Find trivial and non-trivial solutions to a system with a parameter
up vote
1
down vote
favorite
begin{cases} x + (2p−1)y−pz = 0 \
(2−p)x + y−z = 0 \
x + py−z = 0 end{cases}
- For which values of $p$, system has trivial solutions?
- For which values of $p$, system has nontrivial solutions? Write them
Now I have come to $p_1=1, p_2=1, p_3=1/2$ with Sarrus rule. Now how do I get trivial and nontrivial solutions ? I am just looking for instructions
linear-algebra systems-of-equations
edited Nov 22 at 17:46
Lorenzo B.
1,8322520
asked Nov 22 at 17:23
Miko
103
add a comment |
up vote
1
down vote
favorite
begin{cases} x + (2p−1)y−pz = 0 \
(2−p)x + y−z = 0 \
x + py−z = 0 end{cases}
- For which values of $p$, system has trivial solutions?
- For which values of $p$, system has nontrivial solutions? Write them
Now I have come to $p_1=1, p_2=1, p_3=1/2$ with Sarrus rule. Now how do I get trivial and nontrivial solutions ? I am just looking for instructions
linear-algebra systems-of-equations
edited Nov 22 at 17:46
Lorenzo B.
1,8322520
asked Nov 22 at 17:23
Miko
103
Well if $x=y=z = 0$ then any value of $p$ will have trivial solutions. It the equations are linear dependent then those are the only solutions. But if they aren't linearly independent they can have non-trivial solutions. Which can only occur if the determinate is $0$. So this question boils down to nothing more or less than: For what values of $p$ will the determinate be $0$. (Note: there is nothing in the question about finding the solutions).
– fleablood
Nov 22 at 18:03
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
begin{cases} x + (2p−1)y−pz = 0 \
(2−p)x + y−z = 0 \
x + py−z = 0 end{cases}
- For which values of $p$, system has trivial solutions?
- For which values of $p$, system has nontrivial solutions? Write them
Now I have come to $p_1=1, p_2=1, p_3=1/2$ with Sarrus rule. Now how do I get trivial and nontrivial solutions ? I am just looking for instructions
linear-algebra systems-of-equations
edited Nov 22 at 17:46
Lorenzo B.
1,8322520
asked Nov 22 at 17:23
Miko
103
begin{cases} x + (2p−1)y−pz = 0 \
(2−p)x + y−z = 0 \
x + py−z = 0 end{cases}
- For which values of $p$, system has trivial solutions?
- For which values of $p$, system has nontrivial solutions? Write them
Now I have come to $p_1=1, p_2=1, p_3=1/2$ with Sarrus rule. Now how do I get trivial and nontrivial solutions ? I am just looking for instructions
linear-algebra systems-of-equations
linear-algebra systems-of-equations
edited Nov 22 at 17:46
Lorenzo B.
1,8322520
asked Nov 22 at 17:23
Miko
103
edited Nov 22 at 17:46
Lorenzo B.
1,8322520
asked Nov 22 at 17:23
Miko
103
edited Nov 22 at 17:46
Lorenzo B.
1,8322520
edited Nov 22 at 17:46
Lorenzo B.
1,8322520
edited Nov 22 at 17:46
Lorenzo B.
1,8322520
1,8322520
asked Nov 22 at 17:23
Miko
103
asked Nov 22 at 17:23
Miko
103
asked Nov 22 at 17:23
Miko
103
103
Well if $x=y=z = 0$ then any value of $p$ will have trivial solutions. It the equations are linear dependent then those are the only solutions. But if they aren't linearly independent they can have non-trivial solutions. Which can only occur if the determinate is $0$. So this question boils down to nothing more or less than: For what values of $p$ will the determinate be $0$. (Note: there is nothing in the question about finding the solutions).
– fleablood
Nov 22 at 18:03
add a comment |
Well if $x=y=z = 0$ then any value of $p$ will have trivial solutions. It the equations are linear dependent then those are the only solutions. But if they aren't linearly independent they can have non-trivial solutions. Which can only occur if the determinate is $0$. So this question boils down to nothing more or less than: For what values of $p$ will the determinate be $0$. (Note: there is nothing in the question about finding the solutions).
– fleablood
Nov 22 at 18:03
Well if $x=y=z = 0$ then any value of $p$ will have trivial solutions. It the equations are linear dependent then those are the only solutions. But if they aren't linearly independent they can have non-trivial solutions. Which can only occur if the determinate is $0$. So this question boils down to nothing more or less than: For what values of $p$ will the determinate be $0$. (Note: there is nothing in the question about finding the solutions).
– fleablood
Nov 22 at 18:03
Well if $x=y=z = 0$ then any value of $p$ will have trivial solutions. It the equations are linear dependent then those are the only solutions. But if they aren't linearly independent they can have non-trivial solutions. Which can only occur if the determinate is $0$. So this question boils down to nothing more or less than: For what values of $p$ will the determinate be $0$. (Note: there is nothing in the question about finding the solutions).
– fleablood
Nov 22 at 18:03
add a comment |
3 Answers
3
active
oldest
votes
up vote
0
down vote
accepted
Well if $x=y=z = 0$ then any value of $p$ will have trivial solutions.
So the answer to the first is: For all values of $p$ there will be trivial solutions.
It the equations are linear dependent then those are the only solutions. But if they aren't linearly independent they can have non-trivial solutions. Which can only occur if the determinate is $0$. So this question boils down to nothing more or less than: For what values of $p$ will the determinate be $0$. (Note: there is nothing in the question about finding the solutions).
Using rule of sarrus the determinate is:
$-1 - (2p-1)-p^2(2-p) +p + p +(2-p)(2p-1) =$
$ p^3 - 4p^2 +5p-2= (p-1)^2(p-2)$
Which if we set to $0$ and solve... we get
$p =1$ or $p =2$.
And $p = 1$ we get the system:
$x +y - z = 0; x+y-z = 0; x+y-z =0$ or $z= x+y$ which has infinitely many solutions.
And if $p=2$ we get the system:
$x+3y -2z =0; y-z = 0; x+2y - z=0$ or $y=z=-x$ which has infinitely many solutions.
Bear in mind $begin{cases} a_1x + a_2y+a_3z = 0 \
b_1x + b_2y+b_3z = 0 \
c_1x + c_2y+c_3z = 0 end{cases}$
Will either have exactly one trivial solution $x=y=z =0$ if the equation or independent. But if the equations are not independent there will always be an infinite number of solutions.
answered Nov 22 at 18:50
fleablood
67.6k22684
add a comment |
up vote
0
down vote
Hint :
Apply Gauss-Elimination to :
$$left(begin{array}{ccc|c} 1 & 2p-1 & -p &0 \ 2-p & 1 & -1 & 0 \ 1 & p & -1 & 0end{array}right)$$
answered Nov 22 at 17:39
Rebellos
13.9k31243
Is that necessary? I used Sarrus rule to avoid Gauss-Elimination as it can get complicated with those p
– Miko
Nov 22 at 17:45
@Miko By applying Gauss-Elimination, you will yield an expression involving $p$ that you will have to investigate in order to answer to your exercise. This is the most straight-forward and classic way. I don't understand what those $p_i$s are ? The Rule of Sarrus is a memorisation technique involving determinants, certainly not the math tool to use here.
– Rebellos
Nov 22 at 17:51
add a comment |
up vote
0
down vote
Hint:
This system has non-trivial solutions if and only if its determinant is zero . This determinant can be calculated by row and column operations:
begin{align}
begin{vmatrix}
1&2p-1&-p \ 2-p&1&-1 \ 1&p&-1
end{vmatrix} &=
begin{vmatrix}
1&2p-1&-p \ 0&1+(p-2)(2p-1)&-1-p(p-2) \ 0&1-p&p-1
end{vmatrix}
=begin{vmatrix}
1&p-1&-p \ 0&(p-2)(p-1)&-1-p(p-2) \ 0&0&p-1
end{vmatrix}\[1ex]& = (p-1)begin{vmatrix}
1& 1&-p \ 0&p-2 &-1-p(p-2) \0&0&p-1
end{vmatrix} = (p-1)begin{vmatrix}
p-2 &-1-p(p-2) \ 0&p-1
end{vmatrix} \[1ex]
&=(p-1)^2(p-2).
end{align}
answered Nov 22 at 20:53
Bernard
117k637109
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009379%2ffind-trivial-and-non-trivial-solutions-to-a-system-with-a-parameter%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Well if $x=y=z = 0$ then any value of $p$ will have trivial solutions.
So the answer to the first is: For all values of $p$ there will be trivial solutions.
It the equations are linear dependent then those are the only solutions. But if they aren't linearly independent they can have non-trivial solutions. Which can only occur if the determinate is $0$. So this question boils down to nothing more or less than: For what values of $p$ will the determinate be $0$. (Note: there is nothing in the question about finding the solutions).
Using rule of sarrus the determinate is:
$-1 - (2p-1)-p^2(2-p) +p + p +(2-p)(2p-1) =$
$ p^3 - 4p^2 +5p-2= (p-1)^2(p-2)$
Which if we set to $0$ and solve... we get
$p =1$ or $p =2$.
And $p = 1$ we get the system:
$x +y - z = 0; x+y-z = 0; x+y-z =0$ or $z= x+y$ which has infinitely many solutions.
And if $p=2$ we get the system:
$x+3y -2z =0; y-z = 0; x+2y - z=0$ or $y=z=-x$ which has infinitely many solutions.
Bear in mind $begin{cases} a_1x + a_2y+a_3z = 0 \
b_1x + b_2y+b_3z = 0 \
c_1x + c_2y+c_3z = 0 end{cases}$
Will either have exactly one trivial solution $x=y=z =0$ if the equation or independent. But if the equations are not independent there will always be an infinite number of solutions.
answered Nov 22 at 18:50
fleablood
67.6k22684
add a comment |
up vote
0
down vote
accepted
Well if $x=y=z = 0$ then any value of $p$ will have trivial solutions.
So the answer to the first is: For all values of $p$ there will be trivial solutions.
It the equations are linear dependent then those are the only solutions. But if they aren't linearly independent they can have non-trivial solutions. Which can only occur if the determinate is $0$. So this question boils down to nothing more or less than: For what values of $p$ will the determinate be $0$. (Note: there is nothing in the question about finding the solutions).
Using rule of sarrus the determinate is:
$-1 - (2p-1)-p^2(2-p) +p + p +(2-p)(2p-1) =$
$ p^3 - 4p^2 +5p-2= (p-1)^2(p-2)$
Which if we set to $0$ and solve... we get
$p =1$ or $p =2$.
And $p = 1$ we get the system:
$x +y - z = 0; x+y-z = 0; x+y-z =0$ or $z= x+y$ which has infinitely many solutions.
And if $p=2$ we get the system:
$x+3y -2z =0; y-z = 0; x+2y - z=0$ or $y=z=-x$ which has infinitely many solutions.
Bear in mind $begin{cases} a_1x + a_2y+a_3z = 0 \
b_1x + b_2y+b_3z = 0 \
c_1x + c_2y+c_3z = 0 end{cases}$
Will either have exactly one trivial solution $x=y=z =0$ if the equation or independent. But if the equations are not independent there will always be an infinite number of solutions.
answered Nov 22 at 18:50
fleablood
67.6k22684
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Well if $x=y=z = 0$ then any value of $p$ will have trivial solutions.
So the answer to the first is: For all values of $p$ there will be trivial solutions.
It the equations are linear dependent then those are the only solutions. But if they aren't linearly independent they can have non-trivial solutions. Which can only occur if the determinate is $0$. So this question boils down to nothing more or less than: For what values of $p$ will the determinate be $0$. (Note: there is nothing in the question about finding the solutions).
Using rule of sarrus the determinate is:
$-1 - (2p-1)-p^2(2-p) +p + p +(2-p)(2p-1) =$
$ p^3 - 4p^2 +5p-2= (p-1)^2(p-2)$
Which if we set to $0$ and solve... we get
$p =1$ or $p =2$.
And $p = 1$ we get the system:
$x +y - z = 0; x+y-z = 0; x+y-z =0$ or $z= x+y$ which has infinitely many solutions.
And if $p=2$ we get the system:
$x+3y -2z =0; y-z = 0; x+2y - z=0$ or $y=z=-x$ which has infinitely many solutions.
Bear in mind $begin{cases} a_1x + a_2y+a_3z = 0 \
b_1x + b_2y+b_3z = 0 \
c_1x + c_2y+c_3z = 0 end{cases}$
Will either have exactly one trivial solution $x=y=z =0$ if the equation or independent. But if the equations are not independent there will always be an infinite number of solutions.
answered Nov 22 at 18:50
fleablood
67.6k22684
Well if $x=y=z = 0$ then any value of $p$ will have trivial solutions.
So the answer to the first is: For all values of $p$ there will be trivial solutions.
It the equations are linear dependent then those are the only solutions. But if they aren't linearly independent they can have non-trivial solutions. Which can only occur if the determinate is $0$. So this question boils down to nothing more or less than: For what values of $p$ will the determinate be $0$. (Note: there is nothing in the question about finding the solutions).
Using rule of sarrus the determinate is:
$-1 - (2p-1)-p^2(2-p) +p + p +(2-p)(2p-1) =$
$ p^3 - 4p^2 +5p-2= (p-1)^2(p-2)$
Which if we set to $0$ and solve... we get
$p =1$ or $p =2$.
And $p = 1$ we get the system:
$x +y - z = 0; x+y-z = 0; x+y-z =0$ or $z= x+y$ which has infinitely many solutions.
And if $p=2$ we get the system:
$x+3y -2z =0; y-z = 0; x+2y - z=0$ or $y=z=-x$ which has infinitely many solutions.
Bear in mind $begin{cases} a_1x + a_2y+a_3z = 0 \
b_1x + b_2y+b_3z = 0 \
c_1x + c_2y+c_3z = 0 end{cases}$
Will either have exactly one trivial solution $x=y=z =0$ if the equation or independent. But if the equations are not independent there will always be an infinite number of solutions.
answered Nov 22 at 18:50
fleablood
67.6k22684
edited Nov 22 at 19:11
answered Nov 22 at 18:50
fleablood
67.6k22684
answered Nov 22 at 18:50
fleablood
67.6k22684
answered Nov 22 at 18:50
fleablood
67.6k22684
67.6k22684
add a comment |
add a comment |
up vote
0
down vote
Hint :
Apply Gauss-Elimination to :
$$left(begin{array}{ccc|c} 1 & 2p-1 & -p &0 \ 2-p & 1 & -1 & 0 \ 1 & p & -1 & 0end{array}right)$$
answered Nov 22 at 17:39
Rebellos
13.9k31243
Is that necessary? I used Sarrus rule to avoid Gauss-Elimination as it can get complicated with those p
– Miko
Nov 22 at 17:45
@Miko By applying Gauss-Elimination, you will yield an expression involving $p$ that you will have to investigate in order to answer to your exercise. This is the most straight-forward and classic way. I don't understand what those $p_i$s are ? The Rule of Sarrus is a memorisation technique involving determinants, certainly not the math tool to use here.
– Rebellos
Nov 22 at 17:51
add a comment |
up vote
0
down vote
Hint :
Apply Gauss-Elimination to :
$$left(begin{array}{ccc|c} 1 & 2p-1 & -p &0 \ 2-p & 1 & -1 & 0 \ 1 & p & -1 & 0end{array}right)$$
answered Nov 22 at 17:39
Rebellos
13.9k31243
Is that necessary? I used Sarrus rule to avoid Gauss-Elimination as it can get complicated with those p
– Miko
Nov 22 at 17:45
@Miko By applying Gauss-Elimination, you will yield an expression involving $p$ that you will have to investigate in order to answer to your exercise. This is the most straight-forward and classic way. I don't understand what those $p_i$s are ? The Rule of Sarrus is a memorisation technique involving determinants, certainly not the math tool to use here.
– Rebellos
Nov 22 at 17:51
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint :
Apply Gauss-Elimination to :
$$left(begin{array}{ccc|c} 1 & 2p-1 & -p &0 \ 2-p & 1 & -1 & 0 \ 1 & p & -1 & 0end{array}right)$$
answered Nov 22 at 17:39
Rebellos
13.9k31243
Hint :
Apply Gauss-Elimination to :
$$left(begin{array}{ccc|c} 1 & 2p-1 & -p &0 \ 2-p & 1 & -1 & 0 \ 1 & p & -1 & 0end{array}right)$$
answered Nov 22 at 17:39
Rebellos
13.9k31243
answered Nov 22 at 17:39
Rebellos
13.9k31243
answered Nov 22 at 17:39
Rebellos
13.9k31243
answered Nov 22 at 17:39
Rebellos
13.9k31243
13.9k31243
Is that necessary? I used Sarrus rule to avoid Gauss-Elimination as it can get complicated with those p
– Miko
Nov 22 at 17:45
@Miko By applying Gauss-Elimination, you will yield an expression involving $p$ that you will have to investigate in order to answer to your exercise. This is the most straight-forward and classic way. I don't understand what those $p_i$s are ? The Rule of Sarrus is a memorisation technique involving determinants, certainly not the math tool to use here.
– Rebellos
Nov 22 at 17:51
add a comment |
Is that necessary? I used Sarrus rule to avoid Gauss-Elimination as it can get complicated with those p
– Miko
Nov 22 at 17:45
@Miko By applying Gauss-Elimination, you will yield an expression involving $p$ that you will have to investigate in order to answer to your exercise. This is the most straight-forward and classic way. I don't understand what those $p_i$s are ? The Rule of Sarrus is a memorisation technique involving determinants, certainly not the math tool to use here.
– Rebellos
Nov 22 at 17:51
Is that necessary? I used Sarrus rule to avoid Gauss-Elimination as it can get complicated with those p
– Miko
Nov 22 at 17:45
Is that necessary? I used Sarrus rule to avoid Gauss-Elimination as it can get complicated with those p
– Miko
Nov 22 at 17:45
@Miko By applying Gauss-Elimination, you will yield an expression involving $p$ that you will have to investigate in order to answer to your exercise. This is the most straight-forward and classic way. I don't understand what those $p_i$s are ? The Rule of Sarrus is a memorisation technique involving determinants, certainly not the math tool to use here.
– Rebellos
Nov 22 at 17:51
@Miko By applying Gauss-Elimination, you will yield an expression involving $p$ that you will have to investigate in order to answer to your exercise. This is the most straight-forward and classic way. I don't understand what those $p_i$s are ? The Rule of Sarrus is a memorisation technique involving determinants, certainly not the math tool to use here.
– Rebellos
Nov 22 at 17:51
add a comment |
up vote
0
down vote
Hint:
This system has non-trivial solutions if and only if its determinant is zero . This determinant can be calculated by row and column operations:
begin{align}
begin{vmatrix}
1&2p-1&-p \ 2-p&1&-1 \ 1&p&-1
end{vmatrix} &=
begin{vmatrix}
1&2p-1&-p \ 0&1+(p-2)(2p-1)&-1-p(p-2) \ 0&1-p&p-1
end{vmatrix}
=begin{vmatrix}
1&p-1&-p \ 0&(p-2)(p-1)&-1-p(p-2) \ 0&0&p-1
end{vmatrix}\[1ex]& = (p-1)begin{vmatrix}
1& 1&-p \ 0&p-2 &-1-p(p-2) \0&0&p-1
end{vmatrix} = (p-1)begin{vmatrix}
p-2 &-1-p(p-2) \ 0&p-1
end{vmatrix} \[1ex]
&=(p-1)^2(p-2).
end{align}
answered Nov 22 at 20:53
Bernard
117k637109
add a comment |
up vote
0
down vote
Hint:
This system has non-trivial solutions if and only if its determinant is zero . This determinant can be calculated by row and column operations:
begin{align}
begin{vmatrix}
1&2p-1&-p \ 2-p&1&-1 \ 1&p&-1
end{vmatrix} &=
begin{vmatrix}
1&2p-1&-p \ 0&1+(p-2)(2p-1)&-1-p(p-2) \ 0&1-p&p-1
end{vmatrix}
=begin{vmatrix}
1&p-1&-p \ 0&(p-2)(p-1)&-1-p(p-2) \ 0&0&p-1
end{vmatrix}\[1ex]& = (p-1)begin{vmatrix}
1& 1&-p \ 0&p-2 &-1-p(p-2) \0&0&p-1
end{vmatrix} = (p-1)begin{vmatrix}
p-2 &-1-p(p-2) \ 0&p-1
end{vmatrix} \[1ex]
&=(p-1)^2(p-2).
end{align}
answered Nov 22 at 20:53
Bernard
117k637109
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint:
This system has non-trivial solutions if and only if its determinant is zero . This determinant can be calculated by row and column operations:
begin{align}
begin{vmatrix}
1&2p-1&-p \ 2-p&1&-1 \ 1&p&-1
end{vmatrix} &=
begin{vmatrix}
1&2p-1&-p \ 0&1+(p-2)(2p-1)&-1-p(p-2) \ 0&1-p&p-1
end{vmatrix}
=begin{vmatrix}
1&p-1&-p \ 0&(p-2)(p-1)&-1-p(p-2) \ 0&0&p-1
end{vmatrix}\[1ex]& = (p-1)begin{vmatrix}
1& 1&-p \ 0&p-2 &-1-p(p-2) \0&0&p-1
end{vmatrix} = (p-1)begin{vmatrix}
p-2 &-1-p(p-2) \ 0&p-1
end{vmatrix} \[1ex]
&=(p-1)^2(p-2).
end{align}
answered Nov 22 at 20:53
Bernard
117k637109
Hint:
This system has non-trivial solutions if and only if its determinant is zero . This determinant can be calculated by row and column operations:
begin{align}
begin{vmatrix}
1&2p-1&-p \ 2-p&1&-1 \ 1&p&-1
end{vmatrix} &=
begin{vmatrix}
1&2p-1&-p \ 0&1+(p-2)(2p-1)&-1-p(p-2) \ 0&1-p&p-1
end{vmatrix}
=begin{vmatrix}
1&p-1&-p \ 0&(p-2)(p-1)&-1-p(p-2) \ 0&0&p-1
end{vmatrix}\[1ex]& = (p-1)begin{vmatrix}
1& 1&-p \ 0&p-2 &-1-p(p-2) \0&0&p-1
end{vmatrix} = (p-1)begin{vmatrix}
p-2 &-1-p(p-2) \ 0&p-1
end{vmatrix} \[1ex]
&=(p-1)^2(p-2).
end{align}
answered Nov 22 at 20:53
Bernard
117k637109
answered Nov 22 at 20:53
Bernard
117k637109
answered Nov 22 at 20:53
Bernard
117k637109
answered Nov 22 at 20:53
Bernard
117k637109
117k637109
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009379%2ffind-trivial-and-non-trivial-solutions-to-a-system-with-a-parameter%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Well if $x=y=z = 0$ then any value of $p$ will have trivial solutions. It the equations are linear dependent then those are the only solutions. But if they aren't linearly independent they can have non-trivial solutions. Which can only occur if the determinate is $0$. So this question boils down to nothing more or less than: For what values of $p$ will the determinate be $0$. (Note: there is nothing in the question about finding the solutions).
– fleablood
Nov 22 at 18:03