Need help with solving linear transformation












0












$begingroup$


If T: P1 -> P1 is a linear transformation such that
T(1 + 2x) = 4 + 3x and T(5 + 9 x) = -2 - 4x, then
T(4 - 3 x) =?



I started off with expressing (4-3x) as a linear combination of the two other polynomials:



c1(1+2x) + c2(5+9x) = 4-3x.



I then solved the equation with gauss, which gave me: c1 = -42 and c2= 23.
To solve the equation i continued with: T(4-3x) = { T(-42)(1+2x) + 23(5+9x)}.



This is where I am stuck though. How do I proceed to figure out what's on the right hand side of T(4-3x)?






linear-algebra linear-transformations







share|cite|improve this question









$endgroup$












  • $begingroup$
    You use the fact that $T$ is linear. By the way, your computations are wrong.
    $endgroup$
    – José Carlos Santos
    Dec 15 '18 at 14:41










  • $begingroup$
    So I guess that the values I got after solving the equation was wrong, if that's what you mean. No matter how many times I try to gauss-eliminate, I still get -42, 23.
    $endgroup$
    – wznd
    Dec 15 '18 at 14:48










  • $begingroup$
    But you should have got $c_1=-51$ and $c_2=11$.
    $endgroup$
    – José Carlos Santos
    Dec 15 '18 at 14:53










  • $begingroup$
    Since $T$ is linear, you might want to understand it as a 2x2 matrix. In this sense, one has $T(1+2x)=T(1)+2T(x)$, where $1$ could be the unit vector in the first direction and $x$ the unit vector perpendicular to it.. You only need to understand $T(1)$ and $T(x)$. If I am not wrong, you should get $T(1)=-40-35x$ and $T(x)=22+19x$. Now you need to figure out, how to combine this to get $T(4-3x)$.
    $endgroup$
    – Caroline
    Dec 15 '18 at 14:56












  • $begingroup$
    Which I don't exactly understand, I've done it by hand and tried out several websites aswell and they all get the same result. My matrix: ( 1, 2 ,4 ) ( 5, 9 ,-3). The second paranthesis should be below the first, as a regular matrix, but I don't know how to do that in comments.
    $endgroup$
    – wznd
    Dec 15 '18 at 14:56
















0












$begingroup$


If T: P1 -> P1 is a linear transformation such that
T(1 + 2x) = 4 + 3x and T(5 + 9 x) = -2 - 4x, then
T(4 - 3 x) =?



I started off with expressing (4-3x) as a linear combination of the two other polynomials:



c1(1+2x) + c2(5+9x) = 4-3x.



I then solved the equation with gauss, which gave me: c1 = -42 and c2= 23.
To solve the equation i continued with: T(4-3x) = { T(-42)(1+2x) + 23(5+9x)}.



This is where I am stuck though. How do I proceed to figure out what's on the right hand side of T(4-3x)?






linear-algebra linear-transformations







share|cite|improve this question









$endgroup$












  • $begingroup$
    You use the fact that $T$ is linear. By the way, your computations are wrong.
    $endgroup$
    – José Carlos Santos
    Dec 15 '18 at 14:41










  • $begingroup$
    So I guess that the values I got after solving the equation was wrong, if that's what you mean. No matter how many times I try to gauss-eliminate, I still get -42, 23.
    $endgroup$
    – wznd
    Dec 15 '18 at 14:48










  • $begingroup$
    But you should have got $c_1=-51$ and $c_2=11$.
    $endgroup$
    – José Carlos Santos
    Dec 15 '18 at 14:53










  • $begingroup$
    Since $T$ is linear, you might want to understand it as a 2x2 matrix. In this sense, one has $T(1+2x)=T(1)+2T(x)$, where $1$ could be the unit vector in the first direction and $x$ the unit vector perpendicular to it.. You only need to understand $T(1)$ and $T(x)$. If I am not wrong, you should get $T(1)=-40-35x$ and $T(x)=22+19x$. Now you need to figure out, how to combine this to get $T(4-3x)$.
    $endgroup$
    – Caroline
    Dec 15 '18 at 14:56












  • $begingroup$
    Which I don't exactly understand, I've done it by hand and tried out several websites aswell and they all get the same result. My matrix: ( 1, 2 ,4 ) ( 5, 9 ,-3). The second paranthesis should be below the first, as a regular matrix, but I don't know how to do that in comments.
    $endgroup$
    – wznd
    Dec 15 '18 at 14:56














0












0








0





$begingroup$


If T: P1 -> P1 is a linear transformation such that
T(1 + 2x) = 4 + 3x and T(5 + 9 x) = -2 - 4x, then
T(4 - 3 x) =?



I started off with expressing (4-3x) as a linear combination of the two other polynomials:



c1(1+2x) + c2(5+9x) = 4-3x.



I then solved the equation with gauss, which gave me: c1 = -42 and c2= 23.
To solve the equation i continued with: T(4-3x) = { T(-42)(1+2x) + 23(5+9x)}.



This is where I am stuck though. How do I proceed to figure out what's on the right hand side of T(4-3x)?






linear-algebra linear-transformations







share|cite|improve this question









$endgroup$




If T: P1 -> P1 is a linear transformation such that
T(1 + 2x) = 4 + 3x and T(5 + 9 x) = -2 - 4x, then
T(4 - 3 x) =?



I started off with expressing (4-3x) as a linear combination of the two other polynomials:



c1(1+2x) + c2(5+9x) = 4-3x.



I then solved the equation with gauss, which gave me: c1 = -42 and c2= 23.
To solve the equation i continued with: T(4-3x) = { T(-42)(1+2x) + 23(5+9x)}.



This is where I am stuck though. How do I proceed to figure out what's on the right hand side of T(4-3x)?






linear-algebra linear-transformations




linear-algebra linear-transformations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 15 '18 at 14:39









wzndwznd

84




84












  • $begingroup$
    You use the fact that $T$ is linear. By the way, your computations are wrong.
    $endgroup$
    – José Carlos Santos
    Dec 15 '18 at 14:41










  • $begingroup$
    So I guess that the values I got after solving the equation was wrong, if that's what you mean. No matter how many times I try to gauss-eliminate, I still get -42, 23.
    $endgroup$
    – wznd
    Dec 15 '18 at 14:48










  • $begingroup$
    But you should have got $c_1=-51$ and $c_2=11$.
    $endgroup$
    – José Carlos Santos
    Dec 15 '18 at 14:53










  • $begingroup$
    Since $T$ is linear, you might want to understand it as a 2x2 matrix. In this sense, one has $T(1+2x)=T(1)+2T(x)$, where $1$ could be the unit vector in the first direction and $x$ the unit vector perpendicular to it.. You only need to understand $T(1)$ and $T(x)$. If I am not wrong, you should get $T(1)=-40-35x$ and $T(x)=22+19x$. Now you need to figure out, how to combine this to get $T(4-3x)$.
    $endgroup$
    – Caroline
    Dec 15 '18 at 14:56












  • $begingroup$
    Which I don't exactly understand, I've done it by hand and tried out several websites aswell and they all get the same result. My matrix: ( 1, 2 ,4 ) ( 5, 9 ,-3). The second paranthesis should be below the first, as a regular matrix, but I don't know how to do that in comments.
    $endgroup$
    – wznd
    Dec 15 '18 at 14:56


















  • $begingroup$
    You use the fact that $T$ is linear. By the way, your computations are wrong.
    $endgroup$
    – José Carlos Santos
    Dec 15 '18 at 14:41










  • $begingroup$
    So I guess that the values I got after solving the equation was wrong, if that's what you mean. No matter how many times I try to gauss-eliminate, I still get -42, 23.
    $endgroup$
    – wznd
    Dec 15 '18 at 14:48










  • $begingroup$
    But you should have got $c_1=-51$ and $c_2=11$.
    $endgroup$
    – José Carlos Santos
    Dec 15 '18 at 14:53










  • $begingroup$
    Since $T$ is linear, you might want to understand it as a 2x2 matrix. In this sense, one has $T(1+2x)=T(1)+2T(x)$, where $1$ could be the unit vector in the first direction and $x$ the unit vector perpendicular to it.. You only need to understand $T(1)$ and $T(x)$. If I am not wrong, you should get $T(1)=-40-35x$ and $T(x)=22+19x$. Now you need to figure out, how to combine this to get $T(4-3x)$.
    $endgroup$
    – Caroline
    Dec 15 '18 at 14:56












  • $begingroup$
    Which I don't exactly understand, I've done it by hand and tried out several websites aswell and they all get the same result. My matrix: ( 1, 2 ,4 ) ( 5, 9 ,-3). The second paranthesis should be below the first, as a regular matrix, but I don't know how to do that in comments.
    $endgroup$
    – wznd
    Dec 15 '18 at 14:56
















$begingroup$
You use the fact that $T$ is linear. By the way, your computations are wrong.
$endgroup$
– José Carlos Santos
Dec 15 '18 at 14:41




$begingroup$
You use the fact that $T$ is linear. By the way, your computations are wrong.
$endgroup$
– José Carlos Santos
Dec 15 '18 at 14:41












$begingroup$
So I guess that the values I got after solving the equation was wrong, if that's what you mean. No matter how many times I try to gauss-eliminate, I still get -42, 23.
$endgroup$
– wznd
Dec 15 '18 at 14:48




$begingroup$
So I guess that the values I got after solving the equation was wrong, if that's what you mean. No matter how many times I try to gauss-eliminate, I still get -42, 23.
$endgroup$
– wznd
Dec 15 '18 at 14:48












$begingroup$
But you should have got $c_1=-51$ and $c_2=11$.
$endgroup$
– José Carlos Santos
Dec 15 '18 at 14:53




$begingroup$
But you should have got $c_1=-51$ and $c_2=11$.
$endgroup$
– José Carlos Santos
Dec 15 '18 at 14:53












$begingroup$
Since $T$ is linear, you might want to understand it as a 2x2 matrix. In this sense, one has $T(1+2x)=T(1)+2T(x)$, where $1$ could be the unit vector in the first direction and $x$ the unit vector perpendicular to it.. You only need to understand $T(1)$ and $T(x)$. If I am not wrong, you should get $T(1)=-40-35x$ and $T(x)=22+19x$. Now you need to figure out, how to combine this to get $T(4-3x)$.
$endgroup$
– Caroline
Dec 15 '18 at 14:56






$begingroup$
Since $T$ is linear, you might want to understand it as a 2x2 matrix. In this sense, one has $T(1+2x)=T(1)+2T(x)$, where $1$ could be the unit vector in the first direction and $x$ the unit vector perpendicular to it.. You only need to understand $T(1)$ and $T(x)$. If I am not wrong, you should get $T(1)=-40-35x$ and $T(x)=22+19x$. Now you need to figure out, how to combine this to get $T(4-3x)$.
$endgroup$
– Caroline
Dec 15 '18 at 14:56














$begingroup$
Which I don't exactly understand, I've done it by hand and tried out several websites aswell and they all get the same result. My matrix: ( 1, 2 ,4 ) ( 5, 9 ,-3). The second paranthesis should be below the first, as a regular matrix, but I don't know how to do that in comments.
$endgroup$
– wznd
Dec 15 '18 at 14:56




$begingroup$
Which I don't exactly understand, I've done it by hand and tried out several websites aswell and they all get the same result. My matrix: ( 1, 2 ,4 ) ( 5, 9 ,-3). The second paranthesis should be below the first, as a regular matrix, but I don't know how to do that in comments.
$endgroup$
– wznd
Dec 15 '18 at 14:56










1 Answer
1






active

oldest

votes


















0












$begingroup$

The linear map $T$ can be understood as 2x2 matrix in the following sense. Take the canonical basis in $P_1$, that is: $B={1,x}$ (why is this a basis?)
Then the linear map $T$ is fully defined by defining the image of $T$ of the elements of $B$. That is, we need to understand $T(1)$ and $T(x)$. Let me try to clearify this. Since any element of $P_1$ can be written as linear combination of $1$ and $x$ in a unique way, we may identify
begin{align}
p=a+bx = a (1) + b(x) longleftrightarrow p = begin{pmatrix} a\b end{pmatrix}.
end{align}
Now we can write
begin{align}
T(4-3x) = begin{pmatrix} T_{11} & T_{12}\T_{21}&T_{22}end{pmatrix} begin{pmatrix} 4\-3 end{pmatrix}.
end{align}
Now we need to compute all the $T_{ij}$s. Lets start! We are given $T(1+2x)=4+3x$ and $T(5+9x)=-2-4x$. In our notation, this means:
begin{align}
begin{pmatrix} T_{11} & T_{12}\T_{21}&T_{22}end{pmatrix} begin{pmatrix} 1\2 end{pmatrix} = begin{pmatrix} 4\3 end{pmatrix} text{ and }begin{pmatrix} T_{11} & T_{12}\T_{21}&T_{22}end{pmatrix} begin{pmatrix} 5\9 end{pmatrix} = begin{pmatrix} -2\-4 end{pmatrix} qquad qquad (ast)
end{align}
The upper two lines of these equations read
begin{align}
T_{11}+2T_{12}=4 text{ and }5T_{11}+9T_{12}=-2.
end{align}
Do Gauß here, or invert the matrix (whatever you like more)! You should find $T_{11}=-40$ and $T_{12}=22$. Analogously, you proceed to find $T_{21}=-35$ and $T_{22}=19$ (using the lower two equations from ($ast)$. If you have done this, you have full knoledge of $T$. The answer to you question then simply reads
begin{align}
T(4-3x) = begin{pmatrix} T_{11} & T_{12}\T_{21} & T_{22}end{pmatrix}begin{pmatrix}4\-3 end{pmatrix} = (4T_{11} -3T_{12}) + (4T_{21}-3T_{22})x = -226-197x.
end{align}
I left out the Gauß part, so you can practice (I hope that is ok). Note, that one should be more careful with equality signs, but I chose to ignore this ('canonical' isomorphism between $mathbb{R}^2$ and $P_1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much for this. I'll definetly start practicing this. You've been a tremendous help!
    $endgroup$
    – wznd
    Dec 15 '18 at 16:03











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

The linear map $T$ can be understood as 2x2 matrix in the following sense. Take the canonical basis in $P_1$, that is: $B={1,x}$ (why is this a basis?)
Then the linear map $T$ is fully defined by defining the image of $T$ of the elements of $B$. That is, we need to understand $T(1)$ and $T(x)$. Let me try to clearify this. Since any element of $P_1$ can be written as linear combination of $1$ and $x$ in a unique way, we may identify
begin{align}
p=a+bx = a (1) + b(x) longleftrightarrow p = begin{pmatrix} a\b end{pmatrix}.
end{align}
Now we can write
begin{align}
T(4-3x) = begin{pmatrix} T_{11} & T_{12}\T_{21}&T_{22}end{pmatrix} begin{pmatrix} 4\-3 end{pmatrix}.
end{align}
Now we need to compute all the $T_{ij}$s. Lets start! We are given $T(1+2x)=4+3x$ and $T(5+9x)=-2-4x$. In our notation, this means:
begin{align}
begin{pmatrix} T_{11} & T_{12}\T_{21}&T_{22}end{pmatrix} begin{pmatrix} 1\2 end{pmatrix} = begin{pmatrix} 4\3 end{pmatrix} text{ and }begin{pmatrix} T_{11} & T_{12}\T_{21}&T_{22}end{pmatrix} begin{pmatrix} 5\9 end{pmatrix} = begin{pmatrix} -2\-4 end{pmatrix} qquad qquad (ast)
end{align}
The upper two lines of these equations read
begin{align}
T_{11}+2T_{12}=4 text{ and }5T_{11}+9T_{12}=-2.
end{align}
Do Gauß here, or invert the matrix (whatever you like more)! You should find $T_{11}=-40$ and $T_{12}=22$. Analogously, you proceed to find $T_{21}=-35$ and $T_{22}=19$ (using the lower two equations from ($ast)$. If you have done this, you have full knoledge of $T$. The answer to you question then simply reads
begin{align}
T(4-3x) = begin{pmatrix} T_{11} & T_{12}\T_{21} & T_{22}end{pmatrix}begin{pmatrix}4\-3 end{pmatrix} = (4T_{11} -3T_{12}) + (4T_{21}-3T_{22})x = -226-197x.
end{align}
I left out the Gauß part, so you can practice (I hope that is ok). Note, that one should be more careful with equality signs, but I chose to ignore this ('canonical' isomorphism between $mathbb{R}^2$ and $P_1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much for this. I'll definetly start practicing this. You've been a tremendous help!
    $endgroup$
    – wznd
    Dec 15 '18 at 16:03
















0












$begingroup$

The linear map $T$ can be understood as 2x2 matrix in the following sense. Take the canonical basis in $P_1$, that is: $B={1,x}$ (why is this a basis?)
Then the linear map $T$ is fully defined by defining the image of $T$ of the elements of $B$. That is, we need to understand $T(1)$ and $T(x)$. Let me try to clearify this. Since any element of $P_1$ can be written as linear combination of $1$ and $x$ in a unique way, we may identify
begin{align}
p=a+bx = a (1) + b(x) longleftrightarrow p = begin{pmatrix} a\b end{pmatrix}.
end{align}
Now we can write
begin{align}
T(4-3x) = begin{pmatrix} T_{11} & T_{12}\T_{21}&T_{22}end{pmatrix} begin{pmatrix} 4\-3 end{pmatrix}.
end{align}
Now we need to compute all the $T_{ij}$s. Lets start! We are given $T(1+2x)=4+3x$ and $T(5+9x)=-2-4x$. In our notation, this means:
begin{align}
begin{pmatrix} T_{11} & T_{12}\T_{21}&T_{22}end{pmatrix} begin{pmatrix} 1\2 end{pmatrix} = begin{pmatrix} 4\3 end{pmatrix} text{ and }begin{pmatrix} T_{11} & T_{12}\T_{21}&T_{22}end{pmatrix} begin{pmatrix} 5\9 end{pmatrix} = begin{pmatrix} -2\-4 end{pmatrix} qquad qquad (ast)
end{align}
The upper two lines of these equations read
begin{align}
T_{11}+2T_{12}=4 text{ and }5T_{11}+9T_{12}=-2.
end{align}
Do Gauß here, or invert the matrix (whatever you like more)! You should find $T_{11}=-40$ and $T_{12}=22$. Analogously, you proceed to find $T_{21}=-35$ and $T_{22}=19$ (using the lower two equations from ($ast)$. If you have done this, you have full knoledge of $T$. The answer to you question then simply reads
begin{align}
T(4-3x) = begin{pmatrix} T_{11} & T_{12}\T_{21} & T_{22}end{pmatrix}begin{pmatrix}4\-3 end{pmatrix} = (4T_{11} -3T_{12}) + (4T_{21}-3T_{22})x = -226-197x.
end{align}
I left out the Gauß part, so you can practice (I hope that is ok). Note, that one should be more careful with equality signs, but I chose to ignore this ('canonical' isomorphism between $mathbb{R}^2$ and $P_1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much for this. I'll definetly start practicing this. You've been a tremendous help!
    $endgroup$
    – wznd
    Dec 15 '18 at 16:03














0












0








0





$begingroup$

The linear map $T$ can be understood as 2x2 matrix in the following sense. Take the canonical basis in $P_1$, that is: $B={1,x}$ (why is this a basis?)
Then the linear map $T$ is fully defined by defining the image of $T$ of the elements of $B$. That is, we need to understand $T(1)$ and $T(x)$. Let me try to clearify this. Since any element of $P_1$ can be written as linear combination of $1$ and $x$ in a unique way, we may identify
begin{align}
p=a+bx = a (1) + b(x) longleftrightarrow p = begin{pmatrix} a\b end{pmatrix}.
end{align}
Now we can write
begin{align}
T(4-3x) = begin{pmatrix} T_{11} & T_{12}\T_{21}&T_{22}end{pmatrix} begin{pmatrix} 4\-3 end{pmatrix}.
end{align}
Now we need to compute all the $T_{ij}$s. Lets start! We are given $T(1+2x)=4+3x$ and $T(5+9x)=-2-4x$. In our notation, this means:
begin{align}
begin{pmatrix} T_{11} & T_{12}\T_{21}&T_{22}end{pmatrix} begin{pmatrix} 1\2 end{pmatrix} = begin{pmatrix} 4\3 end{pmatrix} text{ and }begin{pmatrix} T_{11} & T_{12}\T_{21}&T_{22}end{pmatrix} begin{pmatrix} 5\9 end{pmatrix} = begin{pmatrix} -2\-4 end{pmatrix} qquad qquad (ast)
end{align}
The upper two lines of these equations read
begin{align}
T_{11}+2T_{12}=4 text{ and }5T_{11}+9T_{12}=-2.
end{align}
Do Gauß here, or invert the matrix (whatever you like more)! You should find $T_{11}=-40$ and $T_{12}=22$. Analogously, you proceed to find $T_{21}=-35$ and $T_{22}=19$ (using the lower two equations from ($ast)$. If you have done this, you have full knoledge of $T$. The answer to you question then simply reads
begin{align}
T(4-3x) = begin{pmatrix} T_{11} & T_{12}\T_{21} & T_{22}end{pmatrix}begin{pmatrix}4\-3 end{pmatrix} = (4T_{11} -3T_{12}) + (4T_{21}-3T_{22})x = -226-197x.
end{align}
I left out the Gauß part, so you can practice (I hope that is ok). Note, that one should be more careful with equality signs, but I chose to ignore this ('canonical' isomorphism between $mathbb{R}^2$ and $P_1$.






share|cite|improve this answer









$endgroup$



The linear map $T$ can be understood as 2x2 matrix in the following sense. Take the canonical basis in $P_1$, that is: $B={1,x}$ (why is this a basis?)
Then the linear map $T$ is fully defined by defining the image of $T$ of the elements of $B$. That is, we need to understand $T(1)$ and $T(x)$. Let me try to clearify this. Since any element of $P_1$ can be written as linear combination of $1$ and $x$ in a unique way, we may identify
begin{align}
p=a+bx = a (1) + b(x) longleftrightarrow p = begin{pmatrix} a\b end{pmatrix}.
end{align}
Now we can write
begin{align}
T(4-3x) = begin{pmatrix} T_{11} & T_{12}\T_{21}&T_{22}end{pmatrix} begin{pmatrix} 4\-3 end{pmatrix}.
end{align}
Now we need to compute all the $T_{ij}$s. Lets start! We are given $T(1+2x)=4+3x$ and $T(5+9x)=-2-4x$. In our notation, this means:
begin{align}
begin{pmatrix} T_{11} & T_{12}\T_{21}&T_{22}end{pmatrix} begin{pmatrix} 1\2 end{pmatrix} = begin{pmatrix} 4\3 end{pmatrix} text{ and }begin{pmatrix} T_{11} & T_{12}\T_{21}&T_{22}end{pmatrix} begin{pmatrix} 5\9 end{pmatrix} = begin{pmatrix} -2\-4 end{pmatrix} qquad qquad (ast)
end{align}
The upper two lines of these equations read
begin{align}
T_{11}+2T_{12}=4 text{ and }5T_{11}+9T_{12}=-2.
end{align}
Do Gauß here, or invert the matrix (whatever you like more)! You should find $T_{11}=-40$ and $T_{12}=22$. Analogously, you proceed to find $T_{21}=-35$ and $T_{22}=19$ (using the lower two equations from ($ast)$. If you have done this, you have full knoledge of $T$. The answer to you question then simply reads
begin{align}
T(4-3x) = begin{pmatrix} T_{11} & T_{12}\T_{21} & T_{22}end{pmatrix}begin{pmatrix}4\-3 end{pmatrix} = (4T_{11} -3T_{12}) + (4T_{21}-3T_{22})x = -226-197x.
end{align}
I left out the Gauß part, so you can practice (I hope that is ok). Note, that one should be more careful with equality signs, but I chose to ignore this ('canonical' isomorphism between $mathbb{R}^2$ and $P_1$.







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answered Dec 15 '18 at 15:52









CarolineCaroline

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  • $begingroup$
    Thank you so much for this. I'll definetly start practicing this. You've been a tremendous help!
    $endgroup$
    – wznd
    Dec 15 '18 at 16:03


















  • $begingroup$
    Thank you so much for this. I'll definetly start practicing this. You've been a tremendous help!
    $endgroup$
    – wznd
    Dec 15 '18 at 16:03
















$begingroup$
Thank you so much for this. I'll definetly start practicing this. You've been a tremendous help!
$endgroup$
– wznd
Dec 15 '18 at 16:03




$begingroup$
Thank you so much for this. I'll definetly start practicing this. You've been a tremendous help!
$endgroup$
– wznd
Dec 15 '18 at 16:03


















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