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Find the matrix A

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,711
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

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,711
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

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,711
Suppose it is correct, then what are the solutions of $Ax=\begin{bmatrix}1\\3\end{bmatrix}$? (Thinking)
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,711
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

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,711
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

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,711
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

Well-known member
MHB Site Helper
Apr 13, 2013
3,718