Mixing
System of Equations-Linear Algebra
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Below is a question from a competitive exam
Let $c_1,....c_n$ be scalars, not all zero, such that $sum_i=1^nc_ia_i=0$
where $a_i$ are the column vectors in $R^n$.
Consider the set of linear equations
$Ax=b$
where $A=[a_1....a_n]$ and $b=sum_i=1^na_i$. The set of equations has
- a unique solution at $x=J_n$ where $J_n$ denotes an n-dimensional vector of all 1.
- No Solution
- Infinitely many solutions
- Finitely many solutions.
**My Attempt :*
From given information, it is clear that the columns of the matrix A are dependent and hence there exists a non-zero vector $x_n$ where it denotes solution to the null space, such that $Ax_n=0$...(a)
Now, it is also given that $b=sum_i=1^na_i$, which means a column vector of all 1,say $x_p$ a particular solution to this b, is such that $Ax_p=b$ ..(b)
From (a) and (b),if c is a constant,then $A(x_p+c.x_n)=b$ and hence this system seems to have infinite solutions.
Is my attempt correct?
linear-algebra
asked Jul 15 at 7:11
user3767495
857
add a comment |Â
up vote
2
down vote
favorite
Below is a question from a competitive exam
Let $c_1,....c_n$ be scalars, not all zero, such that $sum_i=1^nc_ia_i=0$
where $a_i$ are the column vectors in $R^n$.
Consider the set of linear equations
$Ax=b$
where $A=[a_1....a_n]$ and $b=sum_i=1^na_i$. The set of equations has
- a unique solution at $x=J_n$ where $J_n$ denotes an n-dimensional vector of all 1.
- No Solution
- Infinitely many solutions
- Finitely many solutions.
**My Attempt :*
From given information, it is clear that the columns of the matrix A are dependent and hence there exists a non-zero vector $x_n$ where it denotes solution to the null space, such that $Ax_n=0$...(a)
Now, it is also given that $b=sum_i=1^na_i$, which means a column vector of all 1,say $x_p$ a particular solution to this b, is such that $Ax_p=b$ ..(b)
From (a) and (b),if c is a constant,then $A(x_p+c.x_n)=b$ and hence this system seems to have infinite solutions.
Is my attempt correct?
linear-algebra
asked Jul 15 at 7:11
user3767495
857
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Below is a question from a competitive exam
Let $c_1,....c_n$ be scalars, not all zero, such that $sum_i=1^nc_ia_i=0$
where $a_i$ are the column vectors in $R^n$.
Consider the set of linear equations
$Ax=b$
where $A=[a_1....a_n]$ and $b=sum_i=1^na_i$. The set of equations has
- a unique solution at $x=J_n$ where $J_n$ denotes an n-dimensional vector of all 1.
- No Solution
- Infinitely many solutions
- Finitely many solutions.
**My Attempt :*
From given information, it is clear that the columns of the matrix A are dependent and hence there exists a non-zero vector $x_n$ where it denotes solution to the null space, such that $Ax_n=0$...(a)
Now, it is also given that $b=sum_i=1^na_i$, which means a column vector of all 1,say $x_p$ a particular solution to this b, is such that $Ax_p=b$ ..(b)
From (a) and (b),if c is a constant,then $A(x_p+c.x_n)=b$ and hence this system seems to have infinite solutions.
Is my attempt correct?
linear-algebra
asked Jul 15 at 7:11
user3767495
857
Below is a question from a competitive exam
Let $c_1,....c_n$ be scalars, not all zero, such that $sum_i=1^nc_ia_i=0$
where $a_i$ are the column vectors in $R^n$.
Consider the set of linear equations
$Ax=b$
where $A=[a_1....a_n]$ and $b=sum_i=1^na_i$. The set of equations has
- a unique solution at $x=J_n$ where $J_n$ denotes an n-dimensional vector of all 1.
- No Solution
- Infinitely many solutions
- Finitely many solutions.
**My Attempt :*
From given information, it is clear that the columns of the matrix A are dependent and hence there exists a non-zero vector $x_n$ where it denotes solution to the null space, such that $Ax_n=0$...(a)
Now, it is also given that $b=sum_i=1^na_i$, which means a column vector of all 1,say $x_p$ a particular solution to this b, is such that $Ax_p=b$ ..(b)
From (a) and (b),if c is a constant,then $A(x_p+c.x_n)=b$ and hence this system seems to have infinite solutions.
Is my attempt correct?
linear-algebra
asked Jul 15 at 7:11
user3767495
857
asked Jul 15 at 7:11
user3767495
857
asked Jul 15 at 7:11
user3767495
857
asked Jul 15 at 7:11
user3767495
857
857
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
Yes, your attempt is correct.
There are infinitely many solutions. The matrix $A$ is singular and we have illustrated a particular solution, hence there must be infinitely many solutions.
answered Jul 15 at 7:34
Siong Thye Goh
77.8k134796
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Yes, your attempt is correct.
There are infinitely many solutions. The matrix $A$ is singular and we have illustrated a particular solution, hence there must be infinitely many solutions.
answered Jul 15 at 7:34
Siong Thye Goh
77.8k134796
add a comment |Â
up vote
2
down vote
Yes, your attempt is correct.
There are infinitely many solutions. The matrix $A$ is singular and we have illustrated a particular solution, hence there must be infinitely many solutions.
answered Jul 15 at 7:34
Siong Thye Goh
77.8k134796
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Yes, your attempt is correct.
There are infinitely many solutions. The matrix $A$ is singular and we have illustrated a particular solution, hence there must be infinitely many solutions.
answered Jul 15 at 7:34
Siong Thye Goh
77.8k134796
Yes, your attempt is correct.
There are infinitely many solutions. The matrix $A$ is singular and we have illustrated a particular solution, hence there must be infinitely many solutions.
answered Jul 15 at 7:34
Siong Thye Goh
77.8k134796
answered Jul 15 at 7:34
Siong Thye Goh
77.8k134796
answered Jul 15 at 7:34
Siong Thye Goh
77.8k134796
answered Jul 15 at 7:34
Siong Thye Goh
77.8k134796
77.8k134796
add a comment |Â
add a comment |Â
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