Mixing
Find matrix $A$ of $Ax=b$ if $b=begin{bmatrix} 1 \ 2 \ 1 \ end{bmatrix} \$
$begingroup$
Let non-homogeneous system $Ax=b$ nas solution $$x=begin{bmatrix} 1 \ 0 \1 \ 0\ end{bmatrix} + alpha_1begin{bmatrix} 1\ 1\ -1 \ 0\ end{bmatrix} + alpha_2begin{bmatrix} 1\ 0\ 1 \ 1\ end{bmatrix}; space alpha_1 , alpha_2 in Bbb{R} \ $$ a)A is an m × n matrix of rank r. Describe all possible values of m, n, and
r. $$ $$b) Find matrix A if $b=begin{bmatrix} 1 \ 2 \ 1 \ end{bmatrix} \$
$$-$$
a)I found a solution that matrix $A$ has 4 columns (why exactly 4??) , and because homogeneous system result are vectors $begin{bmatrix} 1\ 1\ -1 \ 0 end{bmatrix}, begin{bmatrix} 1\ 0\ 1 \ 1\ end{bmatrix}$ they form null space , so $dim Ker(A)=2$ and from rank-nullity theorem we have that r=2 so then $m ge r$. I don't know how can I find from given solution $x$ how many columns matrix $A$ has and why number of rows is $mge r$ ?
b)If my matrix $A$ has 4 coumns and $rankA=2$ I have to write 2 independent vectors and two non-independent.How to know which two vectors I should use to create column space of $A$ if my $b=begin{bmatrix} 1 \ 2 \ 1 \ end{bmatrix} \$ , and it's obviously that other two have to be linear combination of them to form null space. $$ $$ I found that solution for b) is $$A=begin{bmatrix}
1 & -1 & 0 &-1\
1 & 0 & 1&-2 \
1 & -1 & 0 &-1\
end{bmatrix}$$ First and third column are lineary independent and their sum gives me a vector b, does it mean that linear combination of independent vectors gives me a vector b from $Ax=b$ ?
linear-algebra matrix-equations
asked Feb 2 at 15:37
FiggaroFiggaro
406
$endgroup$
add a comment |
$begingroup$
Let non-homogeneous system $Ax=b$ nas solution $$x=begin{bmatrix} 1 \ 0 \1 \ 0\ end{bmatrix} + alpha_1begin{bmatrix} 1\ 1\ -1 \ 0\ end{bmatrix} + alpha_2begin{bmatrix} 1\ 0\ 1 \ 1\ end{bmatrix}; space alpha_1 , alpha_2 in Bbb{R} \ $$ a)A is an m × n matrix of rank r. Describe all possible values of m, n, and
r. $$ $$b) Find matrix A if $b=begin{bmatrix} 1 \ 2 \ 1 \ end{bmatrix} \$
$$-$$
a)I found a solution that matrix $A$ has 4 columns (why exactly 4??) , and because homogeneous system result are vectors $begin{bmatrix} 1\ 1\ -1 \ 0 end{bmatrix}, begin{bmatrix} 1\ 0\ 1 \ 1\ end{bmatrix}$ they form null space , so $dim Ker(A)=2$ and from rank-nullity theorem we have that r=2 so then $m ge r$. I don't know how can I find from given solution $x$ how many columns matrix $A$ has and why number of rows is $mge r$ ?
b)If my matrix $A$ has 4 coumns and $rankA=2$ I have to write 2 independent vectors and two non-independent.How to know which two vectors I should use to create column space of $A$ if my $b=begin{bmatrix} 1 \ 2 \ 1 \ end{bmatrix} \$ , and it's obviously that other two have to be linear combination of them to form null space. $$ $$ I found that solution for b) is $$A=begin{bmatrix}
1 & -1 & 0 &-1\
1 & 0 & 1&-2 \
1 & -1 & 0 &-1\
end{bmatrix}$$ First and third column are lineary independent and their sum gives me a vector b, does it mean that linear combination of independent vectors gives me a vector b from $Ax=b$ ?
linear-algebra matrix-equations
asked Feb 2 at 15:37
FiggaroFiggaro
406
$endgroup$
$begingroup$
$x$'s are $4$-dimensional vectors, so in order for the multiplication $Ax$ to make sense, the number of columns in $A$ needs to be the same as the dimension of the $x$, i.e., $4$.
$endgroup$
– EuxhenH
Feb 2 at 15:55
add a comment |
$begingroup$
Let non-homogeneous system $Ax=b$ nas solution $$x=begin{bmatrix} 1 \ 0 \1 \ 0\ end{bmatrix} + alpha_1begin{bmatrix} 1\ 1\ -1 \ 0\ end{bmatrix} + alpha_2begin{bmatrix} 1\ 0\ 1 \ 1\ end{bmatrix}; space alpha_1 , alpha_2 in Bbb{R} \ $$ a)A is an m × n matrix of rank r. Describe all possible values of m, n, and
r. $$ $$b) Find matrix A if $b=begin{bmatrix} 1 \ 2 \ 1 \ end{bmatrix} \$
$$-$$
a)I found a solution that matrix $A$ has 4 columns (why exactly 4??) , and because homogeneous system result are vectors $begin{bmatrix} 1\ 1\ -1 \ 0 end{bmatrix}, begin{bmatrix} 1\ 0\ 1 \ 1\ end{bmatrix}$ they form null space , so $dim Ker(A)=2$ and from rank-nullity theorem we have that r=2 so then $m ge r$. I don't know how can I find from given solution $x$ how many columns matrix $A$ has and why number of rows is $mge r$ ?
b)If my matrix $A$ has 4 coumns and $rankA=2$ I have to write 2 independent vectors and two non-independent.How to know which two vectors I should use to create column space of $A$ if my $b=begin{bmatrix} 1 \ 2 \ 1 \ end{bmatrix} \$ , and it's obviously that other two have to be linear combination of them to form null space. $$ $$ I found that solution for b) is $$A=begin{bmatrix}
1 & -1 & 0 &-1\
1 & 0 & 1&-2 \
1 & -1 & 0 &-1\
end{bmatrix}$$ First and third column are lineary independent and their sum gives me a vector b, does it mean that linear combination of independent vectors gives me a vector b from $Ax=b$ ?
linear-algebra matrix-equations
asked Feb 2 at 15:37
FiggaroFiggaro
406
$endgroup$
Let non-homogeneous system $Ax=b$ nas solution $$x=begin{bmatrix} 1 \ 0 \1 \ 0\ end{bmatrix} + alpha_1begin{bmatrix} 1\ 1\ -1 \ 0\ end{bmatrix} + alpha_2begin{bmatrix} 1\ 0\ 1 \ 1\ end{bmatrix}; space alpha_1 , alpha_2 in Bbb{R} \ $$ a)A is an m × n matrix of rank r. Describe all possible values of m, n, and
r. $$ $$b) Find matrix A if $b=begin{bmatrix} 1 \ 2 \ 1 \ end{bmatrix} \$
$$-$$
a)I found a solution that matrix $A$ has 4 columns (why exactly 4??) , and because homogeneous system result are vectors $begin{bmatrix} 1\ 1\ -1 \ 0 end{bmatrix}, begin{bmatrix} 1\ 0\ 1 \ 1\ end{bmatrix}$ they form null space , so $dim Ker(A)=2$ and from rank-nullity theorem we have that r=2 so then $m ge r$. I don't know how can I find from given solution $x$ how many columns matrix $A$ has and why number of rows is $mge r$ ?
b)If my matrix $A$ has 4 coumns and $rankA=2$ I have to write 2 independent vectors and two non-independent.How to know which two vectors I should use to create column space of $A$ if my $b=begin{bmatrix} 1 \ 2 \ 1 \ end{bmatrix} \$ , and it's obviously that other two have to be linear combination of them to form null space. $$ $$ I found that solution for b) is $$A=begin{bmatrix}
1 & -1 & 0 &-1\
1 & 0 & 1&-2 \
1 & -1 & 0 &-1\
end{bmatrix}$$ First and third column are lineary independent and their sum gives me a vector b, does it mean that linear combination of independent vectors gives me a vector b from $Ax=b$ ?
linear-algebra matrix-equations
linear-algebra matrix-equations
asked Feb 2 at 15:37
FiggaroFiggaro
406
asked Feb 2 at 15:37
FiggaroFiggaro
406
asked Feb 2 at 15:37
FiggaroFiggaro
406
asked Feb 2 at 15:37
FiggaroFiggaro
406
asked Feb 2 at 15:37
FiggaroFiggaro
406
406
$begingroup$
$x$'s are $4$-dimensional vectors, so in order for the multiplication $Ax$ to make sense, the number of columns in $A$ needs to be the same as the dimension of the $x$, i.e., $4$.
$endgroup$
– EuxhenH
Feb 2 at 15:55
add a comment |
$begingroup$
$x$'s are $4$-dimensional vectors, so in order for the multiplication $Ax$ to make sense, the number of columns in $A$ needs to be the same as the dimension of the $x$, i.e., $4$.
$endgroup$
– EuxhenH
Feb 2 at 15:55
$begingroup$
$x$'s are $4$-dimensional vectors, so in order for the multiplication $Ax$ to make sense, the number of columns in $A$ needs to be the same as the dimension of the $x$, i.e., $4$.
$endgroup$
– EuxhenH
Feb 2 at 15:55
$begingroup$
$x$'s are $4$-dimensional vectors, so in order for the multiplication $Ax$ to make sense, the number of columns in $A$ needs to be the same as the dimension of the $x$, i.e., $4$.
$endgroup$
– EuxhenH
Feb 2 at 15:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First of all $Ax=b$ with $binmathbb R^3$ and $xinmathbb R^4$ means that $A$ is a $3times 4$ matrix. This does not make sense otherwise. Write the components of $A$ as $a_{ij}$ for $i=1,dots,3$ and $j=1,dots,4$. The equation $Ax=b$ for those specific values of $x$ and $b$ give
$$
(Ax)_i=a_{i1}+a_{i3}+alpha_1 (a_{i1}+a_{i2}-a_{i3})+alpha_2(a_{i1}+a_{i3}+a_{i4})=b_i.
$$
Since this is true for all $alpha_1$ and $alpha_2$ in $mathbb R$, we need the combinations in the brackets to vanish. So we have for $i=1,dots,3$
$$
a_{i1}+a_{i3}=b_i,quad a_{i1}+a_{i2}-a_{i3}=0,quad a_{i1}+a_{i3}+a_{i4}=0.
$$
It is not too hard to see that we then have a three dimensional space of solutions spanned by the values of $a_{11},a_{21}$ and $a_{31}$ as
$$
A=begin{pmatrix}a_{11}&b_1-2a_{11}&b_1-a_{11}&-b_1\a_{21}&b_2-2a_{21}&b_2-a_{21}&-b_2\a_{31}&b_3-2a_{31}&b_3-a_{31}&-b_3end{pmatrix}.
$$
answered Feb 2 at 16:15
Alec B-GAlec B-G
52019
$endgroup$
$begingroup$
I don't understand anything from $A(x)_i$..Why $a_{i1} + a_{i3}+..$ Why $a_{i3}$?
$endgroup$
– Figgaro
Feb 2 at 18:44
$begingroup$
The $i^{text{th}}$ row of $b$ is $b_i$. The $i^{text{th}}$ row of $Ax$ is $sum_{j=1}^4a_{ij}x_j$. Note that these are numbers as $b$ and $Ax$ are vectors.
$endgroup$
– Alec B-G
Feb 2 at 18:49
add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
First of all $Ax=b$ with $binmathbb R^3$ and $xinmathbb R^4$ means that $A$ is a $3times 4$ matrix. This does not make sense otherwise. Write the components of $A$ as $a_{ij}$ for $i=1,dots,3$ and $j=1,dots,4$. The equation $Ax=b$ for those specific values of $x$ and $b$ give
$$
(Ax)_i=a_{i1}+a_{i3}+alpha_1 (a_{i1}+a_{i2}-a_{i3})+alpha_2(a_{i1}+a_{i3}+a_{i4})=b_i.
$$
Since this is true for all $alpha_1$ and $alpha_2$ in $mathbb R$, we need the combinations in the brackets to vanish. So we have for $i=1,dots,3$
$$
a_{i1}+a_{i3}=b_i,quad a_{i1}+a_{i2}-a_{i3}=0,quad a_{i1}+a_{i3}+a_{i4}=0.
$$
It is not too hard to see that we then have a three dimensional space of solutions spanned by the values of $a_{11},a_{21}$ and $a_{31}$ as
$$
A=begin{pmatrix}a_{11}&b_1-2a_{11}&b_1-a_{11}&-b_1\a_{21}&b_2-2a_{21}&b_2-a_{21}&-b_2\a_{31}&b_3-2a_{31}&b_3-a_{31}&-b_3end{pmatrix}.
$$
answered Feb 2 at 16:15
Alec B-GAlec B-G
52019
$endgroup$
$begingroup$
I don't understand anything from $A(x)_i$..Why $a_{i1} + a_{i3}+..$ Why $a_{i3}$?
$endgroup$
– Figgaro
Feb 2 at 18:44
$begingroup$
The $i^{text{th}}$ row of $b$ is $b_i$. The $i^{text{th}}$ row of $Ax$ is $sum_{j=1}^4a_{ij}x_j$. Note that these are numbers as $b$ and $Ax$ are vectors.
$endgroup$
– Alec B-G
Feb 2 at 18:49
add a comment |
$begingroup$
First of all $Ax=b$ with $binmathbb R^3$ and $xinmathbb R^4$ means that $A$ is a $3times 4$ matrix. This does not make sense otherwise. Write the components of $A$ as $a_{ij}$ for $i=1,dots,3$ and $j=1,dots,4$. The equation $Ax=b$ for those specific values of $x$ and $b$ give
$$
(Ax)_i=a_{i1}+a_{i3}+alpha_1 (a_{i1}+a_{i2}-a_{i3})+alpha_2(a_{i1}+a_{i3}+a_{i4})=b_i.
$$
Since this is true for all $alpha_1$ and $alpha_2$ in $mathbb R$, we need the combinations in the brackets to vanish. So we have for $i=1,dots,3$
$$
a_{i1}+a_{i3}=b_i,quad a_{i1}+a_{i2}-a_{i3}=0,quad a_{i1}+a_{i3}+a_{i4}=0.
$$
It is not too hard to see that we then have a three dimensional space of solutions spanned by the values of $a_{11},a_{21}$ and $a_{31}$ as
$$
A=begin{pmatrix}a_{11}&b_1-2a_{11}&b_1-a_{11}&-b_1\a_{21}&b_2-2a_{21}&b_2-a_{21}&-b_2\a_{31}&b_3-2a_{31}&b_3-a_{31}&-b_3end{pmatrix}.
$$
answered Feb 2 at 16:15
Alec B-GAlec B-G
52019
$endgroup$
$begingroup$
I don't understand anything from $A(x)_i$..Why $a_{i1} + a_{i3}+..$ Why $a_{i3}$?
$endgroup$
– Figgaro
Feb 2 at 18:44
$begingroup$
The $i^{text{th}}$ row of $b$ is $b_i$. The $i^{text{th}}$ row of $Ax$ is $sum_{j=1}^4a_{ij}x_j$. Note that these are numbers as $b$ and $Ax$ are vectors.
$endgroup$
– Alec B-G
Feb 2 at 18:49
add a comment |
$begingroup$
First of all $Ax=b$ with $binmathbb R^3$ and $xinmathbb R^4$ means that $A$ is a $3times 4$ matrix. This does not make sense otherwise. Write the components of $A$ as $a_{ij}$ for $i=1,dots,3$ and $j=1,dots,4$. The equation $Ax=b$ for those specific values of $x$ and $b$ give
$$
(Ax)_i=a_{i1}+a_{i3}+alpha_1 (a_{i1}+a_{i2}-a_{i3})+alpha_2(a_{i1}+a_{i3}+a_{i4})=b_i.
$$
Since this is true for all $alpha_1$ and $alpha_2$ in $mathbb R$, we need the combinations in the brackets to vanish. So we have for $i=1,dots,3$
$$
a_{i1}+a_{i3}=b_i,quad a_{i1}+a_{i2}-a_{i3}=0,quad a_{i1}+a_{i3}+a_{i4}=0.
$$
It is not too hard to see that we then have a three dimensional space of solutions spanned by the values of $a_{11},a_{21}$ and $a_{31}$ as
$$
A=begin{pmatrix}a_{11}&b_1-2a_{11}&b_1-a_{11}&-b_1\a_{21}&b_2-2a_{21}&b_2-a_{21}&-b_2\a_{31}&b_3-2a_{31}&b_3-a_{31}&-b_3end{pmatrix}.
$$
answered Feb 2 at 16:15
Alec B-GAlec B-G
52019
$endgroup$
First of all $Ax=b$ with $binmathbb R^3$ and $xinmathbb R^4$ means that $A$ is a $3times 4$ matrix. This does not make sense otherwise. Write the components of $A$ as $a_{ij}$ for $i=1,dots,3$ and $j=1,dots,4$. The equation $Ax=b$ for those specific values of $x$ and $b$ give
$$
(Ax)_i=a_{i1}+a_{i3}+alpha_1 (a_{i1}+a_{i2}-a_{i3})+alpha_2(a_{i1}+a_{i3}+a_{i4})=b_i.
$$
Since this is true for all $alpha_1$ and $alpha_2$ in $mathbb R$, we need the combinations in the brackets to vanish. So we have for $i=1,dots,3$
$$
a_{i1}+a_{i3}=b_i,quad a_{i1}+a_{i2}-a_{i3}=0,quad a_{i1}+a_{i3}+a_{i4}=0.
$$
It is not too hard to see that we then have a three dimensional space of solutions spanned by the values of $a_{11},a_{21}$ and $a_{31}$ as
$$
A=begin{pmatrix}a_{11}&b_1-2a_{11}&b_1-a_{11}&-b_1\a_{21}&b_2-2a_{21}&b_2-a_{21}&-b_2\a_{31}&b_3-2a_{31}&b_3-a_{31}&-b_3end{pmatrix}.
$$
answered Feb 2 at 16:15
Alec B-GAlec B-G
52019
answered Feb 2 at 16:15
Alec B-GAlec B-G
52019
answered Feb 2 at 16:15
Alec B-GAlec B-G
52019
answered Feb 2 at 16:15
Alec B-GAlec B-G
52019
52019
$begingroup$
I don't understand anything from $A(x)_i$..Why $a_{i1} + a_{i3}+..$ Why $a_{i3}$?
$endgroup$
– Figgaro
Feb 2 at 18:44
$begingroup$
The $i^{text{th}}$ row of $b$ is $b_i$. The $i^{text{th}}$ row of $Ax$ is $sum_{j=1}^4a_{ij}x_j$. Note that these are numbers as $b$ and $Ax$ are vectors.
$endgroup$
– Alec B-G
Feb 2 at 18:49
add a comment |
$begingroup$
I don't understand anything from $A(x)_i$..Why $a_{i1} + a_{i3}+..$ Why $a_{i3}$?
$endgroup$
– Figgaro
Feb 2 at 18:44
$begingroup$
The $i^{text{th}}$ row of $b$ is $b_i$. The $i^{text{th}}$ row of $Ax$ is $sum_{j=1}^4a_{ij}x_j$. Note that these are numbers as $b$ and $Ax$ are vectors.
$endgroup$
– Alec B-G
Feb 2 at 18:49
$begingroup$
I don't understand anything from $A(x)_i$..Why $a_{i1} + a_{i3}+..$ Why $a_{i3}$?
$endgroup$
– Figgaro
Feb 2 at 18:44
$begingroup$
I don't understand anything from $A(x)_i$..Why $a_{i1} + a_{i3}+..$ Why $a_{i3}$?
$endgroup$
– Figgaro
Feb 2 at 18:44
$begingroup$
The $i^{text{th}}$ row of $b$ is $b_i$. The $i^{text{th}}$ row of $Ax$ is $sum_{j=1}^4a_{ij}x_j$. Note that these are numbers as $b$ and $Ax$ are vectors.
$endgroup$
– Alec B-G
Feb 2 at 18:49
$begingroup$
The $i^{text{th}}$ row of $b$ is $b_i$. The $i^{text{th}}$ row of $Ax$ is $sum_{j=1}^4a_{ij}x_j$. Note that these are numbers as $b$ and $Ax$ are vectors.
$endgroup$
– Alec B-G
Feb 2 at 18:49
add a comment |
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$begingroup$
$x$'s are $4$-dimensional vectors, so in order for the multiplication $Ax$ to make sense, the number of columns in $A$ needs to be the same as the dimension of the $x$, i.e., $4$.
$endgroup$
– EuxhenH
Feb 2 at 15:55