0 -1 1 1 -3 1 R2+=3R3 1 rearrange 1 3 1 1 1 1 0 0
Elementary Linear Algebra (MindTap Course List)
8th Edition
ISBN:9781305658004
Author:Ron Larson
Publisher:Ron Larson
Chapter7: Eigenvalues And Eigenvectors
Section7.CM: Cumulative Review
Problem 18CM
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Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
Linear Functions
A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
Question
Can anyone explain why x2 and x3 is a free variable in the third line of the solution?
![1
(3) Let A =
and let S be the set of vectors x E R³ such that Ax E Span
Find a basis for S
2 1
which contains the vector
-3
1
{}
Solution 1. Note that Ax e Span
if and only if the first coordinate of Ax is 0, in other words if
[1 1 3] x = [0], thus S = nul([1 1 3]). We find a basis for S by solving for the general solution in
vector form; here the matrix is already in reduced echelon form, and x2, x3 are free variables; we solve for
-x2 – 3x3
xị to get x1 = -x2 – 3x3; thus the general solution is x =
X2
= x2
+ x3
. To find
X3
1
0.
a basis containing -3
note that it is contained in S (since [1 1 3|-3
[0])
so we can apply the
%3D
1
procedure for finding a basis of the
span
of
0 0
3 R*zR2 0 1 3
1 0 1
-1
-3
-1
ㅇ
1
R2**3Rs 0
rearrange
-3
1
1
1
3
1
1
1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc93be347-1a0a-4c78-aca0-c94abc3e4570%2Fdb578b53-24f8-440f-a55c-e1cf17ea3d71%2F5o2vcv4_processed.png&w=3840&q=75)
Transcribed Image Text:1
(3) Let A =
and let S be the set of vectors x E R³ such that Ax E Span
Find a basis for S
2 1
which contains the vector
-3
1
{}
Solution 1. Note that Ax e Span
if and only if the first coordinate of Ax is 0, in other words if
[1 1 3] x = [0], thus S = nul([1 1 3]). We find a basis for S by solving for the general solution in
vector form; here the matrix is already in reduced echelon form, and x2, x3 are free variables; we solve for
-x2 – 3x3
xị to get x1 = -x2 – 3x3; thus the general solution is x =
X2
= x2
+ x3
. To find
X3
1
0.
a basis containing -3
note that it is contained in S (since [1 1 3|-3
[0])
so we can apply the
%3D
1
procedure for finding a basis of the
span
of
0 0
3 R*zR2 0 1 3
1 0 1
-1
-3
-1
ㅇ
1
R2**3Rs 0
rearrange
-3
1
1
1
3
1
1
1
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