Non trivial solution of a system?
$begingroup$
So i wanted to know if this statement is true:
If there is a free variable in a row reduced matrix, does this imply that the system has a non trivial solution ?
E.g. (this excercise is done in order to calculate if λ=4 is an eigenvalue of the matrix A, but i am interested in something else in particular) By definition, 4 is an eigenvalue if and only if the system Ax = 4x has a non trivial solution, i.e., if and only if (A − 4Id)x = 0 has a non trivial solution.
A= $$
begin{pmatrix}
3 & 0 & -1 \
2 & 3 & 1 \
-3 & 4 & 5 \
end{pmatrix}
$$
I compute A-4Id, row reduce it and set it equal to 0 to get
A= $$
begin{pmatrix}
1 & 0& 1 & 0 \
0 & 1 & 1& 0 \
0 & 0 & 0&0 \
end{pmatrix}
$$
From here i can see that x3 is a free variable, and that
x1+x3=0
and x2+x3=0
Now on the answer sheet is states :
x3 is a free variable, and the system has a non-trivial solution.
I also get that one eigenvector corresponding to λ=4 is (-1,-1,0)
So practically my question is: on my excercise, where can i see that the system has a non trivial solution ? By seeing that x3 is a free variable or by stating that (-1,-1,0) is a solution ?
I have dyscalculia and get easily confused . Please help me answer my question in the simplest and clearest way possible ! Thank you ! Greatly appreciate everything
matrices eigenvalues-eigenvectors matrix-equations
asked Dec 11 '18 at 22:02
BM97BM97
758
$endgroup$
add a comment |
$begingroup$
So i wanted to know if this statement is true:
If there is a free variable in a row reduced matrix, does this imply that the system has a non trivial solution ?
E.g. (this excercise is done in order to calculate if λ=4 is an eigenvalue of the matrix A, but i am interested in something else in particular) By definition, 4 is an eigenvalue if and only if the system Ax = 4x has a non trivial solution, i.e., if and only if (A − 4Id)x = 0 has a non trivial solution.
A= $$
begin{pmatrix}
3 & 0 & -1 \
2 & 3 & 1 \
-3 & 4 & 5 \
end{pmatrix}
$$
I compute A-4Id, row reduce it and set it equal to 0 to get
A= $$
begin{pmatrix}
1 & 0& 1 & 0 \
0 & 1 & 1& 0 \
0 & 0 & 0&0 \
end{pmatrix}
$$
From here i can see that x3 is a free variable, and that
x1+x3=0
and x2+x3=0
Now on the answer sheet is states :
x3 is a free variable, and the system has a non-trivial solution.
I also get that one eigenvector corresponding to λ=4 is (-1,-1,0)
So practically my question is: on my excercise, where can i see that the system has a non trivial solution ? By seeing that x3 is a free variable or by stating that (-1,-1,0) is a solution ?
I have dyscalculia and get easily confused . Please help me answer my question in the simplest and clearest way possible ! Thank you ! Greatly appreciate everything
matrices eigenvalues-eigenvectors matrix-equations
asked Dec 11 '18 at 22:02
BM97BM97
758
$endgroup$
add a comment |
$begingroup$
So i wanted to know if this statement is true:
If there is a free variable in a row reduced matrix, does this imply that the system has a non trivial solution ?
E.g. (this excercise is done in order to calculate if λ=4 is an eigenvalue of the matrix A, but i am interested in something else in particular) By definition, 4 is an eigenvalue if and only if the system Ax = 4x has a non trivial solution, i.e., if and only if (A − 4Id)x = 0 has a non trivial solution.
A= $$
begin{pmatrix}
3 & 0 & -1 \
2 & 3 & 1 \
-3 & 4 & 5 \
end{pmatrix}
$$
I compute A-4Id, row reduce it and set it equal to 0 to get
A= $$
begin{pmatrix}
1 & 0& 1 & 0 \
0 & 1 & 1& 0 \
0 & 0 & 0&0 \
end{pmatrix}
$$
From here i can see that x3 is a free variable, and that
x1+x3=0
and x2+x3=0
Now on the answer sheet is states :
x3 is a free variable, and the system has a non-trivial solution.
I also get that one eigenvector corresponding to λ=4 is (-1,-1,0)
So practically my question is: on my excercise, where can i see that the system has a non trivial solution ? By seeing that x3 is a free variable or by stating that (-1,-1,0) is a solution ?
I have dyscalculia and get easily confused . Please help me answer my question in the simplest and clearest way possible ! Thank you ! Greatly appreciate everything
matrices eigenvalues-eigenvectors matrix-equations
asked Dec 11 '18 at 22:02
BM97BM97
758
$endgroup$
So i wanted to know if this statement is true:
If there is a free variable in a row reduced matrix, does this imply that the system has a non trivial solution ?
E.g. (this excercise is done in order to calculate if λ=4 is an eigenvalue of the matrix A, but i am interested in something else in particular) By definition, 4 is an eigenvalue if and only if the system Ax = 4x has a non trivial solution, i.e., if and only if (A − 4Id)x = 0 has a non trivial solution.
A= $$
begin{pmatrix}
3 & 0 & -1 \
2 & 3 & 1 \
-3 & 4 & 5 \
end{pmatrix}
$$
I compute A-4Id, row reduce it and set it equal to 0 to get
A= $$
begin{pmatrix}
1 & 0& 1 & 0 \
0 & 1 & 1& 0 \
0 & 0 & 0&0 \
end{pmatrix}
$$
From here i can see that x3 is a free variable, and that
x1+x3=0
and x2+x3=0
Now on the answer sheet is states :
x3 is a free variable, and the system has a non-trivial solution.
I also get that one eigenvector corresponding to λ=4 is (-1,-1,0)
So practically my question is: on my excercise, where can i see that the system has a non trivial solution ? By seeing that x3 is a free variable or by stating that (-1,-1,0) is a solution ?
I have dyscalculia and get easily confused . Please help me answer my question in the simplest and clearest way possible ! Thank you ! Greatly appreciate everything
matrices eigenvalues-eigenvectors matrix-equations
matrices eigenvalues-eigenvectors matrix-equations
asked Dec 11 '18 at 22:02
BM97BM97
758
asked Dec 11 '18 at 22:02
BM97BM97
758
asked Dec 11 '18 at 22:02
BM97BM97
758
asked Dec 11 '18 at 22:02
BM97BM97
758
asked Dec 11 '18 at 22:02
BM97BM97
758
758
add a comment |
add a comment |
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