How to find a system of equations when the solution is given?

[Lotto]

TL;DR Summary: I have to find a system of equations with this solution ${(1,2,0,3)^T+t(1,1,1,-2)^T+s(1,-1,3,0)^T;s,t \in \mathbb{R}}$ when we know that matrix of this equation has:

1. two non-zero rows
2. 3 non-zero rows.

My idea is that I could somehow use the fact that $t(1,1,1,-2)^T+s(1,-1,3,0)^T$ is the homogenous solution of the system. But what to do? Any hints?

 

[fresh_42]

You have a plane $P$ in 4D space $\mathbb{R}^4.$ The homogenous solution $P_0$ is a parallel plane where the origin has been moved from $\vec{v}_0=(1,2,0,3)^T$ to $(0,0,0,0)^T.$ The other two vectors $\vec{v}_1\, , \,\vec{v}_2$ are the directions that span the plane.
$$
\vec{x}=\vec{v}_0 + P= \vec{v}_0 + \operatorname{span}\{\vec{v}_1,\vec{v}_2\}
$$
Your equation is the parameterized description of the plane. You could for instance choose combinations like $(s,t)\in \{(0,1),(1,0)\}$ to get three points on the plane which also characterizes $P$.

I'm not quite sure whether the problem statement asks you to describe $P$ as a linear transformation of the standard basis $\{\vec{e}_1,\vec{e}_2,\vec{e}_3,\vec{e}_4\}$ in $\mathbb{R}^4$ into a basis $\{\vec{v}_1,\vec{v}_2,\vec{e}_3,\vec{e}_4\}$ (or the other way around. I tend to confuse the directions of basis transformations. I struggle between whether the bases are transformed or the subspaces), ...

... or to describe $P_0$ as the solution of $A\cdot \vec{x}=\vec{0}.$

 

[Lotto]

You have a plane $P$ in 4D space $\mathbb{R}^4.$ The homogenous solution $P_0$ is a parallel plane where the origin has been moved from $\vec{v}_0=(1,2,0,3)^T$ to $(0,0,0,0)^T.$ The other two vectors $\vec{v}_1\, , \,\vec{v}_2$ are the directions that span the plane.
$$
\vec{x}=\vec{v}_0 + P= \vec{v}_0 + \operatorname{span}\{\vec{v}_1,\vec{v}_2\}
$$
Your equation is the parameterized description of the plane. You could for instance choose combinations like $(s,t)\in \{(0,1),(1,0)\}$ to get three points on the plane which also characterizes $P$.

I'm not quite sure whether the problem statement asks you to describe $P$ as a linear transformation of the standard basis $\{\vec{e}_1,\vec{e}_2,\vec{e}_3,\vec{e}_4\}$ in $\mathbb{R}^4$ into a basis $\{\vec{v}_1,\vec{v}_2,\vec{e}_3,\vec{e}_4\}$ (or the other way around. I tend to confuse the directions of basis transformations. I struggle between whether the bases are transformed or the subspaces), ...

... or to describe $P_0$ as the solution of $A\cdot \vec{x}=\vec{0}.$

If I understand it correctly, I am supposed to find a 2 x 4 matrix of the system of equations, whose solution is the solution above. One of the solutions (for the first task) corresponds to

$\begin{pmatrix} -2 & 1 & 1 & 0\\ 1 & 1 & 0 & 1\\ \end{pmatrix} \vec x = \begin{pmatrix} 0 \\ 6 \\ \end {pmatrix}.$

But how to find it quickly?

 

[fresh_42]

But how to find it quickly?

They are both orthogonal to your direction vectors and linearly independent. The plane is spanned by $(1,1,1,-2)$ and $(1,-1,3,0).$ The row vectors are $(-2,1,1,0) \perp (1,1,1,-2)$ and $(1,1,0,1)\perp (1,-1,3,0).$ The vector on the right is
$$
\begin{pmatrix}-2&1&1&0\\1&1&0&1\end{pmatrix}\cdot \begin{pmatrix}1\\2\\0\\3\end{pmatrix}=\begin{pmatrix}0\\6\end{pmatrix}
$$
That's quickly but it doesn't tell you what this is all about.

 

[Steve4Physics]

My idea is that I could somehow use the fact that $t(1,1,1,-2)^T+s(1,-1,3,0)^T$ is the homogenous solution of the system. But what to do? Any hints?

As a non-mathematician, here are some thoughts which which might help if you haven’t yet sorted it. (I'm sure the mathematicians will correct me if I'm wrong!)

$\vec {v_0} =(1,2,0,3)^T,~~ \vec {v_1}= (1,-1,3,0)^T,~~ \vec {v_2}= (1,1,1,-2)^T$
$\vec x = \vec {v_0} + s\vec {v_1} + t\vec {v_2}$

Note I’ve re-ordered/re-named slightly because $s$ comes before $t$ alphabetically!

There are 2 parameters $s$ and $t$, so the system is underdetermined; 2 of the 4 variables ($x_1, x_2, x_3$ and $x_4$) are free. So WLOG one option it to take matrix $\mathbf A$ to be in rref and express it as:

$\mathbf A = \begin{pmatrix}1&0&a&b\\0&1&c&d\end{pmatrix}$

Of course any 2 of the variables could be selected to be the free ones, giving different alternative answers.

Evaluating $\mathbf A\vec {v_1}=\vec 0$ and $\mathbf A\vec {v_2}=\vec 0$ allows $a,b,c$ and $d$ to be found.

$s\vec {v_1} + t\vec {v_2}$ is then the solution for the homogeneous system $\mathbf A \vec x= \vec 0$.

$v_0$ is the given particular solution. If we now calculate $\mathbf A \vec {v_0}$ we get the correponding value for $\vec b$.

It follows that $\mathbf A (\vec {v_0} + s\vec {v_1}+t\vec {v_2}) =\vec b$ as required.

Edited (severely!)

 

[Lotto]

As a non-mathematician, here are some thoughts which which might help if you haven’t yet sorted it. (I'm sure the mathematicians will correct me if I'm wrong!)

$\vec {v_0} =(1,2,0,3)^T,~~ \vec {v_1}= (1,-1,3,0)^T,~~ \vec {v_2}= (1,1,1,-2)^T$
$\vec x = \vec {v_0} + s\vec {v_1} + t\vec {v_2}$

Note I’ve re-ordered/re-named slightly because $s$ comes before $t$ alphabetically!

There are 2 parameters $s$ and $t$, so the system is underdetermined; 2 of the 4 variables ($x_1, x_2, x_3$ and $x_4$) are free. So WLOG one option it to take matrix $\mathbf A$ to be in rref and express it as:

$\mathbf A = \begin{pmatrix}1&0&a&b\\0&1&c&d\end{pmatrix}$

Of course any 2 of the variables could be selected to be the free ones, giving different alternative answers.

Evaluating $\mathbf A\vec {v_1}=\vec 0$ and $\mathbf A\vec {v_2}=\vec 0$ allows $a,b,c$ and $d$ to be found.

$s\vec {v_1} + t\vec {v_2}$ is then the solution for the homogeneous system $\mathbf A \vec x= \vec 0$.

$v_0$ is the given particular solution. If we now calculate $\mathbf A \vec {v_0}$ we get the correponding value for $\vec b$.

It follows that $\mathbf A (\vec {v_0} + s\vec {v_1}+t\vec {v_2}) =\vec b$ as required.

Edited (severely!)

So, if I understand it correctly, we basically say that we want a matrix of a linear transformation because only then we can write $\vec {b}=A(\vec{v_0}+s\vec {v_1} + t\vec {v_2})=A\vec{v_0}+A(s\vec {v_1} + t\vec {v_2})$. We know, that $A(s\vec {v_1} + t\vec {v_2})=\vec 0$, so we can choose a solution where $sA\vec{v_1}=\vec 0$ and $tA\vec{v_2}=\vec 0$. Is this how you would solve it?

So if we want the case when we have a three-line matrix, it would be similar.

In a very brief solution I was today given is this solution for the tasks 1) and 2):

For example, why do we start with writing the two vectors from the task into a matrix?? Moreover, this solution looks much faster Could you or someone else clarify it? Because I don't get it.

 

[Steve4Physics]

So, if I understand it correctly, we basically say that we want a matrix of a linear transformation because only then we can write $\vec {b}=A(\vec{v_0}+s\vec {v_1} + t\vec {v_2})=A\vec{v_0}+A(s\vec {v_1} + t\vec {v_2})$. We know, that $A(s\vec {v_1} + t\vec {v_2})=\vec 0$, so we can choose a solution where $sA\vec{v_1}=\vec 0$ and $tA\vec{v_2}=\vec 0$. Is this how you would solve it?

Yes, that's an implicit part of my Post #5 approach. Of course $sA\vec{v_1}=\vec 0$ and $tA\vec{v_2}=\vec 0$ can be simplified to $A\vec{v_1}=\vec 0$ and $A\vec{v_2}=\vec 0$.

So if we want the case when we have a three-line matrix, it would be similar

To get a 3-row matrix, all that is necessary is to take the 2-row matrix and add a 3rd row which is any linear combination of the first 2 rows. That's because the new row adds no new information. In the solution you posted it appears that the 3rd row is simply a duplicate of the 2nd row, which is fine.

In a very brief solution I was today given is this solution for the tasks 1) and 2):

View attachment 352606
For example, why do we start with writing the two vectors from the task into a matrix?? Moreover, this solution looks much faster Could you or someone else clarify it? Because I don't get it.

I can't follow the solution. It appears to be a fairly randomly arranged collection of workings with no explanations and some bits are illegible. So I can't help - maybe someone else can.

If you are really stuck on this, I recommend making sure that you fully understand the usual process for solving $\mathbf A \vec x = \vec b$ using the given answers. I.e. for $\mathbf A \vec x = \vec b$ find $\vec x$ when:

a) $\mathbf A = \begin{pmatrix}-2&1&1&0\\1&1&0&1\end{pmatrix}$ and $\vec b = (0,6)^T$.

b) $\mathbf A = \begin{pmatrix}-2&1&1&0\\1&1&0&1\\1&1&0&1\end{pmatrix}$ and $\vec b = (0,6,6)^T$

That should give some useful insights.