Infinitely many solutions of a linear system of equations
$begingroup$
Find the value(s) of a real parameter k so that the system of equations has infinitely many solutions.
$x+y+z=1$
$2x+ky+3z=-2$
$3x+5y+kz=-1$
Please help, I don't know how to go about this. Would you have to try every combination of the equations until you get a contradiction?
linear-algebra
asked Dec 11 '18 at 1:36
usernamesAreHardusernamesAreHard
434
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add a comment |
$begingroup$
Find the value(s) of a real parameter k so that the system of equations has infinitely many solutions.
$x+y+z=1$
$2x+ky+3z=-2$
$3x+5y+kz=-1$
Please help, I don't know how to go about this. Would you have to try every combination of the equations until you get a contradiction?
linear-algebra
asked Dec 11 '18 at 1:36
usernamesAreHardusernamesAreHard
434
$endgroup$
$begingroup$
You dont want a contradiction, this gives you empty solution. You want to find a $k$ such that one of the equation can be obtained as a linear combination of the other two equation.
$endgroup$
– quantum
Dec 11 '18 at 1:45
$begingroup$
What conditions have you learned for a system of equations having an infinite number of solutions? Start with that.
$endgroup$
– amd
Dec 11 '18 at 2:52
add a comment |
$begingroup$
Find the value(s) of a real parameter k so that the system of equations has infinitely many solutions.
$x+y+z=1$
$2x+ky+3z=-2$
$3x+5y+kz=-1$
Please help, I don't know how to go about this. Would you have to try every combination of the equations until you get a contradiction?
linear-algebra
asked Dec 11 '18 at 1:36
usernamesAreHardusernamesAreHard
434
$endgroup$
Find the value(s) of a real parameter k so that the system of equations has infinitely many solutions.
$x+y+z=1$
$2x+ky+3z=-2$
$3x+5y+kz=-1$
Please help, I don't know how to go about this. Would you have to try every combination of the equations until you get a contradiction?
linear-algebra
linear-algebra
asked Dec 11 '18 at 1:36
usernamesAreHardusernamesAreHard
434
asked Dec 11 '18 at 1:36
usernamesAreHardusernamesAreHard
434
asked Dec 11 '18 at 1:36
usernamesAreHardusernamesAreHard
434
asked Dec 11 '18 at 1:36
usernamesAreHardusernamesAreHard
434
asked Dec 11 '18 at 1:36
usernamesAreHardusernamesAreHard
434
434
$begingroup$
You dont want a contradiction, this gives you empty solution. You want to find a $k$ such that one of the equation can be obtained as a linear combination of the other two equation.
$endgroup$
– quantum
Dec 11 '18 at 1:45
$begingroup$
What conditions have you learned for a system of equations having an infinite number of solutions? Start with that.
$endgroup$
– amd
Dec 11 '18 at 2:52
add a comment |
$begingroup$
You dont want a contradiction, this gives you empty solution. You want to find a $k$ such that one of the equation can be obtained as a linear combination of the other two equation.
$endgroup$
– quantum
Dec 11 '18 at 1:45
$begingroup$
What conditions have you learned for a system of equations having an infinite number of solutions? Start with that.
$endgroup$
– amd
Dec 11 '18 at 2:52
$begingroup$
You dont want a contradiction, this gives you empty solution. You want to find a $k$ such that one of the equation can be obtained as a linear combination of the other two equation.
$endgroup$
– quantum
Dec 11 '18 at 1:45
$begingroup$
You dont want a contradiction, this gives you empty solution. You want to find a $k$ such that one of the equation can be obtained as a linear combination of the other two equation.
$endgroup$
– quantum
Dec 11 '18 at 1:45
$begingroup$
What conditions have you learned for a system of equations having an infinite number of solutions? Start with that.
$endgroup$
– amd
Dec 11 '18 at 2:52
$begingroup$
What conditions have you learned for a system of equations having an infinite number of solutions? Start with that.
$endgroup$
– amd
Dec 11 '18 at 2:52
add a comment |
2 Answers
2
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oldest
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Add the first and second and subtract the third equation we have: $(k-4)(y-z) = 0$. Thus we require $k = 4$ to have infinite solutions.
answered Dec 11 '18 at 1:40
DeepSeaDeepSea
71.3k54487
$endgroup$
add a comment |
$begingroup$
Firs you rewrite the system of equation as a matrix equation $Ax=b$. The system has a unique solution if and only if $det(A)neq0$. But since $det A=(k-4)(k-1)=0$. So the solution is unique if and only if $kneq 4$ and $kneq 1$.
But if you check it further, when $k=1$ the system has no solution while when $k=4$ the system has infinitely many solutions. So $k=4$ is the only answer.
answered Dec 11 '18 at 1:56
user9077user9077
1,239612
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add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
Add the first and second and subtract the third equation we have: $(k-4)(y-z) = 0$. Thus we require $k = 4$ to have infinite solutions.
answered Dec 11 '18 at 1:40
DeepSeaDeepSea
71.3k54487
$endgroup$
add a comment |
$begingroup$
Add the first and second and subtract the third equation we have: $(k-4)(y-z) = 0$. Thus we require $k = 4$ to have infinite solutions.
answered Dec 11 '18 at 1:40
DeepSeaDeepSea
71.3k54487
$endgroup$
add a comment |
$begingroup$
Add the first and second and subtract the third equation we have: $(k-4)(y-z) = 0$. Thus we require $k = 4$ to have infinite solutions.
answered Dec 11 '18 at 1:40
DeepSeaDeepSea
71.3k54487
$endgroup$
Add the first and second and subtract the third equation we have: $(k-4)(y-z) = 0$. Thus we require $k = 4$ to have infinite solutions.
answered Dec 11 '18 at 1:40
DeepSeaDeepSea
71.3k54487
answered Dec 11 '18 at 1:40
DeepSeaDeepSea
71.3k54487
answered Dec 11 '18 at 1:40
DeepSeaDeepSea
71.3k54487
answered Dec 11 '18 at 1:40
DeepSeaDeepSea
71.3k54487
71.3k54487
add a comment |
add a comment |
$begingroup$
Firs you rewrite the system of equation as a matrix equation $Ax=b$. The system has a unique solution if and only if $det(A)neq0$. But since $det A=(k-4)(k-1)=0$. So the solution is unique if and only if $kneq 4$ and $kneq 1$.
But if you check it further, when $k=1$ the system has no solution while when $k=4$ the system has infinitely many solutions. So $k=4$ is the only answer.
answered Dec 11 '18 at 1:56
user9077user9077
1,239612
$endgroup$
add a comment |
$begingroup$
Firs you rewrite the system of equation as a matrix equation $Ax=b$. The system has a unique solution if and only if $det(A)neq0$. But since $det A=(k-4)(k-1)=0$. So the solution is unique if and only if $kneq 4$ and $kneq 1$.
But if you check it further, when $k=1$ the system has no solution while when $k=4$ the system has infinitely many solutions. So $k=4$ is the only answer.
answered Dec 11 '18 at 1:56
user9077user9077
1,239612
$endgroup$
add a comment |
$begingroup$
Firs you rewrite the system of equation as a matrix equation $Ax=b$. The system has a unique solution if and only if $det(A)neq0$. But since $det A=(k-4)(k-1)=0$. So the solution is unique if and only if $kneq 4$ and $kneq 1$.
But if you check it further, when $k=1$ the system has no solution while when $k=4$ the system has infinitely many solutions. So $k=4$ is the only answer.
answered Dec 11 '18 at 1:56
user9077user9077
1,239612
$endgroup$
Firs you rewrite the system of equation as a matrix equation $Ax=b$. The system has a unique solution if and only if $det(A)neq0$. But since $det A=(k-4)(k-1)=0$. So the solution is unique if and only if $kneq 4$ and $kneq 1$.
But if you check it further, when $k=1$ the system has no solution while when $k=4$ the system has infinitely many solutions. So $k=4$ is the only answer.
answered Dec 11 '18 at 1:56
user9077user9077
1,239612
answered Dec 11 '18 at 1:56
user9077user9077
1,239612
answered Dec 11 '18 at 1:56
user9077user9077
1,239612
answered Dec 11 '18 at 1:56
user9077user9077
1,239612
1,239612
add a comment |
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$begingroup$
You dont want a contradiction, this gives you empty solution. You want to find a $k$ such that one of the equation can be obtained as a linear combination of the other two equation.
$endgroup$
– quantum
Dec 11 '18 at 1:45
$begingroup$
What conditions have you learned for a system of equations having an infinite number of solutions? Start with that.
$endgroup$
– amd
Dec 11 '18 at 2:52