Mixing
Find all values of a for which the system has trivial solutions
$begingroup$
$$begin{pmatrix}1+a & 1 & 1 & 1\ 1 & 1-a & 1 & 1\1 & 1 & 1 & a\1 & 2 & 2 & 1\end{pmatrix}$$
As the title says, find all values of a for which the system has trivial solutions.
I have tried row reduction but I haven't gotten anywhere with it. So what I have done is calculate the determinant, which is:
2 a^3 - a
From what I understand system has trivial solutions if and only if det(A) != 0.
So the solution for this problem would be every ${rm I!R}$ number except where det(A) == 0 ?
matrices
asked Feb 3 at 1:21
IronCanTacoIronCanTaco
134
$endgroup$
add a comment |
$begingroup$
$$begin{pmatrix}1+a & 1 & 1 & 1\ 1 & 1-a & 1 & 1\1 & 1 & 1 & a\1 & 2 & 2 & 1\end{pmatrix}$$
As the title says, find all values of a for which the system has trivial solutions.
I have tried row reduction but I haven't gotten anywhere with it. So what I have done is calculate the determinant, which is:
2 a^3 - a
From what I understand system has trivial solutions if and only if det(A) != 0.
So the solution for this problem would be every ${rm I!R}$ number except where det(A) == 0 ?
matrices
asked Feb 3 at 1:21
IronCanTacoIronCanTaco
134
$endgroup$
1
$begingroup$
Your reasoning seems correct.
$endgroup$
– Hanul Jeon
Feb 3 at 1:29
$begingroup$
What do you mean by “trivial solutions?” That usually means that the zero vector is a solution, which is true for every homogeneous system. Perhaps you meant “unique” instead of “trivial?”
$endgroup$
– amd
Feb 3 at 2:08
add a comment |
$begingroup$
$$begin{pmatrix}1+a & 1 & 1 & 1\ 1 & 1-a & 1 & 1\1 & 1 & 1 & a\1 & 2 & 2 & 1\end{pmatrix}$$
As the title says, find all values of a for which the system has trivial solutions.
I have tried row reduction but I haven't gotten anywhere with it. So what I have done is calculate the determinant, which is:
2 a^3 - a
From what I understand system has trivial solutions if and only if det(A) != 0.
So the solution for this problem would be every ${rm I!R}$ number except where det(A) == 0 ?
matrices
asked Feb 3 at 1:21
IronCanTacoIronCanTaco
134
$endgroup$
$$begin{pmatrix}1+a & 1 & 1 & 1\ 1 & 1-a & 1 & 1\1 & 1 & 1 & a\1 & 2 & 2 & 1\end{pmatrix}$$
As the title says, find all values of a for which the system has trivial solutions.
I have tried row reduction but I haven't gotten anywhere with it. So what I have done is calculate the determinant, which is:
2 a^3 - a
From what I understand system has trivial solutions if and only if det(A) != 0.
So the solution for this problem would be every ${rm I!R}$ number except where det(A) == 0 ?
matrices
matrices
asked Feb 3 at 1:21
IronCanTacoIronCanTaco
134
asked Feb 3 at 1:21
IronCanTacoIronCanTaco
134
asked Feb 3 at 1:21
IronCanTacoIronCanTaco
134
asked Feb 3 at 1:21
IronCanTacoIronCanTaco
134
asked Feb 3 at 1:21
IronCanTacoIronCanTaco
134
134
1
$begingroup$
Your reasoning seems correct.
$endgroup$
– Hanul Jeon
Feb 3 at 1:29
$begingroup$
What do you mean by “trivial solutions?” That usually means that the zero vector is a solution, which is true for every homogeneous system. Perhaps you meant “unique” instead of “trivial?”
$endgroup$
– amd
Feb 3 at 2:08
add a comment |
1
$begingroup$
Your reasoning seems correct.
$endgroup$
– Hanul Jeon
Feb 3 at 1:29
$begingroup$
What do you mean by “trivial solutions?” That usually means that the zero vector is a solution, which is true for every homogeneous system. Perhaps you meant “unique” instead of “trivial?”
$endgroup$
– amd
Feb 3 at 2:08
1
1
$begingroup$
Your reasoning seems correct.
$endgroup$
– Hanul Jeon
Feb 3 at 1:29
$begingroup$
Your reasoning seems correct.
$endgroup$
– Hanul Jeon
Feb 3 at 1:29
$begingroup$
What do you mean by “trivial solutions?” That usually means that the zero vector is a solution, which is true for every homogeneous system. Perhaps you meant “unique” instead of “trivial?”
$endgroup$
– amd
Feb 3 at 2:08
$begingroup$
What do you mean by “trivial solutions?” That usually means that the zero vector is a solution, which is true for every homogeneous system. Perhaps you meant “unique” instead of “trivial?”
$endgroup$
– amd
Feb 3 at 2:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your determinant is correct, so we get some values to check
$$det A = a left(2 a^2-1right) implies a = 0, pm dfrac{1}{sqrt{2}}$$
If $a = 0$, the RREF is
$$begin{bmatrix}
1 & 0 & 0 & 0 \
0 & 1 & 1 & 0 \
0 & 0 & 0 & 1 \
0 & 0 & 0 & 0 \
end{bmatrix}$$
If $a = dfrac{1}{sqrt{2}}$, the RREF is
$$begin{bmatrix}
1 & 0 & 0 & sqrt{2}-1 \
0 & 1 & 0 & 1-sqrt{2} \
0 & 0 & 1 & frac{1}{sqrt{2}} \
0 & 0 & 0 & 0 \
end{bmatrix}$$
If $a = -dfrac{1}{sqrt{2}}$, the RREF is
$$begin{bmatrix}
1 & 0 & 0 & -sqrt{2}-1 \
0 & 1 & 0 & sqrt{2}+1 \
0 & 0 & 1 & -frac{1}{sqrt{2}} \
0 & 0 & 0 & 0 \
end{bmatrix}$$
If $a ne 0, a ne pm dfrac{1}{sqrt{2}}$, the RREF is
$$begin{bmatrix}
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 \
end{bmatrix}$$
Hopefully you can complete the question now.
answered Feb 3 at 6:26
MooMoo
5,64031020
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
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oldest
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votes
$begingroup$
Your determinant is correct, so we get some values to check
$$det A = a left(2 a^2-1right) implies a = 0, pm dfrac{1}{sqrt{2}}$$
If $a = 0$, the RREF is
$$begin{bmatrix}
1 & 0 & 0 & 0 \
0 & 1 & 1 & 0 \
0 & 0 & 0 & 1 \
0 & 0 & 0 & 0 \
end{bmatrix}$$
If $a = dfrac{1}{sqrt{2}}$, the RREF is
$$begin{bmatrix}
1 & 0 & 0 & sqrt{2}-1 \
0 & 1 & 0 & 1-sqrt{2} \
0 & 0 & 1 & frac{1}{sqrt{2}} \
0 & 0 & 0 & 0 \
end{bmatrix}$$
If $a = -dfrac{1}{sqrt{2}}$, the RREF is
$$begin{bmatrix}
1 & 0 & 0 & -sqrt{2}-1 \
0 & 1 & 0 & sqrt{2}+1 \
0 & 0 & 1 & -frac{1}{sqrt{2}} \
0 & 0 & 0 & 0 \
end{bmatrix}$$
If $a ne 0, a ne pm dfrac{1}{sqrt{2}}$, the RREF is
$$begin{bmatrix}
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 \
end{bmatrix}$$
Hopefully you can complete the question now.
answered Feb 3 at 6:26
MooMoo
5,64031020
$endgroup$
add a comment |
$begingroup$
Your determinant is correct, so we get some values to check
$$det A = a left(2 a^2-1right) implies a = 0, pm dfrac{1}{sqrt{2}}$$
If $a = 0$, the RREF is
$$begin{bmatrix}
1 & 0 & 0 & 0 \
0 & 1 & 1 & 0 \
0 & 0 & 0 & 1 \
0 & 0 & 0 & 0 \
end{bmatrix}$$
If $a = dfrac{1}{sqrt{2}}$, the RREF is
$$begin{bmatrix}
1 & 0 & 0 & sqrt{2}-1 \
0 & 1 & 0 & 1-sqrt{2} \
0 & 0 & 1 & frac{1}{sqrt{2}} \
0 & 0 & 0 & 0 \
end{bmatrix}$$
If $a = -dfrac{1}{sqrt{2}}$, the RREF is
$$begin{bmatrix}
1 & 0 & 0 & -sqrt{2}-1 \
0 & 1 & 0 & sqrt{2}+1 \
0 & 0 & 1 & -frac{1}{sqrt{2}} \
0 & 0 & 0 & 0 \
end{bmatrix}$$
If $a ne 0, a ne pm dfrac{1}{sqrt{2}}$, the RREF is
$$begin{bmatrix}
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 \
end{bmatrix}$$
Hopefully you can complete the question now.
answered Feb 3 at 6:26
MooMoo
5,64031020
$endgroup$
add a comment |
$begingroup$
Your determinant is correct, so we get some values to check
$$det A = a left(2 a^2-1right) implies a = 0, pm dfrac{1}{sqrt{2}}$$
If $a = 0$, the RREF is
$$begin{bmatrix}
1 & 0 & 0 & 0 \
0 & 1 & 1 & 0 \
0 & 0 & 0 & 1 \
0 & 0 & 0 & 0 \
end{bmatrix}$$
If $a = dfrac{1}{sqrt{2}}$, the RREF is
$$begin{bmatrix}
1 & 0 & 0 & sqrt{2}-1 \
0 & 1 & 0 & 1-sqrt{2} \
0 & 0 & 1 & frac{1}{sqrt{2}} \
0 & 0 & 0 & 0 \
end{bmatrix}$$
If $a = -dfrac{1}{sqrt{2}}$, the RREF is
$$begin{bmatrix}
1 & 0 & 0 & -sqrt{2}-1 \
0 & 1 & 0 & sqrt{2}+1 \
0 & 0 & 1 & -frac{1}{sqrt{2}} \
0 & 0 & 0 & 0 \
end{bmatrix}$$
If $a ne 0, a ne pm dfrac{1}{sqrt{2}}$, the RREF is
$$begin{bmatrix}
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 \
end{bmatrix}$$
Hopefully you can complete the question now.
answered Feb 3 at 6:26
MooMoo
5,64031020
$endgroup$
Your determinant is correct, so we get some values to check
$$det A = a left(2 a^2-1right) implies a = 0, pm dfrac{1}{sqrt{2}}$$
If $a = 0$, the RREF is
$$begin{bmatrix}
1 & 0 & 0 & 0 \
0 & 1 & 1 & 0 \
0 & 0 & 0 & 1 \
0 & 0 & 0 & 0 \
end{bmatrix}$$
If $a = dfrac{1}{sqrt{2}}$, the RREF is
$$begin{bmatrix}
1 & 0 & 0 & sqrt{2}-1 \
0 & 1 & 0 & 1-sqrt{2} \
0 & 0 & 1 & frac{1}{sqrt{2}} \
0 & 0 & 0 & 0 \
end{bmatrix}$$
If $a = -dfrac{1}{sqrt{2}}$, the RREF is
$$begin{bmatrix}
1 & 0 & 0 & -sqrt{2}-1 \
0 & 1 & 0 & sqrt{2}+1 \
0 & 0 & 1 & -frac{1}{sqrt{2}} \
0 & 0 & 0 & 0 \
end{bmatrix}$$
If $a ne 0, a ne pm dfrac{1}{sqrt{2}}$, the RREF is
$$begin{bmatrix}
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 \
end{bmatrix}$$
Hopefully you can complete the question now.
answered Feb 3 at 6:26
MooMoo
5,64031020
answered Feb 3 at 6:26
MooMoo
5,64031020
answered Feb 3 at 6:26
MooMoo
5,64031020
answered Feb 3 at 6:26
MooMoo
5,64031020
5,64031020
add a comment |
add a comment |
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1
$begingroup$
Your reasoning seems correct.
$endgroup$
– Hanul Jeon
Feb 3 at 1:29
$begingroup$
What do you mean by “trivial solutions?” That usually means that the zero vector is a solution, which is true for every homogeneous system. Perhaps you meant “unique” instead of “trivial?”
$endgroup$
– amd
Feb 3 at 2:08