1 System of Linear Equations Given matrix A € Rmxn and column vector b Rmx1 describe the set X of all v € R¹x1 that solves Av= 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(A), 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, 3ER. (a) A = b= 5 (b) A = [51], b= [8] (c) 6 A = - [8] 6 A = - [8] 1 6 4= [§ ³], A (d) (e) b= b= b= A
1 System of Linear Equations Given matrix A € Rmxn and column vector b Rmx1 describe the set X of all v € R¹x1 that solves Av= 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(A), 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, 3ER. (a) A = b= 5 (b) A = [51], b= [8] (c) 6 A = - [8] 6 A = - [8] 1 6 4= [§ ³], A (d) (e) b= b= b= A
College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter6: Matrices And Determinants
Section: Chapter Questions
Problem 66E: Distribution of Cash An ATM at a bank in Qualicum Beach, British Columbia, dispenses $20 and $50...
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part G and H
![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%2F1fa6da54-4e28-4e8d-979d-89b29b7ed7e0%2F1801adbf-008a-4269-984c-274ca5cd18f7%2Ftdmt2h_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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