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Given the characteristic equation of a $2 times 2$ matrix: $lambda^2-10 lambda+12=0$. How do you find the two different matrices that both include the equation?

For convenience of referencing, let's say our $2times2$ matrix is $M=beginbmatrixa&b\c&dendbmatrix$.

Then its characteristic polynomial is
$$p(lambda)=detbeginbmatrixa-lambda&b\c&d-lambdaendbmatrix=(a-lambda)(d-lambda)-bc=lambda^2-(operatornametrM)lambda+(det M),$$
where $operatornametrM=a+d$ is the trace and $det M=ad-bc$ is the determinant of the matrix $M$.

From here you can create lots of matrices with the desired characteristic equation. First, you want $operatornametrM=a+d=10$, so pick any two numbers $a$ and $d$ that add up to $10$. Once you've placed those, pick $c$ and $d$ to make $det M=ad-bc=12$.

Given any polynomial

$p(x) = displaystyle sum_0^n p_i x^i in Bbb F[x] tag 1$

which is satisfied by a matrix $A in M_n(Bbb F)$, that is, such that

$p(A) = displaystyle sum_0^n p_i A^i = 0, tag 2$

then any matrix $B$ similar to $A$,

$B = QAQ^-1 tag 3$

also satisfies $p(x)$ since

$p(B) = displaystyle sum_0^n p_i B^i = sum_0^n p_i (QAQ^-1)^i = sum_0^n p_i QA^iQ^-1 = Q left ( sum_0^n p_i A^i right ) Q^-1, tag 4$

where we have used the fact that $(QAQ^-1)^i = QA^iQ^-1$.

Thus there are in general very many matrices which satisfy $p(x)$, given that at least one does. The exceptional case is $A = mu I$ for some $mu in F$; then $B = QAQ^-1 = Q mu I Q^-1 = mu I$ and we don't get anything new. But if we add a strictly upper triangular matrix to $A$, the resulting matrix $A'$ has the same characteristic polynomial but $QA'Q^-1 ne A'$ for at least some $Q$; in this way we may treat this exceptional case.

Can we then find a matrix which satisfies the polynomial $p(x)$? With $deg p = n$, we will look for matrices in $M_n(Bbb F)$. Suppose for the moment that $p(x)$ is monic, that is, that $p_n = 1$; it is then well-known that the companion matrix of $p(x)$, viz.

$C(p) = beginbmatrix 0 & 0 & ldots & 0 & -p_0 \ 1 & 0 & ldots & 0 & -p_1 \ 0 & 1 & ldots & 0 & -p_2 \ vdots & vdots & ddots & vdots & vdots \ 0 & 0 & ldots & 1 & -p_n - 1 endbmatrix, tag 5$

has characteristic polynomial $p(x)$; therefore,

$p(C(p)) = 0; tag 6$

we may clearly remove the restriction $p_n = 1$ from (6) by simply multiplying by $p_n$; therefore we have shown that any polynomial of degree $n$ is satisfied by some matrix in $M_n(Bbb F)$; and via the similarity operation (4) or its variant discussed above, we may construct other matrices satisfying $p(x)$.

Of course we can apply all of this to the polynomial $x^2 - 10x + 12$, and see that there are many $2 times 2$ matrices such that $A^2 - 10 A + 12 = 0$.

Note that $lambda^2 - 10lambda +12 =(lambda^2-10lambda + 9)+3 = (lambda-9)(lambda-1)+3$

So then one such matrix is
$$beginbmatrix 9&1\ -3&1endbmatrix$$

Do you see a quick way to permute the elements of this matrix and then end up with the same polynomial?