# Find the matrix A

#### evinda

##### Well-known member
MHB Site Helper
Hello!!!

The general solution of the system $Ax=\begin{bmatrix} 1\\ 3 \end{bmatrix}$ is $x=\begin{bmatrix} 1\\ 0 \end{bmatrix}+ \lambda \begin{bmatrix} 0\\ 1 \end{bmatrix}$. I want to find the matrix $A$.

I have done the following so far:

$$x=\begin{bmatrix} 1\\ 0 \end{bmatrix}+ \lambda \begin{bmatrix} 0\\ 1 \end{bmatrix}=\begin{bmatrix} 1\\ \lambda \end{bmatrix}.$$

Let $A=\begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix}$.

$$\begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix} \begin{bmatrix} 1\\ \lambda \end{bmatrix}=\begin{bmatrix} 1\\ 3 \end{bmatrix} \Leftrightarrow \begin{bmatrix} a_{11}+ \lambda a_{12}\\ a_{21}+\lambda a_{22} \end{bmatrix}=\begin{bmatrix} 1\\ 3 \end{bmatrix} \Leftrightarrow \begin{Bmatrix} a_{11}+\lambda a_{12}=1 \\ a_{21}+\lambda a_{22}=3 \end{Bmatrix} \Leftrightarrow \begin{Bmatrix} a_{11}=1-\lambda a_{12} \\ a_{21}=3-\lambda a_{22} \end{Bmatrix}.$$

So $A$ is the following matrix:

$$A=\begin{bmatrix} 1-\lambda a_{12} & a_{12}\\ 3-\lambda a_{22} & a_{22} \end{bmatrix}.$$

Is everything right? Can we get more information or is this sufficient?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hey evinda !!

$A$ is a fixed matrix, isn't it?
It shouldn't depend on $\lambda$ should it?

The lambda is only supposed to describe the set of the solutions for $x$, isn't it?

#### evinda

##### Well-known member
MHB Site Helper
Hey evinda !!

$A$ is a fixed matrix, isn't it?
It shouldn't depend on $\lambda$ should it?

The lambda is only supposed to describe the set of the solutions for $x$, isn't it?
Oh yes, right... Even if we pick, for example, $\lambda=0$, we do not get a specific matrix $A$.

How can we get further information?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Oh yes, right... Even if we pick, for example, $\lambda=0$, we do not get a specific matrix $A$.

How can we get further information?
Since we should get the same $A$ regardless of the value of $\lambda$, I think that for instance $a_{12}$ must be $0$.

#### evinda

##### Well-known member
MHB Site Helper
Since we should get the same $A$ regardless of the value of $\lambda$, I think that for instance $a_{12}$ must be $0$.
So whichever value $\lambda$ has, we should get the same matrix $A$.

This is only possible when $a_{12}=a_{22}=0$.

Thus, $A=\begin{bmatrix} 1 & 0\\ 3 & 0 \end{bmatrix}$.

Is this right?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Suppose it is correct, then what are the solutions of $Ax=\begin{bmatrix}1\\3\end{bmatrix}$?

#### evinda

##### Well-known member
MHB Site Helper
Suppose it is correct, then what are the solutions of $Ax=\begin{bmatrix}1\\3\end{bmatrix}$?
This is wrong, since we would have the following:

$$\begin{bmatrix} 1 & 0\\ 3 & 0 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix}= \begin{bmatrix} 1 \\ 3 \end{bmatrix} \Rightarrow x_{1}=1, x_{1}=\frac{1}{3}$$

The solution has to hold for any $\lambda$ and thus we set $a_{12}=0$.

Then $Ax=\begin{bmatrix} 1 \\ 3 \end{bmatrix} \Rightarrow \begin{bmatrix} 1 & 0\\ 3+\lambda a_{22} & a_{22} \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix}= \begin{bmatrix} 1 \\ 3 \end{bmatrix} \Rightarrow x_1=1, (3+\lambda a_{22}) x_1+ a_{22} x_2=3 \Rightarrow 3+\lambda a_{22}+a_{22} x_2=3 \Rightarrow a_{22} (\lambda+x_2)=0$.

Above we showed that it cannot hold that $a_{22}=0$ and so we get that $x_2=-\lambda$, and this is an acceptable solution, according to the hypothesis.

Thus, $A=\begin{bmatrix} 1 & 0\\ 3+\lambda a_{22} & a_{22}\end{bmatrix}$.

Now $A$ again is not fixed, but from $Ax=\begin{bmatrix} 1 \\ 3 \end{bmatrix}$ we get that $x_1=1$ and $x_2=-\lambda$.

At the hypothesis, $x_2=\lambda$. But is this the desired solution since $\lambda$ is arbitrary? Or is something wrong?

By setting $a_{22}=0$, we get $x_1=\frac{1}{3}$, which is rejected.

Is it right? Could something be improved?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
This is wrong, since we would have the following:

$$\begin{bmatrix} 1 & 0\\ 3 & 0 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix}= \begin{bmatrix} 1 \\ 3 \end{bmatrix} \Rightarrow x_{1}=1, x_{1}=\frac{1}{3}$$
How did you get $x_{1}=\frac{1}{3}$?

#### evinda

##### Well-known member
MHB Site Helper
How did you get $x_{1}=\frac{1}{3}$?
Oh sorry... We get that $3 \cdot x_1+ 0 \cdot x_2=3 \Rightarrow x_1=1$, and $x_2$ is arbitrary, so it is indeed a solution...

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Oh sorry... We get that $3 \cdot x_1+ 0 \cdot x_2=3 \Rightarrow x_1=1$, and $x_2$ is arbitrary, so it is indeed a solution...
Ah good.

Does the solution for $x$ match the given general solution?

#### evinda

##### Well-known member
MHB Site Helper
Ah good.

Does the solution for $x$ match the given general solution?
Yes, it does...

So is this sufficient to say that since it has to hold for every $\lambda$, it has to hold that $a_{12}=a_{22}=0$ and so the matrix $A$ is $A=\begin{bmatrix} 1 & 0 \\ 3 & 0 \end{bmatrix}$, since this is confirmed by replacing the matrix that we got at $Ax= \begin{bmatrix} 1 \\ 3 \end{bmatrix}$ ?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Yes, it does...

So is this sufficient to say that since it has to hold for every $\lambda$, it has to hold that $a_{12}=a_{22}=0$ and so the matrix $A$ is $A=\begin{bmatrix} 1 & 0 \\ 3 & 0 \end{bmatrix}$, since this is confirmed by replacing the matrix that we got at $Ax= \begin{bmatrix} 1 \\ 3 \end{bmatrix}$ ?
Yep.

I'd say 'and this is confirmed...' rather than 'since this is confimed...' though.

#### evinda

##### Well-known member
MHB Site Helper
Yep.

I'd say 'and this is confirmed...' rather than 'since this is confimed...' though.
Nice... Thank you very much!!!