# Thread: Find the matrix A

Originally Posted by Klaas van Aarsen
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 $Ax= \begin{bmatrix} 1 \\ 3 \end{bmatrix}$ ?

2.

3. Originally Posted by evinda
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.

Originally Posted by Klaas van Aarsen
Yep.

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

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