= [B] 4= 1 A [1 2 3] 4 5 6 32 [1 2 3] [14] A b= 48 a ples of vectors v = [r, y, z] that solve the homogeneous eq ib b=
= [B] 4= 1 A [1 2 3] 4 5 6 32 [1 2 3] [14] A b= 48 a ples of vectors v = [r, y, z] that solve the homogeneous eq ib b=
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
part G and H needed
![(f)
09
A =
b =
A =
[1 2 3]
4 5 6
b=
(h)
1 2 3
A =
b=
[]
7
48 a
Hint: To find examples of vectors v = [x, y, z] that solve the homogeneous equation
Av = 0, you can deliberately set some of the variables to be equal 0 or 1:
if rank(A) = 2, then the dimension of null(A) is 1, so you only need to find one
vector ₁ and then null(A) = span(v₁). Try to put r = 1 and solve the rest of the
equation. If the equation did not have a solutions, then you would put y = 1 or
z = 1 instead.
• if rank(A) = 1, then you can pick values of two of the variables x, y, z. Remember
that now you need to find two independent vectors to span null(A). For example,
put (x, y) = (1,0) and try find z; and then put (x, y) = (0, 1) and try find z.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd1a47737-b22a-410a-a4a6-b365084bb91b%2F272072dd-9000-4485-8d12-bc0853c77490%2Fydcftye_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(f)
09
A =
b =
A =
[1 2 3]
4 5 6
b=
(h)
1 2 3
A =
b=
[]
7
48 a
Hint: To find examples of vectors v = [x, y, z] that solve the homogeneous equation
Av = 0, you can deliberately set some of the variables to be equal 0 or 1:
if rank(A) = 2, then the dimension of null(A) is 1, so you only need to find one
vector ₁ and then null(A) = span(v₁). Try to put r = 1 and solve the rest of the
equation. If the equation did not have a solutions, then you would put y = 1 or
z = 1 instead.
• if rank(A) = 1, then you can pick values of two of the variables x, y, z. Remember
that now you need to find two independent vectors to span null(A). For example,
put (x, y) = (1,0) and try find z; and then put (x, y) = (0, 1) and try find z.
![1 System of Linear Equations
Given matrix A € Rmxn and column vector bRmx1 describe the set X of all v € R¹x1
that solves Au = b. Remember that X = 0 (empty set, meaning there is no v that solves
the system of equations) iff b span(A.,₁,..., A.), otherwise there exists a solution. If a
solution to exists, then the set of all the solutions can be expressed as X = 2o+null(A), where
null(A) = span (v₁,..., Uk), and U₁,..., U is a maximal set of linearly independent solutions
of the homogeneous equation Av = 0. The dimension k of null(A) equals to n - rank(4),
i.e. the number of variables minus the number of independent rows/columns of A.
You can use a formula for A-¹, e.g. https://www.cuemath.com/algebra/inverse-of-2x2-
matrix/.
Don't forget to consider all the cases based on the values of the parameters a, 3 € R.
(a)
A =
b=
5
A = [51], b= [8]
6
A =
- [8]
6
A =
- [8]
1
6
4-B 3 --A
A =
b=
2
(b)
(c)
(d)
(e)
b=
b=](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd1a47737-b22a-410a-a4a6-b365084bb91b%2F272072dd-9000-4485-8d12-bc0853c77490%2Flmrnofr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1 System of Linear Equations
Given matrix A € Rmxn and column vector bRmx1 describe the set X of all v € R¹x1
that solves Au = b. Remember that X = 0 (empty set, meaning there is no v that solves
the system of equations) iff b span(A.,₁,..., A.), otherwise there exists a solution. If a
solution to exists, then the set of all the solutions can be expressed as X = 2o+null(A), where
null(A) = span (v₁,..., Uk), and U₁,..., U is a maximal set of linearly independent solutions
of the homogeneous equation Av = 0. The dimension k of null(A) equals to n - rank(4),
i.e. the number of variables minus the number of independent rows/columns of A.
You can use a formula for A-¹, e.g. https://www.cuemath.com/algebra/inverse-of-2x2-
matrix/.
Don't forget to consider all the cases based on the values of the parameters a, 3 € R.
(a)
A =
b=
5
A = [51], b= [8]
6
A =
- [8]
6
A =
- [8]
1
6
4-B 3 --A
A =
b=
2
(b)
(c)
(d)
(e)
b=
b=
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