MHB Find Matrix A for System Ax=1, 3

[evinda]

Hello! (Wave)

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? (Thinking)

 

[I like Serena]

Hey evinda!

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

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

 

[evinda]

Hey evinda!

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

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

Oh yes, right... (Thinking) Even if we pick, for example, $\lambda=0$, we do not get a specific matrix $A$.

How can we get further information? (Thinking)

 

[I like Serena]

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 $0$. (Thinking)

 

[evinda]

Since we should get the same $A$ regardless of the value of $\lambda$, I think that for instance $a_{12}$ must be $0$. (Thinking)

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? (Thinking)

 

[I like Serena]

Suppose it is correct, then what are the solutions of $Ax=\begin{bmatrix}1\\3\end{bmatrix}$? (Thinking)

 

[evinda]

Suppose it is correct, then what are the solutions of $Ax=\begin{bmatrix}1\\3\end{bmatrix}$? (Thinking)

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

Is it right? Could something be improved? (Thinking)

 

[I like Serena]

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 $x_{1}=\frac{1}{3}$? (Wondering)

 

[evinda]

How did you get $x_{1}=\frac{1}{3}$? (Wondering)

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... (Wasntme)

 

[I like Serena]

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. (Whew)

Does the solution for $x$ match the given general solution? (Wondering)

 

[evinda]

Ah good. (Whew)

Does the solution for $x$ match the given general solution? (Wondering)

Yes, it does... (Nod)

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 $Ax= \begin{bmatrix} 1 \\ 3 \end{bmatrix}$ ? (Thinking)

 

[I like Serena]

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. (Nod)

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

 

[evinda]

Yep. (Nod)

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

Nice... (Smirk) Thank you very much! (Blush)