How to find values that make a matrix solvable and unsolvable?
$begingroup$
I need to find two more $b$'s other than $b=(2,5,7)$, such that the equation can be solved and two more such that the equation can't be solved. $$ubegin{bmatrix}1\2\3end{bmatrix}+vbegin{bmatrix}1\0\1end{bmatrix}+wbegin{bmatrix}1\3\4end{bmatrix}=b$$
How do I find those values? What should I do to find both, values of b that make the system solvable and values that make the system unsolvable?
I tried to set $b$ as $(b_1,b_2,b_3)$, but I ended up with this matrix:
$$left[begin{array}{l}1&1&1&b_1\0&2&-1&2b_1-b_2\0&0&0&b_1-b_3+b_2end{array}right]$$
Sorry if it is something obvious, I'm knew with linear algebra.
linear-algebra matrices
asked Jan 25 at 23:44
davidllerenavdavidllerenav
1888
$endgroup$
add a comment |
$begingroup$
I need to find two more $b$'s other than $b=(2,5,7)$, such that the equation can be solved and two more such that the equation can't be solved. $$ubegin{bmatrix}1\2\3end{bmatrix}+vbegin{bmatrix}1\0\1end{bmatrix}+wbegin{bmatrix}1\3\4end{bmatrix}=b$$
How do I find those values? What should I do to find both, values of b that make the system solvable and values that make the system unsolvable?
I tried to set $b$ as $(b_1,b_2,b_3)$, but I ended up with this matrix:
$$left[begin{array}{l}1&1&1&b_1\0&2&-1&2b_1-b_2\0&0&0&b_1-b_3+b_2end{array}right]$$
Sorry if it is something obvious, I'm knew with linear algebra.
linear-algebra matrices
asked Jan 25 at 23:44
davidllerenavdavidllerenav
1888
$endgroup$
$begingroup$
You're almost there. The last step is to use the row echelon form matrix you computed to the see if you can pick values of $b_1,b_2,b_3$ so that you do or don't have solutions. For instance, if $b_1=b_2=0$ and $b_3=1$ can you have a solution?
$endgroup$
– tch
Jan 26 at 0:32
$begingroup$
So I just need to give random values to$ b_1$,$ b_2$ and $b_3$? And if $b_1=b_2$ and $b_3=1$, I think that there is no solution, because in the last row there will be a 0=-1, right?
$endgroup$
– davidllerenav
Jan 26 at 0:43
$begingroup$
Exactly! The problem is asking you to find some values of $b$ where there is a solution, and some where there isn't.
$endgroup$
– tch
Jan 26 at 0:46
add a comment |
$begingroup$
I need to find two more $b$'s other than $b=(2,5,7)$, such that the equation can be solved and two more such that the equation can't be solved. $$ubegin{bmatrix}1\2\3end{bmatrix}+vbegin{bmatrix}1\0\1end{bmatrix}+wbegin{bmatrix}1\3\4end{bmatrix}=b$$
How do I find those values? What should I do to find both, values of b that make the system solvable and values that make the system unsolvable?
I tried to set $b$ as $(b_1,b_2,b_3)$, but I ended up with this matrix:
$$left[begin{array}{l}1&1&1&b_1\0&2&-1&2b_1-b_2\0&0&0&b_1-b_3+b_2end{array}right]$$
Sorry if it is something obvious, I'm knew with linear algebra.
linear-algebra matrices
asked Jan 25 at 23:44
davidllerenavdavidllerenav
1888
$endgroup$
I need to find two more $b$'s other than $b=(2,5,7)$, such that the equation can be solved and two more such that the equation can't be solved. $$ubegin{bmatrix}1\2\3end{bmatrix}+vbegin{bmatrix}1\0\1end{bmatrix}+wbegin{bmatrix}1\3\4end{bmatrix}=b$$
How do I find those values? What should I do to find both, values of b that make the system solvable and values that make the system unsolvable?
I tried to set $b$ as $(b_1,b_2,b_3)$, but I ended up with this matrix:
$$left[begin{array}{l}1&1&1&b_1\0&2&-1&2b_1-b_2\0&0&0&b_1-b_3+b_2end{array}right]$$
Sorry if it is something obvious, I'm knew with linear algebra.
linear-algebra matrices
linear-algebra matrices
asked Jan 25 at 23:44
davidllerenavdavidllerenav
1888
asked Jan 25 at 23:44
davidllerenavdavidllerenav
1888
asked Jan 25 at 23:44
davidllerenavdavidllerenav
1888
asked Jan 25 at 23:44
davidllerenavdavidllerenav
1888
asked Jan 25 at 23:44
davidllerenavdavidllerenav
1888
1888
$begingroup$
You're almost there. The last step is to use the row echelon form matrix you computed to the see if you can pick values of $b_1,b_2,b_3$ so that you do or don't have solutions. For instance, if $b_1=b_2=0$ and $b_3=1$ can you have a solution?
$endgroup$
– tch
Jan 26 at 0:32
$begingroup$
So I just need to give random values to$ b_1$,$ b_2$ and $b_3$? And if $b_1=b_2$ and $b_3=1$, I think that there is no solution, because in the last row there will be a 0=-1, right?
$endgroup$
– davidllerenav
Jan 26 at 0:43
$begingroup$
Exactly! The problem is asking you to find some values of $b$ where there is a solution, and some where there isn't.
$endgroup$
– tch
Jan 26 at 0:46
add a comment |
$begingroup$
You're almost there. The last step is to use the row echelon form matrix you computed to the see if you can pick values of $b_1,b_2,b_3$ so that you do or don't have solutions. For instance, if $b_1=b_2=0$ and $b_3=1$ can you have a solution?
$endgroup$
– tch
Jan 26 at 0:32
$begingroup$
So I just need to give random values to$ b_1$,$ b_2$ and $b_3$? And if $b_1=b_2$ and $b_3=1$, I think that there is no solution, because in the last row there will be a 0=-1, right?
$endgroup$
– davidllerenav
Jan 26 at 0:43
$begingroup$
Exactly! The problem is asking you to find some values of $b$ where there is a solution, and some where there isn't.
$endgroup$
– tch
Jan 26 at 0:46
$begingroup$
You're almost there. The last step is to use the row echelon form matrix you computed to the see if you can pick values of $b_1,b_2,b_3$ so that you do or don't have solutions. For instance, if $b_1=b_2=0$ and $b_3=1$ can you have a solution?
$endgroup$
– tch
Jan 26 at 0:32
$begingroup$
You're almost there. The last step is to use the row echelon form matrix you computed to the see if you can pick values of $b_1,b_2,b_3$ so that you do or don't have solutions. For instance, if $b_1=b_2=0$ and $b_3=1$ can you have a solution?
$endgroup$
– tch
Jan 26 at 0:32
$begingroup$
So I just need to give random values to$ b_1$,$ b_2$ and $b_3$? And if $b_1=b_2$ and $b_3=1$, I think that there is no solution, because in the last row there will be a 0=-1, right?
$endgroup$
– davidllerenav
Jan 26 at 0:43
$begingroup$
So I just need to give random values to$ b_1$,$ b_2$ and $b_3$? And if $b_1=b_2$ and $b_3=1$, I think that there is no solution, because in the last row there will be a 0=-1, right?
$endgroup$
– davidllerenav
Jan 26 at 0:43
$begingroup$
Exactly! The problem is asking you to find some values of $b$ where there is a solution, and some where there isn't.
$endgroup$
– tch
Jan 26 at 0:46
$begingroup$
Exactly! The problem is asking you to find some values of $b$ where there is a solution, and some where there isn't.
$endgroup$
– tch
Jan 26 at 0:46
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
After you apply your row reduction, look at the last row of your matrix.
It shows you that
$0u + 0v + 0 w = b_1 - b_3 + b_2$
This is only possible if $b_1 - b_3 + b_2 = 0$
So if you choose $b_1, b_2, b_3$ that satisfy this condition, your matrix has a solution (in this case infinite), otherwise it has no solutions.
for example if $b = (2, 5, 7)$ as you suggested, then it is solvable because $2 - 7 + 5 = 0$
Since you are new to linear algebra I did not mention determinants, dimensions, ranks etc. You don't need them in this case, just recall how a matrix represents a system of equations.
answered Jan 26 at 22:05
kkshourykkshoury
812
$endgroup$
$begingroup$
Thanks! After reading you answer I was able to find the solution.
$endgroup$
– davidllerenav
Jan 27 at 0:34
add a comment |
$begingroup$
Hint:
Use the criterion for the system to be (not) solvable:
$Ax=b$ has a solution if and only if $operatorname{rank}A$ is the same as the rank of the augmented matrix $(A|b)$.
So, as $operatorname{rank}A=2$, you have to find $b$ such that $(A|b)$ has the maximum rank ($3$).
answered Jan 26 at 0:35
BernardBernard
122k741116
$endgroup$
$begingroup$
Sorry, what does $rank$ mean?
$endgroup$
– davidllerenav
Jan 26 at 0:41
$begingroup$
The number of independent rows or columns. It can be computed via row reduction or with the maximum size of a non-zero subdeterminant in the matrix.
$endgroup$
– Bernard
Jan 26 at 0:45
add a comment |
$begingroup$
You can compute
$$det left(begin{matrix}
1 & 1 & 1 \
2 & 0 & 3 \
3 & 1 & 4
end{matrix}
right) = 9+2-3-8=0$$
which means the three vectors $(1,2,3), (1,0,1), (1,3,4)$ are linearly dependent. However the first two are linearly independent, so they generate a plane in $mathbb{R}^3$ which contains $(1,3,4)$. The plane is
$$t(1,2,3)+s(1,0,1) = (t+s,2t,3t+s)$$
which you can write in Cartesian coordinates as follows: define $x,y,z$ so that
$$begin{cases}
x = t+s \
y = 2t \
z = 3t+s.
end{cases}
$$
Hence $x+y-z=0$, and this is the Cartesian equation of your plane. So every $b=(x,y,z)$ satisfying this equation works. Notice that your $b = (2,5,7)$ works, and also $b=(0,0,0)$ is a trivial choice - which just means that $(1,3,4)$ can be expressed as a linear combination of the first two vectors, as we know. The $b$'s which do not lie on this plane do not work, e.g. $b=(2,5,8)$.
answered Jan 26 at 0:40
GibbsGibbs
5,4103827
$endgroup$
$begingroup$
I didn't understand that well what you said at the beginning, but I understood that, I could choose any other $b$ that is a multiple of $b=(2,5,7)$ then it would lie on the same plane, right?
$endgroup$
– davidllerenav
Jan 26 at 1:13
$begingroup$
Right. My point at the beginning is essentially that $(1,3,4)$ can be expressed as $alpha(1,2,3)+beta(1,0,1)$, so your starting equation can be rewritten as $b = c(1,2,3)+d(1,0,1)$. Then $b$ lies in the plane generated by $(1,2,3)$ and $(1,0,1)$.
$endgroup$
– Gibbs
Jan 26 at 1:14
$begingroup$
Ok, I get it. How did you realize that the three vectors where linearly dependent?
$endgroup$
– davidllerenav
Jan 26 at 2:35
$begingroup$
I just tried to compute the determinant of the matrix they form, as I showed. It turns out to be zero, so the three vectors are linearly dependent.
$endgroup$
– Gibbs
Jan 26 at 10:08
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
After you apply your row reduction, look at the last row of your matrix.
It shows you that
$0u + 0v + 0 w = b_1 - b_3 + b_2$
This is only possible if $b_1 - b_3 + b_2 = 0$
So if you choose $b_1, b_2, b_3$ that satisfy this condition, your matrix has a solution (in this case infinite), otherwise it has no solutions.
for example if $b = (2, 5, 7)$ as you suggested, then it is solvable because $2 - 7 + 5 = 0$
Since you are new to linear algebra I did not mention determinants, dimensions, ranks etc. You don't need them in this case, just recall how a matrix represents a system of equations.
answered Jan 26 at 22:05
kkshourykkshoury
812
$endgroup$
$begingroup$
Thanks! After reading you answer I was able to find the solution.
$endgroup$
– davidllerenav
Jan 27 at 0:34
add a comment |
$begingroup$
After you apply your row reduction, look at the last row of your matrix.
It shows you that
$0u + 0v + 0 w = b_1 - b_3 + b_2$
This is only possible if $b_1 - b_3 + b_2 = 0$
So if you choose $b_1, b_2, b_3$ that satisfy this condition, your matrix has a solution (in this case infinite), otherwise it has no solutions.
for example if $b = (2, 5, 7)$ as you suggested, then it is solvable because $2 - 7 + 5 = 0$
Since you are new to linear algebra I did not mention determinants, dimensions, ranks etc. You don't need them in this case, just recall how a matrix represents a system of equations.
answered Jan 26 at 22:05
kkshourykkshoury
812
$endgroup$
$begingroup$
Thanks! After reading you answer I was able to find the solution.
$endgroup$
– davidllerenav
Jan 27 at 0:34
add a comment |
$begingroup$
After you apply your row reduction, look at the last row of your matrix.
It shows you that
$0u + 0v + 0 w = b_1 - b_3 + b_2$
This is only possible if $b_1 - b_3 + b_2 = 0$
So if you choose $b_1, b_2, b_3$ that satisfy this condition, your matrix has a solution (in this case infinite), otherwise it has no solutions.
for example if $b = (2, 5, 7)$ as you suggested, then it is solvable because $2 - 7 + 5 = 0$
Since you are new to linear algebra I did not mention determinants, dimensions, ranks etc. You don't need them in this case, just recall how a matrix represents a system of equations.
answered Jan 26 at 22:05
kkshourykkshoury
812
$endgroup$
After you apply your row reduction, look at the last row of your matrix.
It shows you that
$0u + 0v + 0 w = b_1 - b_3 + b_2$
This is only possible if $b_1 - b_3 + b_2 = 0$
So if you choose $b_1, b_2, b_3$ that satisfy this condition, your matrix has a solution (in this case infinite), otherwise it has no solutions.
for example if $b = (2, 5, 7)$ as you suggested, then it is solvable because $2 - 7 + 5 = 0$
Since you are new to linear algebra I did not mention determinants, dimensions, ranks etc. You don't need them in this case, just recall how a matrix represents a system of equations.
answered Jan 26 at 22:05
kkshourykkshoury
812
edited Jan 26 at 22:11
answered Jan 26 at 22:05
kkshourykkshoury
812
answered Jan 26 at 22:05
kkshourykkshoury
812
answered Jan 26 at 22:05
kkshourykkshoury
812
812
$begingroup$
Thanks! After reading you answer I was able to find the solution.
$endgroup$
– davidllerenav
Jan 27 at 0:34
add a comment |
$begingroup$
Thanks! After reading you answer I was able to find the solution.
$endgroup$
– davidllerenav
Jan 27 at 0:34
$begingroup$
Thanks! After reading you answer I was able to find the solution.
$endgroup$
– davidllerenav
Jan 27 at 0:34
$begingroup$
Thanks! After reading you answer I was able to find the solution.
$endgroup$
– davidllerenav
Jan 27 at 0:34
add a comment |
$begingroup$
Hint:
Use the criterion for the system to be (not) solvable:
$Ax=b$ has a solution if and only if $operatorname{rank}A$ is the same as the rank of the augmented matrix $(A|b)$.
So, as $operatorname{rank}A=2$, you have to find $b$ such that $(A|b)$ has the maximum rank ($3$).
answered Jan 26 at 0:35
BernardBernard
122k741116
$endgroup$
$begingroup$
Sorry, what does $rank$ mean?
$endgroup$
– davidllerenav
Jan 26 at 0:41
$begingroup$
The number of independent rows or columns. It can be computed via row reduction or with the maximum size of a non-zero subdeterminant in the matrix.
$endgroup$
– Bernard
Jan 26 at 0:45
add a comment |
$begingroup$
Hint:
Use the criterion for the system to be (not) solvable:
$Ax=b$ has a solution if and only if $operatorname{rank}A$ is the same as the rank of the augmented matrix $(A|b)$.
So, as $operatorname{rank}A=2$, you have to find $b$ such that $(A|b)$ has the maximum rank ($3$).
answered Jan 26 at 0:35
BernardBernard
122k741116
$endgroup$
$begingroup$
Sorry, what does $rank$ mean?
$endgroup$
– davidllerenav
Jan 26 at 0:41
$begingroup$
The number of independent rows or columns. It can be computed via row reduction or with the maximum size of a non-zero subdeterminant in the matrix.
$endgroup$
– Bernard
Jan 26 at 0:45
add a comment |
$begingroup$
Hint:
Use the criterion for the system to be (not) solvable:
$Ax=b$ has a solution if and only if $operatorname{rank}A$ is the same as the rank of the augmented matrix $(A|b)$.
So, as $operatorname{rank}A=2$, you have to find $b$ such that $(A|b)$ has the maximum rank ($3$).
answered Jan 26 at 0:35
BernardBernard
122k741116
$endgroup$
Hint:
Use the criterion for the system to be (not) solvable:
$Ax=b$ has a solution if and only if $operatorname{rank}A$ is the same as the rank of the augmented matrix $(A|b)$.
So, as $operatorname{rank}A=2$, you have to find $b$ such that $(A|b)$ has the maximum rank ($3$).
answered Jan 26 at 0:35
BernardBernard
122k741116
answered Jan 26 at 0:35
BernardBernard
122k741116
answered Jan 26 at 0:35
BernardBernard
122k741116
answered Jan 26 at 0:35
BernardBernard
122k741116
122k741116
$begingroup$
Sorry, what does $rank$ mean?
$endgroup$
– davidllerenav
Jan 26 at 0:41
$begingroup$
The number of independent rows or columns. It can be computed via row reduction or with the maximum size of a non-zero subdeterminant in the matrix.
$endgroup$
– Bernard
Jan 26 at 0:45
add a comment |
$begingroup$
Sorry, what does $rank$ mean?
$endgroup$
– davidllerenav
Jan 26 at 0:41
$begingroup$
The number of independent rows or columns. It can be computed via row reduction or with the maximum size of a non-zero subdeterminant in the matrix.
$endgroup$
– Bernard
Jan 26 at 0:45
$begingroup$
Sorry, what does $rank$ mean?
$endgroup$
– davidllerenav
Jan 26 at 0:41
$begingroup$
Sorry, what does $rank$ mean?
$endgroup$
– davidllerenav
Jan 26 at 0:41
$begingroup$
The number of independent rows or columns. It can be computed via row reduction or with the maximum size of a non-zero subdeterminant in the matrix.
$endgroup$
– Bernard
Jan 26 at 0:45
$begingroup$
The number of independent rows or columns. It can be computed via row reduction or with the maximum size of a non-zero subdeterminant in the matrix.
$endgroup$
– Bernard
Jan 26 at 0:45
add a comment |
$begingroup$
You can compute
$$det left(begin{matrix}
1 & 1 & 1 \
2 & 0 & 3 \
3 & 1 & 4
end{matrix}
right) = 9+2-3-8=0$$
which means the three vectors $(1,2,3), (1,0,1), (1,3,4)$ are linearly dependent. However the first two are linearly independent, so they generate a plane in $mathbb{R}^3$ which contains $(1,3,4)$. The plane is
$$t(1,2,3)+s(1,0,1) = (t+s,2t,3t+s)$$
which you can write in Cartesian coordinates as follows: define $x,y,z$ so that
$$begin{cases}
x = t+s \
y = 2t \
z = 3t+s.
end{cases}
$$
Hence $x+y-z=0$, and this is the Cartesian equation of your plane. So every $b=(x,y,z)$ satisfying this equation works. Notice that your $b = (2,5,7)$ works, and also $b=(0,0,0)$ is a trivial choice - which just means that $(1,3,4)$ can be expressed as a linear combination of the first two vectors, as we know. The $b$'s which do not lie on this plane do not work, e.g. $b=(2,5,8)$.
answered Jan 26 at 0:40
GibbsGibbs
5,4103827
$endgroup$
$begingroup$
I didn't understand that well what you said at the beginning, but I understood that, I could choose any other $b$ that is a multiple of $b=(2,5,7)$ then it would lie on the same plane, right?
$endgroup$
– davidllerenav
Jan 26 at 1:13
$begingroup$
Right. My point at the beginning is essentially that $(1,3,4)$ can be expressed as $alpha(1,2,3)+beta(1,0,1)$, so your starting equation can be rewritten as $b = c(1,2,3)+d(1,0,1)$. Then $b$ lies in the plane generated by $(1,2,3)$ and $(1,0,1)$.
$endgroup$
– Gibbs
Jan 26 at 1:14
$begingroup$
Ok, I get it. How did you realize that the three vectors where linearly dependent?
$endgroup$
– davidllerenav
Jan 26 at 2:35
$begingroup$
I just tried to compute the determinant of the matrix they form, as I showed. It turns out to be zero, so the three vectors are linearly dependent.
$endgroup$
– Gibbs
Jan 26 at 10:08
add a comment |
$begingroup$
You can compute
$$det left(begin{matrix}
1 & 1 & 1 \
2 & 0 & 3 \
3 & 1 & 4
end{matrix}
right) = 9+2-3-8=0$$
which means the three vectors $(1,2,3), (1,0,1), (1,3,4)$ are linearly dependent. However the first two are linearly independent, so they generate a plane in $mathbb{R}^3$ which contains $(1,3,4)$. The plane is
$$t(1,2,3)+s(1,0,1) = (t+s,2t,3t+s)$$
which you can write in Cartesian coordinates as follows: define $x,y,z$ so that
$$begin{cases}
x = t+s \
y = 2t \
z = 3t+s.
end{cases}
$$
Hence $x+y-z=0$, and this is the Cartesian equation of your plane. So every $b=(x,y,z)$ satisfying this equation works. Notice that your $b = (2,5,7)$ works, and also $b=(0,0,0)$ is a trivial choice - which just means that $(1,3,4)$ can be expressed as a linear combination of the first two vectors, as we know. The $b$'s which do not lie on this plane do not work, e.g. $b=(2,5,8)$.
answered Jan 26 at 0:40
GibbsGibbs
5,4103827
$endgroup$
$begingroup$
I didn't understand that well what you said at the beginning, but I understood that, I could choose any other $b$ that is a multiple of $b=(2,5,7)$ then it would lie on the same plane, right?
$endgroup$
– davidllerenav
Jan 26 at 1:13
$begingroup$
Right. My point at the beginning is essentially that $(1,3,4)$ can be expressed as $alpha(1,2,3)+beta(1,0,1)$, so your starting equation can be rewritten as $b = c(1,2,3)+d(1,0,1)$. Then $b$ lies in the plane generated by $(1,2,3)$ and $(1,0,1)$.
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– Gibbs
Jan 26 at 1:14
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Ok, I get it. How did you realize that the three vectors where linearly dependent?
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– davidllerenav
Jan 26 at 2:35
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I just tried to compute the determinant of the matrix they form, as I showed. It turns out to be zero, so the three vectors are linearly dependent.
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– Gibbs
Jan 26 at 10:08
add a comment |
$begingroup$
You can compute
$$det left(begin{matrix}
1 & 1 & 1 \
2 & 0 & 3 \
3 & 1 & 4
end{matrix}
right) = 9+2-3-8=0$$
which means the three vectors $(1,2,3), (1,0,1), (1,3,4)$ are linearly dependent. However the first two are linearly independent, so they generate a plane in $mathbb{R}^3$ which contains $(1,3,4)$. The plane is
$$t(1,2,3)+s(1,0,1) = (t+s,2t,3t+s)$$
which you can write in Cartesian coordinates as follows: define $x,y,z$ so that
$$begin{cases}
x = t+s \
y = 2t \
z = 3t+s.
end{cases}
$$
Hence $x+y-z=0$, and this is the Cartesian equation of your plane. So every $b=(x,y,z)$ satisfying this equation works. Notice that your $b = (2,5,7)$ works, and also $b=(0,0,0)$ is a trivial choice - which just means that $(1,3,4)$ can be expressed as a linear combination of the first two vectors, as we know. The $b$'s which do not lie on this plane do not work, e.g. $b=(2,5,8)$.
answered Jan 26 at 0:40
GibbsGibbs
5,4103827
$endgroup$
You can compute
$$det left(begin{matrix}
1 & 1 & 1 \
2 & 0 & 3 \
3 & 1 & 4
end{matrix}
right) = 9+2-3-8=0$$
which means the three vectors $(1,2,3), (1,0,1), (1,3,4)$ are linearly dependent. However the first two are linearly independent, so they generate a plane in $mathbb{R}^3$ which contains $(1,3,4)$. The plane is
$$t(1,2,3)+s(1,0,1) = (t+s,2t,3t+s)$$
which you can write in Cartesian coordinates as follows: define $x,y,z$ so that
$$begin{cases}
x = t+s \
y = 2t \
z = 3t+s.
end{cases}
$$
Hence $x+y-z=0$, and this is the Cartesian equation of your plane. So every $b=(x,y,z)$ satisfying this equation works. Notice that your $b = (2,5,7)$ works, and also $b=(0,0,0)$ is a trivial choice - which just means that $(1,3,4)$ can be expressed as a linear combination of the first two vectors, as we know. The $b$'s which do not lie on this plane do not work, e.g. $b=(2,5,8)$.
answered Jan 26 at 0:40
GibbsGibbs
5,4103827
answered Jan 26 at 0:40
GibbsGibbs
5,4103827
answered Jan 26 at 0:40
GibbsGibbs
5,4103827
answered Jan 26 at 0:40
GibbsGibbs
5,4103827
5,4103827
$begingroup$
I didn't understand that well what you said at the beginning, but I understood that, I could choose any other $b$ that is a multiple of $b=(2,5,7)$ then it would lie on the same plane, right?
$endgroup$
– davidllerenav
Jan 26 at 1:13
$begingroup$
Right. My point at the beginning is essentially that $(1,3,4)$ can be expressed as $alpha(1,2,3)+beta(1,0,1)$, so your starting equation can be rewritten as $b = c(1,2,3)+d(1,0,1)$. Then $b$ lies in the plane generated by $(1,2,3)$ and $(1,0,1)$.
$endgroup$
– Gibbs
Jan 26 at 1:14
$begingroup$
Ok, I get it. How did you realize that the three vectors where linearly dependent?
$endgroup$
– davidllerenav
Jan 26 at 2:35
$begingroup$
I just tried to compute the determinant of the matrix they form, as I showed. It turns out to be zero, so the three vectors are linearly dependent.
$endgroup$
– Gibbs
Jan 26 at 10:08
add a comment |
$begingroup$
I didn't understand that well what you said at the beginning, but I understood that, I could choose any other $b$ that is a multiple of $b=(2,5,7)$ then it would lie on the same plane, right?
$endgroup$
– davidllerenav
Jan 26 at 1:13
$begingroup$
Right. My point at the beginning is essentially that $(1,3,4)$ can be expressed as $alpha(1,2,3)+beta(1,0,1)$, so your starting equation can be rewritten as $b = c(1,2,3)+d(1,0,1)$. Then $b$ lies in the plane generated by $(1,2,3)$ and $(1,0,1)$.
$endgroup$
– Gibbs
Jan 26 at 1:14
$begingroup$
Ok, I get it. How did you realize that the three vectors where linearly dependent?
$endgroup$
– davidllerenav
Jan 26 at 2:35
$begingroup$
I just tried to compute the determinant of the matrix they form, as I showed. It turns out to be zero, so the three vectors are linearly dependent.
$endgroup$
– Gibbs
Jan 26 at 10:08
$begingroup$
I didn't understand that well what you said at the beginning, but I understood that, I could choose any other $b$ that is a multiple of $b=(2,5,7)$ then it would lie on the same plane, right?
$endgroup$
– davidllerenav
Jan 26 at 1:13
$begingroup$
I didn't understand that well what you said at the beginning, but I understood that, I could choose any other $b$ that is a multiple of $b=(2,5,7)$ then it would lie on the same plane, right?
$endgroup$
– davidllerenav
Jan 26 at 1:13
$begingroup$
Right. My point at the beginning is essentially that $(1,3,4)$ can be expressed as $alpha(1,2,3)+beta(1,0,1)$, so your starting equation can be rewritten as $b = c(1,2,3)+d(1,0,1)$. Then $b$ lies in the plane generated by $(1,2,3)$ and $(1,0,1)$.
$endgroup$
– Gibbs
Jan 26 at 1:14
$begingroup$
Right. My point at the beginning is essentially that $(1,3,4)$ can be expressed as $alpha(1,2,3)+beta(1,0,1)$, so your starting equation can be rewritten as $b = c(1,2,3)+d(1,0,1)$. Then $b$ lies in the plane generated by $(1,2,3)$ and $(1,0,1)$.
$endgroup$
– Gibbs
Jan 26 at 1:14
$begingroup$
Ok, I get it. How did you realize that the three vectors where linearly dependent?
$endgroup$
– davidllerenav
Jan 26 at 2:35
$begingroup$
Ok, I get it. How did you realize that the three vectors where linearly dependent?
$endgroup$
– davidllerenav
Jan 26 at 2:35
$begingroup$
I just tried to compute the determinant of the matrix they form, as I showed. It turns out to be zero, so the three vectors are linearly dependent.
$endgroup$
– Gibbs
Jan 26 at 10:08
$begingroup$
I just tried to compute the determinant of the matrix they form, as I showed. It turns out to be zero, so the three vectors are linearly dependent.
$endgroup$
– Gibbs
Jan 26 at 10:08
add a comment |
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$begingroup$
You're almost there. The last step is to use the row echelon form matrix you computed to the see if you can pick values of $b_1,b_2,b_3$ so that you do or don't have solutions. For instance, if $b_1=b_2=0$ and $b_3=1$ can you have a solution?
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– tch
Jan 26 at 0:32
$begingroup$
So I just need to give random values to$ b_1$,$ b_2$ and $b_3$? And if $b_1=b_2$ and $b_3=1$, I think that there is no solution, because in the last row there will be a 0=-1, right?
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– davidllerenav
Jan 26 at 0:43
$begingroup$
Exactly! The problem is asking you to find some values of $b$ where there is a solution, and some where there isn't.
$endgroup$
– tch
Jan 26 at 0:46