
MHB Master
#11
March 25th, 2020,
17:24
Thread Author
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 $Ax= \begin{bmatrix} 1 \\ 3 \end{bmatrix}$ ?

March 25th, 2020 17:24
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MHB Seeker
#12
March 25th, 2020,
17:42
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.

MHB Master
#13
March 26th, 2020,
14:09
Thread Author
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!!!