b= A = [51], b= [8] 6 A = b= -B -2 1 α B A = -1/2 1 6 3 -2 -1 A = b=

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ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter6: Matrices And Determinants
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Problem 66E: Distribution of Cash An ATM at a bank in Qualicum Beach, British Columbia, dispenses $20 and $50...
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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=
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=
(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.
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.
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