Mixing
How can I tell if a matrix is singular or nonsingular?
$begingroup$
When solving a problem about linearly independent/dependent, I don't know what the answer means.
Question: Determine whether the collection of vectors is linearly independent in $R^3$:
$(1,2,4)^T$ , $(2,1,3)^T$ , $(4,-1,1)^T$
Answer:
If $$c_1(1,2,4)^T+c_2(2,1,3)^T+c_3(4,-1,1)^T=(0,0,0)^T$$
Then,
$$begin{array}{rcl}
c_1+2c_2+4c_3=0\
2c_1+c_2-c_3=0\
4c_1+3c_2+c_3=0\
end{array}$$
The coefficient matrix of the system is singular and hence the system has nontrivial solutions. Therefore the vectors are linearly dependent.
I know the method to determine whether the matrix is linearly independent or not by computing $c_1v_1+c_2v_2+...c_nv_n=0$, but here I don't know why the system is singular, so how can I tell whether it's singular or not? Are there any ways to know?
linear-algebra vector-spaces
edited Jan 3 at 3:59
EuxhenH
482210
asked Jan 3 at 3:46
Shadow ZShadow Z
346
$endgroup$
add a comment |
$begingroup$
When solving a problem about linearly independent/dependent, I don't know what the answer means.
Question: Determine whether the collection of vectors is linearly independent in $R^3$:
$(1,2,4)^T$ , $(2,1,3)^T$ , $(4,-1,1)^T$
Answer:
If $$c_1(1,2,4)^T+c_2(2,1,3)^T+c_3(4,-1,1)^T=(0,0,0)^T$$
Then,
$$begin{array}{rcl}
c_1+2c_2+4c_3=0\
2c_1+c_2-c_3=0\
4c_1+3c_2+c_3=0\
end{array}$$
The coefficient matrix of the system is singular and hence the system has nontrivial solutions. Therefore the vectors are linearly dependent.
I know the method to determine whether the matrix is linearly independent or not by computing $c_1v_1+c_2v_2+...c_nv_n=0$, but here I don't know why the system is singular, so how can I tell whether it's singular or not? Are there any ways to know?
linear-algebra vector-spaces
edited Jan 3 at 3:59
EuxhenH
482210
asked Jan 3 at 3:46
Shadow ZShadow Z
346
$endgroup$
1
$begingroup$
Pretty much the whole point of determinants.
$endgroup$
– Randall
Jan 3 at 3:49
add a comment |
$begingroup$
When solving a problem about linearly independent/dependent, I don't know what the answer means.
Question: Determine whether the collection of vectors is linearly independent in $R^3$:
$(1,2,4)^T$ , $(2,1,3)^T$ , $(4,-1,1)^T$
Answer:
If $$c_1(1,2,4)^T+c_2(2,1,3)^T+c_3(4,-1,1)^T=(0,0,0)^T$$
Then,
$$begin{array}{rcl}
c_1+2c_2+4c_3=0\
2c_1+c_2-c_3=0\
4c_1+3c_2+c_3=0\
end{array}$$
The coefficient matrix of the system is singular and hence the system has nontrivial solutions. Therefore the vectors are linearly dependent.
I know the method to determine whether the matrix is linearly independent or not by computing $c_1v_1+c_2v_2+...c_nv_n=0$, but here I don't know why the system is singular, so how can I tell whether it's singular or not? Are there any ways to know?
linear-algebra vector-spaces
edited Jan 3 at 3:59
EuxhenH
482210
asked Jan 3 at 3:46
Shadow ZShadow Z
346
$endgroup$
When solving a problem about linearly independent/dependent, I don't know what the answer means.
Question: Determine whether the collection of vectors is linearly independent in $R^3$:
$(1,2,4)^T$ , $(2,1,3)^T$ , $(4,-1,1)^T$
Answer:
If $$c_1(1,2,4)^T+c_2(2,1,3)^T+c_3(4,-1,1)^T=(0,0,0)^T$$
Then,
$$begin{array}{rcl}
c_1+2c_2+4c_3=0\
2c_1+c_2-c_3=0\
4c_1+3c_2+c_3=0\
end{array}$$
The coefficient matrix of the system is singular and hence the system has nontrivial solutions. Therefore the vectors are linearly dependent.
I know the method to determine whether the matrix is linearly independent or not by computing $c_1v_1+c_2v_2+...c_nv_n=0$, but here I don't know why the system is singular, so how can I tell whether it's singular or not? Are there any ways to know?
linear-algebra vector-spaces
linear-algebra vector-spaces
edited Jan 3 at 3:59
EuxhenH
482210
asked Jan 3 at 3:46
Shadow ZShadow Z
346
edited Jan 3 at 3:59
EuxhenH
482210
asked Jan 3 at 3:46
Shadow ZShadow Z
346
edited Jan 3 at 3:59
EuxhenH
482210
edited Jan 3 at 3:59
EuxhenH
482210
edited Jan 3 at 3:59
EuxhenH
482210
482210
asked Jan 3 at 3:46
Shadow ZShadow Z
346
asked Jan 3 at 3:46
Shadow ZShadow Z
346
asked Jan 3 at 3:46
Shadow ZShadow Z
346
346
1
$begingroup$
Pretty much the whole point of determinants.
$endgroup$
– Randall
Jan 3 at 3:49
add a comment |
1
$begingroup$
Pretty much the whole point of determinants.
$endgroup$
– Randall
Jan 3 at 3:49
1
1
$begingroup$
Pretty much the whole point of determinants.
$endgroup$
– Randall
Jan 3 at 3:49
$begingroup$
Pretty much the whole point of determinants.
$endgroup$
– Randall
Jan 3 at 3:49
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If the determinant of the coefficient matrix is zero, then the matrix is singular and the system in dependent. The homogeneous system in this case has a non-zero solution as well as the trivial zero solution.
Otherwise the matrix is non-singular and the system has a unique solution which in case of homogeneous system is $(0,0,0)^T$
answered Jan 3 at 3:57
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.4k42061
$endgroup$
add a comment |
$begingroup$
There are many approaches that can be take (some have already been provided). One other is to apply Gaussian Elimination so that the matrix is in row echelon form (commonly denoted as $operatorname{REF}(A)$).
If the $operatorname{rank}left(operatorname{REF}(A)right) neq n$ then $A$ is not invertible and the system of equations is singular.
answered Jan 3 at 4:06
DavidGDavidG
2,047720
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If the determinant of the coefficient matrix is zero, then the matrix is singular and the system in dependent. The homogeneous system in this case has a non-zero solution as well as the trivial zero solution.
Otherwise the matrix is non-singular and the system has a unique solution which in case of homogeneous system is $(0,0,0)^T$
answered Jan 3 at 3:57
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.4k42061
$endgroup$
add a comment |
$begingroup$
If the determinant of the coefficient matrix is zero, then the matrix is singular and the system in dependent. The homogeneous system in this case has a non-zero solution as well as the trivial zero solution.
Otherwise the matrix is non-singular and the system has a unique solution which in case of homogeneous system is $(0,0,0)^T$
answered Jan 3 at 3:57
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.4k42061
$endgroup$
add a comment |
$begingroup$
If the determinant of the coefficient matrix is zero, then the matrix is singular and the system in dependent. The homogeneous system in this case has a non-zero solution as well as the trivial zero solution.
Otherwise the matrix is non-singular and the system has a unique solution which in case of homogeneous system is $(0,0,0)^T$
answered Jan 3 at 3:57
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.4k42061
$endgroup$
If the determinant of the coefficient matrix is zero, then the matrix is singular and the system in dependent. The homogeneous system in this case has a non-zero solution as well as the trivial zero solution.
Otherwise the matrix is non-singular and the system has a unique solution which in case of homogeneous system is $(0,0,0)^T$
answered Jan 3 at 3:57
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.4k42061
answered Jan 3 at 3:57
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.4k42061
answered Jan 3 at 3:57
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.4k42061
answered Jan 3 at 3:57
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.4k42061
41.4k42061
add a comment |
add a comment |
$begingroup$
There are many approaches that can be take (some have already been provided). One other is to apply Gaussian Elimination so that the matrix is in row echelon form (commonly denoted as $operatorname{REF}(A)$).
If the $operatorname{rank}left(operatorname{REF}(A)right) neq n$ then $A$ is not invertible and the system of equations is singular.
answered Jan 3 at 4:06
DavidGDavidG
2,047720
$endgroup$
add a comment |
$begingroup$
There are many approaches that can be take (some have already been provided). One other is to apply Gaussian Elimination so that the matrix is in row echelon form (commonly denoted as $operatorname{REF}(A)$).
If the $operatorname{rank}left(operatorname{REF}(A)right) neq n$ then $A$ is not invertible and the system of equations is singular.
answered Jan 3 at 4:06
DavidGDavidG
2,047720
$endgroup$
add a comment |
$begingroup$
There are many approaches that can be take (some have already been provided). One other is to apply Gaussian Elimination so that the matrix is in row echelon form (commonly denoted as $operatorname{REF}(A)$).
If the $operatorname{rank}left(operatorname{REF}(A)right) neq n$ then $A$ is not invertible and the system of equations is singular.
answered Jan 3 at 4:06
DavidGDavidG
2,047720
$endgroup$
There are many approaches that can be take (some have already been provided). One other is to apply Gaussian Elimination so that the matrix is in row echelon form (commonly denoted as $operatorname{REF}(A)$).
If the $operatorname{rank}left(operatorname{REF}(A)right) neq n$ then $A$ is not invertible and the system of equations is singular.
answered Jan 3 at 4:06
DavidGDavidG
2,047720
edited Jan 3 at 7:16
answered Jan 3 at 4:06
DavidGDavidG
2,047720
answered Jan 3 at 4:06
DavidGDavidG
2,047720
answered Jan 3 at 4:06
DavidGDavidG
2,047720
2,047720
add a comment |
add a comment |
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1
$begingroup$
Pretty much the whole point of determinants.
$endgroup$
– Randall
Jan 3 at 3:49