Solution of linear equation with variables
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For what values of a and b the following system of linear equations has
2x + 3y + 5z = 9
7x + 3y - 2z = 8
2x + 3y + az = b
a) no solution.
b) a unique solution.
c) an infinite number of solutions.
[I have found the properties of a and b as (a=5, b!=9) for no solution by its determinant. I have tried to find the others but couldn't find a way to write them in logic considering there are two variables. Am I going to write like if a=... then b=... or is there an other way? Is it going to be a=5 b!=9 for a) and a=5 b=9 for c) ?][I am mostly sucked with finding the unique solution set.]
vector-spaces
asked Nov 15 at 15:36
E.Canberk
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0
down vote
favorite
For what values of a and b the following system of linear equations has
2x + 3y + 5z = 9
7x + 3y - 2z = 8
2x + 3y + az = b
a) no solution.
b) a unique solution.
c) an infinite number of solutions.
[I have found the properties of a and b as (a=5, b!=9) for no solution by its determinant. I have tried to find the others but couldn't find a way to write them in logic considering there are two variables. Am I going to write like if a=... then b=... or is there an other way? Is it going to be a=5 b!=9 for a) and a=5 b=9 for c) ?][I am mostly sucked with finding the unique solution set.]
vector-spaces
asked Nov 15 at 15:36
E.Canberk
13
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
For what values of a and b the following system of linear equations has
2x + 3y + 5z = 9
7x + 3y - 2z = 8
2x + 3y + az = b
a) no solution.
b) a unique solution.
c) an infinite number of solutions.
[I have found the properties of a and b as (a=5, b!=9) for no solution by its determinant. I have tried to find the others but couldn't find a way to write them in logic considering there are two variables. Am I going to write like if a=... then b=... or is there an other way? Is it going to be a=5 b!=9 for a) and a=5 b=9 for c) ?][I am mostly sucked with finding the unique solution set.]
vector-spaces
asked Nov 15 at 15:36
E.Canberk
13
For what values of a and b the following system of linear equations has
2x + 3y + 5z = 9
7x + 3y - 2z = 8
2x + 3y + az = b
a) no solution.
b) a unique solution.
c) an infinite number of solutions.
[I have found the properties of a and b as (a=5, b!=9) for no solution by its determinant. I have tried to find the others but couldn't find a way to write them in logic considering there are two variables. Am I going to write like if a=... then b=... or is there an other way? Is it going to be a=5 b!=9 for a) and a=5 b=9 for c) ?][I am mostly sucked with finding the unique solution set.]
vector-spaces
vector-spaces
asked Nov 15 at 15:36
E.Canberk
13
asked Nov 15 at 15:36
E.Canberk
13
edited Nov 15 at 15:42
asked Nov 15 at 15:36
E.Canberk
13
asked Nov 15 at 15:36
E.Canberk
13
asked Nov 15 at 15:36
E.Canberk
13
13
add a comment |
add a comment |
1 Answer
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The system has a unique solution if and only if the determinant of the matrix of coefficients is non-zero. In all other cases, it has either no solution or infinitely many solutions. To distinguish the two, find the images of $$left(array{1\0\0}right),left(array{0\1\0}right),mbox{ and } left(array{0\0\1}right)$$ under (the linear map corresponding to) the matrix of coefficients. There are infinitely many solutions exactly when you can write $$left(array{9\8\b}right)$$ as a linear combination of those image vectors, otherwise there are no solutions.
answered Nov 15 at 15:45
user3482749
1,709411
When a=5 the determinant is zero therefore no solution but when b=9, does it still count as no solution because of the determinant or can we say unique solution because 0 = 0 in the given values
– E.Canberk
Nov 15 at 15:49
The row equivalet form of the matrix = R1(1, -6, -17, -19), R2(0, 15, 39, 47), R3(0, 0, a-5, b-9)
– E.Canberk
Nov 15 at 15:51
Why would there be a unique solution? You're trying to solve the whole system, not just one dimension of it. And, again, the determinant being zero does not mean that there is no solution. It means that either their is no solution, or there are infinitely many solutions.
– user3482749
Nov 15 at 15:55
Your last sentence was the answer which I was waiting to confirm my question. And for the first sentence, it asks me the values of a and b such that system has unique solution.
– E.Canberk
Nov 15 at 16:01
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The system has a unique solution if and only if the determinant of the matrix of coefficients is non-zero. In all other cases, it has either no solution or infinitely many solutions. To distinguish the two, find the images of $$left(array{1\0\0}right),left(array{0\1\0}right),mbox{ and } left(array{0\0\1}right)$$ under (the linear map corresponding to) the matrix of coefficients. There are infinitely many solutions exactly when you can write $$left(array{9\8\b}right)$$ as a linear combination of those image vectors, otherwise there are no solutions.
answered Nov 15 at 15:45
user3482749
1,709411
When a=5 the determinant is zero therefore no solution but when b=9, does it still count as no solution because of the determinant or can we say unique solution because 0 = 0 in the given values
– E.Canberk
Nov 15 at 15:49
The row equivalet form of the matrix = R1(1, -6, -17, -19), R2(0, 15, 39, 47), R3(0, 0, a-5, b-9)
– E.Canberk
Nov 15 at 15:51
Why would there be a unique solution? You're trying to solve the whole system, not just one dimension of it. And, again, the determinant being zero does not mean that there is no solution. It means that either their is no solution, or there are infinitely many solutions.
– user3482749
Nov 15 at 15:55
Your last sentence was the answer which I was waiting to confirm my question. And for the first sentence, it asks me the values of a and b such that system has unique solution.
– E.Canberk
Nov 15 at 16:01
add a comment |
up vote
0
down vote
The system has a unique solution if and only if the determinant of the matrix of coefficients is non-zero. In all other cases, it has either no solution or infinitely many solutions. To distinguish the two, find the images of $$left(array{1\0\0}right),left(array{0\1\0}right),mbox{ and } left(array{0\0\1}right)$$ under (the linear map corresponding to) the matrix of coefficients. There are infinitely many solutions exactly when you can write $$left(array{9\8\b}right)$$ as a linear combination of those image vectors, otherwise there are no solutions.
answered Nov 15 at 15:45
user3482749
1,709411
When a=5 the determinant is zero therefore no solution but when b=9, does it still count as no solution because of the determinant or can we say unique solution because 0 = 0 in the given values
– E.Canberk
Nov 15 at 15:49
The row equivalet form of the matrix = R1(1, -6, -17, -19), R2(0, 15, 39, 47), R3(0, 0, a-5, b-9)
– E.Canberk
Nov 15 at 15:51
Why would there be a unique solution? You're trying to solve the whole system, not just one dimension of it. And, again, the determinant being zero does not mean that there is no solution. It means that either their is no solution, or there are infinitely many solutions.
– user3482749
Nov 15 at 15:55
Your last sentence was the answer which I was waiting to confirm my question. And for the first sentence, it asks me the values of a and b such that system has unique solution.
– E.Canberk
Nov 15 at 16:01
add a comment |
up vote
0
down vote
up vote
0
down vote
The system has a unique solution if and only if the determinant of the matrix of coefficients is non-zero. In all other cases, it has either no solution or infinitely many solutions. To distinguish the two, find the images of $$left(array{1\0\0}right),left(array{0\1\0}right),mbox{ and } left(array{0\0\1}right)$$ under (the linear map corresponding to) the matrix of coefficients. There are infinitely many solutions exactly when you can write $$left(array{9\8\b}right)$$ as a linear combination of those image vectors, otherwise there are no solutions.
answered Nov 15 at 15:45
user3482749
1,709411
The system has a unique solution if and only if the determinant of the matrix of coefficients is non-zero. In all other cases, it has either no solution or infinitely many solutions. To distinguish the two, find the images of $$left(array{1\0\0}right),left(array{0\1\0}right),mbox{ and } left(array{0\0\1}right)$$ under (the linear map corresponding to) the matrix of coefficients. There are infinitely many solutions exactly when you can write $$left(array{9\8\b}right)$$ as a linear combination of those image vectors, otherwise there are no solutions.
answered Nov 15 at 15:45
user3482749
1,709411
answered Nov 15 at 15:45
user3482749
1,709411
answered Nov 15 at 15:45
user3482749
1,709411
answered Nov 15 at 15:45
user3482749
1,709411
1,709411
When a=5 the determinant is zero therefore no solution but when b=9, does it still count as no solution because of the determinant or can we say unique solution because 0 = 0 in the given values
– E.Canberk
Nov 15 at 15:49
The row equivalet form of the matrix = R1(1, -6, -17, -19), R2(0, 15, 39, 47), R3(0, 0, a-5, b-9)
– E.Canberk
Nov 15 at 15:51
Why would there be a unique solution? You're trying to solve the whole system, not just one dimension of it. And, again, the determinant being zero does not mean that there is no solution. It means that either their is no solution, or there are infinitely many solutions.
– user3482749
Nov 15 at 15:55
Your last sentence was the answer which I was waiting to confirm my question. And for the first sentence, it asks me the values of a and b such that system has unique solution.
– E.Canberk
Nov 15 at 16:01
add a comment |
When a=5 the determinant is zero therefore no solution but when b=9, does it still count as no solution because of the determinant or can we say unique solution because 0 = 0 in the given values
– E.Canberk
Nov 15 at 15:49
The row equivalet form of the matrix = R1(1, -6, -17, -19), R2(0, 15, 39, 47), R3(0, 0, a-5, b-9)
– E.Canberk
Nov 15 at 15:51
Why would there be a unique solution? You're trying to solve the whole system, not just one dimension of it. And, again, the determinant being zero does not mean that there is no solution. It means that either their is no solution, or there are infinitely many solutions.
– user3482749
Nov 15 at 15:55
Your last sentence was the answer which I was waiting to confirm my question. And for the first sentence, it asks me the values of a and b such that system has unique solution.
– E.Canberk
Nov 15 at 16:01
When a=5 the determinant is zero therefore no solution but when b=9, does it still count as no solution because of the determinant or can we say unique solution because 0 = 0 in the given values
– E.Canberk
Nov 15 at 15:49
When a=5 the determinant is zero therefore no solution but when b=9, does it still count as no solution because of the determinant or can we say unique solution because 0 = 0 in the given values
– E.Canberk
Nov 15 at 15:49
The row equivalet form of the matrix = R1(1, -6, -17, -19), R2(0, 15, 39, 47), R3(0, 0, a-5, b-9)
– E.Canberk
Nov 15 at 15:51
The row equivalet form of the matrix = R1(1, -6, -17, -19), R2(0, 15, 39, 47), R3(0, 0, a-5, b-9)
– E.Canberk
Nov 15 at 15:51
Why would there be a unique solution? You're trying to solve the whole system, not just one dimension of it. And, again, the determinant being zero does not mean that there is no solution. It means that either their is no solution, or there are infinitely many solutions.
– user3482749
Nov 15 at 15:55
Why would there be a unique solution? You're trying to solve the whole system, not just one dimension of it. And, again, the determinant being zero does not mean that there is no solution. It means that either their is no solution, or there are infinitely many solutions.
– user3482749
Nov 15 at 15:55
Your last sentence was the answer which I was waiting to confirm my question. And for the first sentence, it asks me the values of a and b such that system has unique solution.
– E.Canberk
Nov 15 at 16:01
Your last sentence was the answer which I was waiting to confirm my question. And for the first sentence, it asks me the values of a and b such that system has unique solution.
– E.Canberk
Nov 15 at 16:01
add a comment |
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